find the pattern and dynamically build regex to match the string











up vote
0
down vote

favorite












If asterisk * is present in the pattern, then it means a sequence of the same character of length 3 unless it is followed by {N} which represents how many characters should appear in the sequence where N will be at least 1. My goal is to determine if the second string exactly matches the pattern of the first string in the input. I'm having trouble building the Regex pattern



*{2}* mmRRR should return TRUE
*{2}* mRRR should return FALSE


https://jsfiddle.net/82smw9zx/



sample code::



pattern1 = /'queryStrSubStr.charAt(0){patternCount}'/;
var patternMatch = new RegExp(pattern1);
if(queryStrSubStr.match(patternMatch)) {
result = true;
} else result = false;









share|improve this question
























  • Do you try to call JS in your pattern or it's part of pattern string?
    – Justinas
    Nov 12 at 14:57










  • Take a look at the Regex Object Constructor if you want to dynamically create patterns.
    – AutomatedChaos
    Nov 12 at 15:02










  • @Justinas -- jsfiddle.net/82smw9zx
    – dragonfly
    Nov 12 at 16:02















up vote
0
down vote

favorite












If asterisk * is present in the pattern, then it means a sequence of the same character of length 3 unless it is followed by {N} which represents how many characters should appear in the sequence where N will be at least 1. My goal is to determine if the second string exactly matches the pattern of the first string in the input. I'm having trouble building the Regex pattern



*{2}* mmRRR should return TRUE
*{2}* mRRR should return FALSE


https://jsfiddle.net/82smw9zx/



sample code::



pattern1 = /'queryStrSubStr.charAt(0){patternCount}'/;
var patternMatch = new RegExp(pattern1);
if(queryStrSubStr.match(patternMatch)) {
result = true;
} else result = false;









share|improve this question
























  • Do you try to call JS in your pattern or it's part of pattern string?
    – Justinas
    Nov 12 at 14:57










  • Take a look at the Regex Object Constructor if you want to dynamically create patterns.
    – AutomatedChaos
    Nov 12 at 15:02










  • @Justinas -- jsfiddle.net/82smw9zx
    – dragonfly
    Nov 12 at 16:02













up vote
0
down vote

favorite









up vote
0
down vote

favorite











If asterisk * is present in the pattern, then it means a sequence of the same character of length 3 unless it is followed by {N} which represents how many characters should appear in the sequence where N will be at least 1. My goal is to determine if the second string exactly matches the pattern of the first string in the input. I'm having trouble building the Regex pattern



*{2}* mmRRR should return TRUE
*{2}* mRRR should return FALSE


https://jsfiddle.net/82smw9zx/



sample code::



pattern1 = /'queryStrSubStr.charAt(0){patternCount}'/;
var patternMatch = new RegExp(pattern1);
if(queryStrSubStr.match(patternMatch)) {
result = true;
} else result = false;









share|improve this question















If asterisk * is present in the pattern, then it means a sequence of the same character of length 3 unless it is followed by {N} which represents how many characters should appear in the sequence where N will be at least 1. My goal is to determine if the second string exactly matches the pattern of the first string in the input. I'm having trouble building the Regex pattern



*{2}* mmRRR should return TRUE
*{2}* mRRR should return FALSE


https://jsfiddle.net/82smw9zx/



sample code::



pattern1 = /'queryStrSubStr.charAt(0){patternCount}'/;
var patternMatch = new RegExp(pattern1);
if(queryStrSubStr.match(patternMatch)) {
result = true;
} else result = false;






javascript jquery regex






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Nov 12 at 14:52









georg

143k33193290




143k33193290










asked Nov 12 at 14:39









dragonfly

97861539




97861539












  • Do you try to call JS in your pattern or it's part of pattern string?
    – Justinas
    Nov 12 at 14:57










  • Take a look at the Regex Object Constructor if you want to dynamically create patterns.
    – AutomatedChaos
    Nov 12 at 15:02










  • @Justinas -- jsfiddle.net/82smw9zx
    – dragonfly
    Nov 12 at 16:02


















  • Do you try to call JS in your pattern or it's part of pattern string?
    – Justinas
    Nov 12 at 14:57










  • Take a look at the Regex Object Constructor if you want to dynamically create patterns.
    – AutomatedChaos
    Nov 12 at 15:02










  • @Justinas -- jsfiddle.net/82smw9zx
    – dragonfly
    Nov 12 at 16:02
















Do you try to call JS in your pattern or it's part of pattern string?
– Justinas
Nov 12 at 14:57




Do you try to call JS in your pattern or it's part of pattern string?
– Justinas
Nov 12 at 14:57












Take a look at the Regex Object Constructor if you want to dynamically create patterns.
– AutomatedChaos
Nov 12 at 15:02




Take a look at the Regex Object Constructor if you want to dynamically create patterns.
– AutomatedChaos
Nov 12 at 15:02












@Justinas -- jsfiddle.net/82smw9zx
– dragonfly
Nov 12 at 16:02




@Justinas -- jsfiddle.net/82smw9zx
– dragonfly
Nov 12 at 16:02












2 Answers
2






active

oldest

votes

















up vote
1
down vote



accepted










You need to use new RegExp() to construct your regex pattern with variables (rather than attempting to include a variable directly in your regular expression literal).



You are trying to include variables queryStrSubStr.charAt(0) and patternCount in a regular expression literal like: /'queryStrSubStr.charAt(0){patternCount}'/, but JavaScript does not interpret those strings as variables inside the literal.



Following example demonstrates how to construct your regex pattern with variables as well as incorporating the html input from your fiddle so that you can test various patterns. Code comments explain how the code works.






$('.btn').click(function() {
let result = wildcards($('.enter_pattern').val());
console.log(result);
});

const wildcards = (s) => {
if (s.startsWith('*')) { // if input string starts with *

let pattern;
let [count, text] = s.split(' '); // split input string into count and text
count = count.match(/{d+}/); // match count pattern like {n}

if (count) { // if there is a count
pattern = new RegExp(text.charAt(0) + count); // regex: first character + matched count pattern
} else { // if there is no count
pattern = new RegExp(text.charAt(0) + '{3}'); // regex: first character + default pattern {3}
}

return s.match(pattern) ? true : false; // return true if text matches pattern or false if not

} else { // if input string does not start with *
return 'No pattern';
}
}

<input type="text" class="enter_pattern" />
<button type="submit" class="btn">Click</button>

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>





/*
Example test output:

Input: *{2}* mmRRR
Log: true

Input: *{2}* mRRR
Log: false

Input: * mmmRRR
Log: true

Input: * mmRRR
Log: false

Input: mmRRR
Log: No pattern
*/





share|improve this answer























  • I entered * {5} * aaaaaBBB'. Your code doesn't return desired output
    – dragonfly
    Nov 12 at 17:20












  • yes I did. When I added comments * is disappearing. Try it
    – dragonfly
    Nov 12 at 17:23












  • I have new RegExp in my code as well.
    – dragonfly
    Nov 12 at 17:24












  • can you please explain else { patt = new RegExp(text.charAt(0) + '{3}'); }
    – dragonfly
    Nov 12 at 17:28










  • @dragonfly - answer edited with more explanation that hopefully clears it up.
    – benvc
    Nov 12 at 17:35


















up vote
0
down vote













First you need to calulate the pattern using a regex:



/*{(d+)}*/


It matches a star, a left Square bracket, followed by one or more digits and ending with a right Square bracket and a star.



How to use:



var text = 'mmRRR';
var char = text.charAt(0);
var pattern = '*{2}*';
var counter = /*{(d+)}*/.exec(pattern)[1] || '3';
var regex = new RegeExp('^' + char + '{' + counter + '}$');
var result = text.match(regex);





share|improve this answer





















    Your Answer






    StackExchange.ifUsing("editor", function () {
    StackExchange.using("externalEditor", function () {
    StackExchange.using("snippets", function () {
    StackExchange.snippets.init();
    });
    });
    }, "code-snippets");

    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "1"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














     

    draft saved


    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53264463%2ffind-the-pattern-and-dynamically-build-regex-to-match-the-string%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    1
    down vote



    accepted










    You need to use new RegExp() to construct your regex pattern with variables (rather than attempting to include a variable directly in your regular expression literal).



    You are trying to include variables queryStrSubStr.charAt(0) and patternCount in a regular expression literal like: /'queryStrSubStr.charAt(0){patternCount}'/, but JavaScript does not interpret those strings as variables inside the literal.



    Following example demonstrates how to construct your regex pattern with variables as well as incorporating the html input from your fiddle so that you can test various patterns. Code comments explain how the code works.






    $('.btn').click(function() {
    let result = wildcards($('.enter_pattern').val());
    console.log(result);
    });

    const wildcards = (s) => {
    if (s.startsWith('*')) { // if input string starts with *

    let pattern;
    let [count, text] = s.split(' '); // split input string into count and text
    count = count.match(/{d+}/); // match count pattern like {n}

    if (count) { // if there is a count
    pattern = new RegExp(text.charAt(0) + count); // regex: first character + matched count pattern
    } else { // if there is no count
    pattern = new RegExp(text.charAt(0) + '{3}'); // regex: first character + default pattern {3}
    }

    return s.match(pattern) ? true : false; // return true if text matches pattern or false if not

    } else { // if input string does not start with *
    return 'No pattern';
    }
    }

    <input type="text" class="enter_pattern" />
    <button type="submit" class="btn">Click</button>

    <script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>





    /*
    Example test output:

    Input: *{2}* mmRRR
    Log: true

    Input: *{2}* mRRR
    Log: false

    Input: * mmmRRR
    Log: true

    Input: * mmRRR
    Log: false

    Input: mmRRR
    Log: No pattern
    */





    share|improve this answer























    • I entered * {5} * aaaaaBBB'. Your code doesn't return desired output
      – dragonfly
      Nov 12 at 17:20












    • yes I did. When I added comments * is disappearing. Try it
      – dragonfly
      Nov 12 at 17:23












    • I have new RegExp in my code as well.
      – dragonfly
      Nov 12 at 17:24












    • can you please explain else { patt = new RegExp(text.charAt(0) + '{3}'); }
      – dragonfly
      Nov 12 at 17:28










    • @dragonfly - answer edited with more explanation that hopefully clears it up.
      – benvc
      Nov 12 at 17:35















    up vote
    1
    down vote



    accepted










    You need to use new RegExp() to construct your regex pattern with variables (rather than attempting to include a variable directly in your regular expression literal).



    You are trying to include variables queryStrSubStr.charAt(0) and patternCount in a regular expression literal like: /'queryStrSubStr.charAt(0){patternCount}'/, but JavaScript does not interpret those strings as variables inside the literal.



    Following example demonstrates how to construct your regex pattern with variables as well as incorporating the html input from your fiddle so that you can test various patterns. Code comments explain how the code works.






    $('.btn').click(function() {
    let result = wildcards($('.enter_pattern').val());
    console.log(result);
    });

    const wildcards = (s) => {
    if (s.startsWith('*')) { // if input string starts with *

    let pattern;
    let [count, text] = s.split(' '); // split input string into count and text
    count = count.match(/{d+}/); // match count pattern like {n}

    if (count) { // if there is a count
    pattern = new RegExp(text.charAt(0) + count); // regex: first character + matched count pattern
    } else { // if there is no count
    pattern = new RegExp(text.charAt(0) + '{3}'); // regex: first character + default pattern {3}
    }

    return s.match(pattern) ? true : false; // return true if text matches pattern or false if not

    } else { // if input string does not start with *
    return 'No pattern';
    }
    }

    <input type="text" class="enter_pattern" />
    <button type="submit" class="btn">Click</button>

    <script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>





    /*
    Example test output:

    Input: *{2}* mmRRR
    Log: true

    Input: *{2}* mRRR
    Log: false

    Input: * mmmRRR
    Log: true

    Input: * mmRRR
    Log: false

    Input: mmRRR
    Log: No pattern
    */





    share|improve this answer























    • I entered * {5} * aaaaaBBB'. Your code doesn't return desired output
      – dragonfly
      Nov 12 at 17:20












    • yes I did. When I added comments * is disappearing. Try it
      – dragonfly
      Nov 12 at 17:23












    • I have new RegExp in my code as well.
      – dragonfly
      Nov 12 at 17:24












    • can you please explain else { patt = new RegExp(text.charAt(0) + '{3}'); }
      – dragonfly
      Nov 12 at 17:28










    • @dragonfly - answer edited with more explanation that hopefully clears it up.
      – benvc
      Nov 12 at 17:35













    up vote
    1
    down vote



    accepted







    up vote
    1
    down vote



    accepted






    You need to use new RegExp() to construct your regex pattern with variables (rather than attempting to include a variable directly in your regular expression literal).



    You are trying to include variables queryStrSubStr.charAt(0) and patternCount in a regular expression literal like: /'queryStrSubStr.charAt(0){patternCount}'/, but JavaScript does not interpret those strings as variables inside the literal.



    Following example demonstrates how to construct your regex pattern with variables as well as incorporating the html input from your fiddle so that you can test various patterns. Code comments explain how the code works.






    $('.btn').click(function() {
    let result = wildcards($('.enter_pattern').val());
    console.log(result);
    });

    const wildcards = (s) => {
    if (s.startsWith('*')) { // if input string starts with *

    let pattern;
    let [count, text] = s.split(' '); // split input string into count and text
    count = count.match(/{d+}/); // match count pattern like {n}

    if (count) { // if there is a count
    pattern = new RegExp(text.charAt(0) + count); // regex: first character + matched count pattern
    } else { // if there is no count
    pattern = new RegExp(text.charAt(0) + '{3}'); // regex: first character + default pattern {3}
    }

    return s.match(pattern) ? true : false; // return true if text matches pattern or false if not

    } else { // if input string does not start with *
    return 'No pattern';
    }
    }

    <input type="text" class="enter_pattern" />
    <button type="submit" class="btn">Click</button>

    <script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>





    /*
    Example test output:

    Input: *{2}* mmRRR
    Log: true

    Input: *{2}* mRRR
    Log: false

    Input: * mmmRRR
    Log: true

    Input: * mmRRR
    Log: false

    Input: mmRRR
    Log: No pattern
    */





    share|improve this answer














    You need to use new RegExp() to construct your regex pattern with variables (rather than attempting to include a variable directly in your regular expression literal).



    You are trying to include variables queryStrSubStr.charAt(0) and patternCount in a regular expression literal like: /'queryStrSubStr.charAt(0){patternCount}'/, but JavaScript does not interpret those strings as variables inside the literal.



    Following example demonstrates how to construct your regex pattern with variables as well as incorporating the html input from your fiddle so that you can test various patterns. Code comments explain how the code works.






    $('.btn').click(function() {
    let result = wildcards($('.enter_pattern').val());
    console.log(result);
    });

    const wildcards = (s) => {
    if (s.startsWith('*')) { // if input string starts with *

    let pattern;
    let [count, text] = s.split(' '); // split input string into count and text
    count = count.match(/{d+}/); // match count pattern like {n}

    if (count) { // if there is a count
    pattern = new RegExp(text.charAt(0) + count); // regex: first character + matched count pattern
    } else { // if there is no count
    pattern = new RegExp(text.charAt(0) + '{3}'); // regex: first character + default pattern {3}
    }

    return s.match(pattern) ? true : false; // return true if text matches pattern or false if not

    } else { // if input string does not start with *
    return 'No pattern';
    }
    }

    <input type="text" class="enter_pattern" />
    <button type="submit" class="btn">Click</button>

    <script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>





    /*
    Example test output:

    Input: *{2}* mmRRR
    Log: true

    Input: *{2}* mRRR
    Log: false

    Input: * mmmRRR
    Log: true

    Input: * mmRRR
    Log: false

    Input: mmRRR
    Log: No pattern
    */





    $('.btn').click(function() {
    let result = wildcards($('.enter_pattern').val());
    console.log(result);
    });

    const wildcards = (s) => {
    if (s.startsWith('*')) { // if input string starts with *

    let pattern;
    let [count, text] = s.split(' '); // split input string into count and text
    count = count.match(/{d+}/); // match count pattern like {n}

    if (count) { // if there is a count
    pattern = new RegExp(text.charAt(0) + count); // regex: first character + matched count pattern
    } else { // if there is no count
    pattern = new RegExp(text.charAt(0) + '{3}'); // regex: first character + default pattern {3}
    }

    return s.match(pattern) ? true : false; // return true if text matches pattern or false if not

    } else { // if input string does not start with *
    return 'No pattern';
    }
    }

    <input type="text" class="enter_pattern" />
    <button type="submit" class="btn">Click</button>

    <script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>





    $('.btn').click(function() {
    let result = wildcards($('.enter_pattern').val());
    console.log(result);
    });

    const wildcards = (s) => {
    if (s.startsWith('*')) { // if input string starts with *

    let pattern;
    let [count, text] = s.split(' '); // split input string into count and text
    count = count.match(/{d+}/); // match count pattern like {n}

    if (count) { // if there is a count
    pattern = new RegExp(text.charAt(0) + count); // regex: first character + matched count pattern
    } else { // if there is no count
    pattern = new RegExp(text.charAt(0) + '{3}'); // regex: first character + default pattern {3}
    }

    return s.match(pattern) ? true : false; // return true if text matches pattern or false if not

    } else { // if input string does not start with *
    return 'No pattern';
    }
    }

    <input type="text" class="enter_pattern" />
    <button type="submit" class="btn">Click</button>

    <script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>






    share|improve this answer














    share|improve this answer



    share|improve this answer








    edited Nov 12 at 19:58

























    answered Nov 12 at 16:01









    benvc

    3,2821319




    3,2821319












    • I entered * {5} * aaaaaBBB'. Your code doesn't return desired output
      – dragonfly
      Nov 12 at 17:20












    • yes I did. When I added comments * is disappearing. Try it
      – dragonfly
      Nov 12 at 17:23












    • I have new RegExp in my code as well.
      – dragonfly
      Nov 12 at 17:24












    • can you please explain else { patt = new RegExp(text.charAt(0) + '{3}'); }
      – dragonfly
      Nov 12 at 17:28










    • @dragonfly - answer edited with more explanation that hopefully clears it up.
      – benvc
      Nov 12 at 17:35


















    • I entered * {5} * aaaaaBBB'. Your code doesn't return desired output
      – dragonfly
      Nov 12 at 17:20












    • yes I did. When I added comments * is disappearing. Try it
      – dragonfly
      Nov 12 at 17:23












    • I have new RegExp in my code as well.
      – dragonfly
      Nov 12 at 17:24












    • can you please explain else { patt = new RegExp(text.charAt(0) + '{3}'); }
      – dragonfly
      Nov 12 at 17:28










    • @dragonfly - answer edited with more explanation that hopefully clears it up.
      – benvc
      Nov 12 at 17:35
















    I entered * {5} * aaaaaBBB'. Your code doesn't return desired output
    – dragonfly
    Nov 12 at 17:20






    I entered * {5} * aaaaaBBB'. Your code doesn't return desired output
    – dragonfly
    Nov 12 at 17:20














    yes I did. When I added comments * is disappearing. Try it
    – dragonfly
    Nov 12 at 17:23






    yes I did. When I added comments * is disappearing. Try it
    – dragonfly
    Nov 12 at 17:23














    I have new RegExp in my code as well.
    – dragonfly
    Nov 12 at 17:24






    I have new RegExp in my code as well.
    – dragonfly
    Nov 12 at 17:24














    can you please explain else { patt = new RegExp(text.charAt(0) + '{3}'); }
    – dragonfly
    Nov 12 at 17:28




    can you please explain else { patt = new RegExp(text.charAt(0) + '{3}'); }
    – dragonfly
    Nov 12 at 17:28












    @dragonfly - answer edited with more explanation that hopefully clears it up.
    – benvc
    Nov 12 at 17:35




    @dragonfly - answer edited with more explanation that hopefully clears it up.
    – benvc
    Nov 12 at 17:35












    up vote
    0
    down vote













    First you need to calulate the pattern using a regex:



    /*{(d+)}*/


    It matches a star, a left Square bracket, followed by one or more digits and ending with a right Square bracket and a star.



    How to use:



    var text = 'mmRRR';
    var char = text.charAt(0);
    var pattern = '*{2}*';
    var counter = /*{(d+)}*/.exec(pattern)[1] || '3';
    var regex = new RegeExp('^' + char + '{' + counter + '}$');
    var result = text.match(regex);





    share|improve this answer

























      up vote
      0
      down vote













      First you need to calulate the pattern using a regex:



      /*{(d+)}*/


      It matches a star, a left Square bracket, followed by one or more digits and ending with a right Square bracket and a star.



      How to use:



      var text = 'mmRRR';
      var char = text.charAt(0);
      var pattern = '*{2}*';
      var counter = /*{(d+)}*/.exec(pattern)[1] || '3';
      var regex = new RegeExp('^' + char + '{' + counter + '}$');
      var result = text.match(regex);





      share|improve this answer























        up vote
        0
        down vote










        up vote
        0
        down vote









        First you need to calulate the pattern using a regex:



        /*{(d+)}*/


        It matches a star, a left Square bracket, followed by one or more digits and ending with a right Square bracket and a star.



        How to use:



        var text = 'mmRRR';
        var char = text.charAt(0);
        var pattern = '*{2}*';
        var counter = /*{(d+)}*/.exec(pattern)[1] || '3';
        var regex = new RegeExp('^' + char + '{' + counter + '}$');
        var result = text.match(regex);





        share|improve this answer












        First you need to calulate the pattern using a regex:



        /*{(d+)}*/


        It matches a star, a left Square bracket, followed by one or more digits and ending with a right Square bracket and a star.



        How to use:



        var text = 'mmRRR';
        var char = text.charAt(0);
        var pattern = '*{2}*';
        var counter = /*{(d+)}*/.exec(pattern)[1] || '3';
        var regex = new RegeExp('^' + char + '{' + counter + '}$');
        var result = text.match(regex);






        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered Nov 12 at 17:34









        Poul Bak

        5,1633832




        5,1633832






























             

            draft saved


            draft discarded



















































             


            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53264463%2ffind-the-pattern-and-dynamically-build-regex-to-match-the-string%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            Popular posts from this blog

            How to change which sound is reproduced for terminal bell?

            Title Spacing in Bjornstrup Chapter, Removing Chapter Number From Contents

            Can I use Tabulator js library in my java Spring + Thymeleaf project?