Extracting data in ES6

up vote
1
down vote

favorite

I have this array const idArray = ["12", "231", "73", "4"] and an object

const blueprints = {

  12: {color: red, views: [{name: "front}, {name: "back}, {name: "top}, {name: "bottom}]},

  231: {color: white, views: [{name: "front}, {name: "back}]},

  73: {color: black, views: [{name: "front}, {name: "back}, {name: "top}, {name: "bottom}]},

  4: {color: silver, views: [{name: "front}, {name: "back}, {name: "top}, {name: "bottom}]},

}

How can I return an array of the following objects that have all front, back, top, and bottom using ES6 map/filter/some and etc?:

result =[

    {colorId: "12", views: [{name: "front}, {name: "back}, {name: "top}, {name: "bottom}]}

    {colorId: "73", views: [{name: "front}, {name: "back}, {name: "top}, {name: "bottom}]}

    {colorId: "4", views: [{name: "front}, {name: "back}, {name: "top}, {name: "bottom}]}

]

I did it here but I feel like it is too messy and hard to read. Anybody could recommend on how to shorten it and make it easier to read using the ES6 functions (map, filter ...)?

 const result = idArray.map(id => {

      const bluePrint = bluePrints[id];

      const exists = blurPrint.views.some(view => view.name === 'top' || view.name === 'bottom');



      if (exists) {

        return {

          colorId: id,

          views: bluePrint.views

        }

      }

    }).filter(bluePrint => !bluePrint);

asked Nov 15 at 17:57

j doe

2

Questions on how to optimize working code are better suited to be asked on codereview.stackexchange.com . You could use one reduce() instead of chaining map() and filter()
– charlietfl
Nov 15 at 18:05

You have lots of missing quotes in your original array and desired result.
– Barmar
Nov 15 at 18:14

Another typo: blurPrint should be bluePrint
– Barmar
Nov 15 at 18:19

bluePrint => !bluePrint should be bluePrint => bluePrint, otherwise you'll just get an array of undefined.
– Barmar
Nov 15 at 18:20

add a comment |

up vote
1
down vote

favorite

I have this array const idArray = ["12", "231", "73", "4"] and an object

const blueprints = {

  12: {color: red, views: [{name: "front}, {name: "back}, {name: "top}, {name: "bottom}]},

  231: {color: white, views: [{name: "front}, {name: "back}]},

  73: {color: black, views: [{name: "front}, {name: "back}, {name: "top}, {name: "bottom}]},

  4: {color: silver, views: [{name: "front}, {name: "back}, {name: "top}, {name: "bottom}]},

}

How can I return an array of the following objects that have all front, back, top, and bottom using ES6 map/filter/some and etc?:

result =[

    {colorId: "12", views: [{name: "front}, {name: "back}, {name: "top}, {name: "bottom}]}

    {colorId: "73", views: [{name: "front}, {name: "back}, {name: "top}, {name: "bottom}]}

    {colorId: "4", views: [{name: "front}, {name: "back}, {name: "top}, {name: "bottom}]}

]

I did it here but I feel like it is too messy and hard to read. Anybody could recommend on how to shorten it and make it easier to read using the ES6 functions (map, filter ...)?

 const result = idArray.map(id => {

      const bluePrint = bluePrints[id];

      const exists = blurPrint.views.some(view => view.name === 'top' || view.name === 'bottom');



      if (exists) {

        return {

          colorId: id,

          views: bluePrint.views

        }

      }

    }).filter(bluePrint => !bluePrint);

asked Nov 15 at 17:57

j doe

2

Questions on how to optimize working code are better suited to be asked on codereview.stackexchange.com . You could use one reduce() instead of chaining map() and filter()
– charlietfl
Nov 15 at 18:05

You have lots of missing quotes in your original array and desired result.
– Barmar
Nov 15 at 18:14

Another typo: blurPrint should be bluePrint
– Barmar
Nov 15 at 18:19

bluePrint => !bluePrint should be bluePrint => bluePrint, otherwise you'll just get an array of undefined.
– Barmar
Nov 15 at 18:20

add a comment |

up vote
1
down vote

favorite

I have this array const idArray = ["12", "231", "73", "4"] and an object

const blueprints = {

  12: {color: red, views: [{name: "front}, {name: "back}, {name: "top}, {name: "bottom}]},

  231: {color: white, views: [{name: "front}, {name: "back}]},

  73: {color: black, views: [{name: "front}, {name: "back}, {name: "top}, {name: "bottom}]},

  4: {color: silver, views: [{name: "front}, {name: "back}, {name: "top}, {name: "bottom}]},

}

How can I return an array of the following objects that have all front, back, top, and bottom using ES6 map/filter/some and etc?:

result =[

    {colorId: "12", views: [{name: "front}, {name: "back}, {name: "top}, {name: "bottom}]}

    {colorId: "73", views: [{name: "front}, {name: "back}, {name: "top}, {name: "bottom}]}

    {colorId: "4", views: [{name: "front}, {name: "back}, {name: "top}, {name: "bottom}]}

]

I did it here but I feel like it is too messy and hard to read. Anybody could recommend on how to shorten it and make it easier to read using the ES6 functions (map, filter ...)?

 const result = idArray.map(id => {

      const bluePrint = bluePrints[id];

      const exists = blurPrint.views.some(view => view.name === 'top' || view.name === 'bottom');



      if (exists) {

        return {

          colorId: id,

          views: bluePrint.views

        }

      }

    }).filter(bluePrint => !bluePrint);

asked Nov 15 at 17:57

j doe

I have this array const idArray = ["12", "231", "73", "4"] and an object

const blueprints = {

  12: {color: red, views: [{name: "front}, {name: "back}, {name: "top}, {name: "bottom}]},

  231: {color: white, views: [{name: "front}, {name: "back}]},

  73: {color: black, views: [{name: "front}, {name: "back}, {name: "top}, {name: "bottom}]},

  4: {color: silver, views: [{name: "front}, {name: "back}, {name: "top}, {name: "bottom}]},

}

How can I return an array of the following objects that have all front, back, top, and bottom using ES6 map/filter/some and etc?:

result =[

    {colorId: "12", views: [{name: "front}, {name: "back}, {name: "top}, {name: "bottom}]}

    {colorId: "73", views: [{name: "front}, {name: "back}, {name: "top}, {name: "bottom}]}

    {colorId: "4", views: [{name: "front}, {name: "back}, {name: "top}, {name: "bottom}]}

]

I did it here but I feel like it is too messy and hard to read. Anybody could recommend on how to shorten it and make it easier to read using the ES6 functions (map, filter ...)?

 const result = idArray.map(id => {

      const bluePrint = bluePrints[id];

      const exists = blurPrint.views.some(view => view.name === 'top' || view.name === 'bottom');



      if (exists) {

        return {

          colorId: id,

          views: bluePrint.views

        }

      }

    }).filter(bluePrint => !bluePrint);

javascript ecmascript-6

asked Nov 15 at 17:57

j doe

asked Nov 15 at 17:57

j doe

asked Nov 15 at 17:57

j doe

asked Nov 15 at 17:57

j doe

asked Nov 15 at 17:57

j doe

2

Questions on how to optimize working code are better suited to be asked on codereview.stackexchange.com . You could use one reduce() instead of chaining map() and filter()
– charlietfl
Nov 15 at 18:05

You have lots of missing quotes in your original array and desired result.
– Barmar
Nov 15 at 18:14

Another typo: blurPrint should be bluePrint
– Barmar
Nov 15 at 18:19

bluePrint => !bluePrint should be bluePrint => bluePrint, otherwise you'll just get an array of undefined.
– Barmar
Nov 15 at 18:20

add a comment |

2

Questions on how to optimize working code are better suited to be asked on codereview.stackexchange.com . You could use one reduce() instead of chaining map() and filter()
– charlietfl
Nov 15 at 18:05

You have lots of missing quotes in your original array and desired result.
– Barmar
Nov 15 at 18:14

Another typo: blurPrint should be bluePrint
– Barmar
Nov 15 at 18:19

bluePrint => !bluePrint should be bluePrint => bluePrint, otherwise you'll just get an array of undefined.
– Barmar
Nov 15 at 18:20

Questions on how to optimize working code are better suited to be asked on codereview.stackexchange.com . You could use one reduce() instead of chaining map() and filter()
– charlietfl
Nov 15 at 18:05

You have lots of missing quotes in your original array and desired result.
– Barmar
Nov 15 at 18:14

Another typo: blurPrint should be bluePrint
– Barmar
Nov 15 at 18:19

bluePrint => !bluePrint should be bluePrint => bluePrint, otherwise you'll just get an array of undefined.
– Barmar
Nov 15 at 18:20

add a comment |

3 Answers
3

active

oldest

votes

up vote
0
down vote

You can filter ids so that every color in your target set of colors is in that id's blueprint.views and then map those ids to your desired result object:

const idArray = ["12", "231", "73", "4"];

const blueprints = {

  12: {color: 'red', views: [{name: "front"}, {name: "back"}, {name: "top"}, {name: "bottom"}]},

  231: {color: 'white', views: [{name: "front"}, {name: "back"}]},

  73: {color: 'black', views: [{name: "front"}, {name: "back"}, {name: "top"}, {name: "bottom"}]},

  4: {color: 'silver', views: [{name: "front"}, {name: "back"}, {name: "top"}, {name: "bottom"}]},

};



const result = idArray

                .filter(id => {

                  const colors = blueprints[id].views.map(e => e.name);

                  return ['front', 'back', 'top', 'bottom'].every(s => colors.includes(s));

                })

                .map(id => ({colorId: id, views: blueprints[id].views}))



console.log(result);

edited Nov 15 at 18:30

answered Nov 15 at 18:21

slider

7,8751129

add a comment |

up vote
0
down vote

You can map() over your idArray to create the array in the desired format, then use filter() to test if every() required string is in the view array and filter out the incomplete items:

const blueprints = {

    12: {color: "red", views: [{name: "front"}, {name: "back"}, {name: "top"}, {name: "bottom"}]},

    231: {color: "white", views: [{name: "front"}, {name: "back"}]},

    73: {color: "black", views: [{name: "front"}, {name: "back"}, {name: "top"}, {name: "bottom"}]},

    4: {color: "silver", views: [{name: "front"}, {name: "back"}, {name: "top"}, {name: "bottom"}]},

  }



const idArray = ["12", "231", "73", "4"] 

const required = ['front', 'back', 'top', 'bottom']



let newArry = idArray

        .map(colorID => ({colorID, views: blueprints[colorID].views }) )

        .filter(item => required.every(direction => item.views.some(v => v.name == direction) ))



console.log(newArry)

edited Nov 15 at 18:34

answered Nov 15 at 18:24

Mark Meyer

33k32854

add a comment |

up vote
0
down vote

Object.keys(blueprints)

  .map(k => ({colorId:k, views:blueprints[k].views}))

  .filter(el =>

    ['front', 'back', 'top', 'bottom'].every(it => 

      el.views.some(s => s.name === it)

  )

)

edited Nov 16 at 6:17

answered Nov 15 at 18:39

dee zg

4,34531330

add a comment |

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

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53325368%2fextracting-data-in-es6%23new-answer', 'question_page');
}
);

Post as a guest

Name

Required, but never shown

3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

up vote
0
down vote

You can filter ids so that every color in your target set of colors is in that id's blueprint.views and then map those ids to your desired result object:

const idArray = ["12", "231", "73", "4"];

const blueprints = {

  12: {color: 'red', views: [{name: "front"}, {name: "back"}, {name: "top"}, {name: "bottom"}]},

  231: {color: 'white', views: [{name: "front"}, {name: "back"}]},

  73: {color: 'black', views: [{name: "front"}, {name: "back"}, {name: "top"}, {name: "bottom"}]},

  4: {color: 'silver', views: [{name: "front"}, {name: "back"}, {name: "top"}, {name: "bottom"}]},

};



const result = idArray

                .filter(id => {

                  const colors = blueprints[id].views.map(e => e.name);

                  return ['front', 'back', 'top', 'bottom'].every(s => colors.includes(s));

                })

                .map(id => ({colorId: id, views: blueprints[id].views}))



console.log(result);

edited Nov 15 at 18:30

answered Nov 15 at 18:21

slider

7,8751129

add a comment |

up vote
0
down vote

You can filter ids so that every color in your target set of colors is in that id's blueprint.views and then map those ids to your desired result object:

const idArray = ["12", "231", "73", "4"];

const blueprints = {

  12: {color: 'red', views: [{name: "front"}, {name: "back"}, {name: "top"}, {name: "bottom"}]},

  231: {color: 'white', views: [{name: "front"}, {name: "back"}]},

  73: {color: 'black', views: [{name: "front"}, {name: "back"}, {name: "top"}, {name: "bottom"}]},

  4: {color: 'silver', views: [{name: "front"}, {name: "back"}, {name: "top"}, {name: "bottom"}]},

};



const result = idArray

                .filter(id => {

                  const colors = blueprints[id].views.map(e => e.name);

                  return ['front', 'back', 'top', 'bottom'].every(s => colors.includes(s));

                })

                .map(id => ({colorId: id, views: blueprints[id].views}))



console.log(result);

edited Nov 15 at 18:30

answered Nov 15 at 18:21

slider

7,8751129

add a comment |

up vote
0
down vote

You can filter ids so that every color in your target set of colors is in that id's blueprint.views and then map those ids to your desired result object:

const idArray = ["12", "231", "73", "4"];

const blueprints = {

  12: {color: 'red', views: [{name: "front"}, {name: "back"}, {name: "top"}, {name: "bottom"}]},

  231: {color: 'white', views: [{name: "front"}, {name: "back"}]},

  73: {color: 'black', views: [{name: "front"}, {name: "back"}, {name: "top"}, {name: "bottom"}]},

  4: {color: 'silver', views: [{name: "front"}, {name: "back"}, {name: "top"}, {name: "bottom"}]},

};



const result = idArray

                .filter(id => {

                  const colors = blueprints[id].views.map(e => e.name);

                  return ['front', 'back', 'top', 'bottom'].every(s => colors.includes(s));

                })

                .map(id => ({colorId: id, views: blueprints[id].views}))



console.log(result);

edited Nov 15 at 18:30

answered Nov 15 at 18:21

slider

7,8751129

You can filter ids so that every color in your target set of colors is in that id's blueprint.views and then map those ids to your desired result object:

const idArray = ["12", "231", "73", "4"];

const blueprints = {

  12: {color: 'red', views: [{name: "front"}, {name: "back"}, {name: "top"}, {name: "bottom"}]},

  231: {color: 'white', views: [{name: "front"}, {name: "back"}]},

  73: {color: 'black', views: [{name: "front"}, {name: "back"}, {name: "top"}, {name: "bottom"}]},

  4: {color: 'silver', views: [{name: "front"}, {name: "back"}, {name: "top"}, {name: "bottom"}]},

};



const result = idArray

                .filter(id => {

                  const colors = blueprints[id].views.map(e => e.name);

                  return ['front', 'back', 'top', 'bottom'].every(s => colors.includes(s));

                })

                .map(id => ({colorId: id, views: blueprints[id].views}))



console.log(result);

const idArray = ["12", "231", "73", "4"];

const blueprints = {

  12: {color: 'red', views: [{name: "front"}, {name: "back"}, {name: "top"}, {name: "bottom"}]},

  231: {color: 'white', views: [{name: "front"}, {name: "back"}]},

  73: {color: 'black', views: [{name: "front"}, {name: "back"}, {name: "top"}, {name: "bottom"}]},

  4: {color: 'silver', views: [{name: "front"}, {name: "back"}, {name: "top"}, {name: "bottom"}]},

};



const result = idArray

                .filter(id => {

                  const colors = blueprints[id].views.map(e => e.name);

                  return ['front', 'back', 'top', 'bottom'].every(s => colors.includes(s));

                })

                .map(id => ({colorId: id, views: blueprints[id].views}))



console.log(result);

const idArray = ["12", "231", "73", "4"];

const blueprints = {

  12: {color: 'red', views: [{name: "front"}, {name: "back"}, {name: "top"}, {name: "bottom"}]},

  231: {color: 'white', views: [{name: "front"}, {name: "back"}]},

  73: {color: 'black', views: [{name: "front"}, {name: "back"}, {name: "top"}, {name: "bottom"}]},

  4: {color: 'silver', views: [{name: "front"}, {name: "back"}, {name: "top"}, {name: "bottom"}]},

};



const result = idArray

                .filter(id => {

                  const colors = blueprints[id].views.map(e => e.name);

                  return ['front', 'back', 'top', 'bottom'].every(s => colors.includes(s));

                })

                .map(id => ({colorId: id, views: blueprints[id].views}))



console.log(result);

edited Nov 15 at 18:30

answered Nov 15 at 18:21

slider

7,8751129

edited Nov 15 at 18:30

answered Nov 15 at 18:21

slider

7,8751129

answered Nov 15 at 18:21

slider

7,8751129

answered Nov 15 at 18:21

slider

7,8751129

add a comment |

up vote
0
down vote

You can map() over your idArray to create the array in the desired format, then use filter() to test if every() required string is in the view array and filter out the incomplete items:

const blueprints = {

    12: {color: "red", views: [{name: "front"}, {name: "back"}, {name: "top"}, {name: "bottom"}]},

    231: {color: "white", views: [{name: "front"}, {name: "back"}]},

    73: {color: "black", views: [{name: "front"}, {name: "back"}, {name: "top"}, {name: "bottom"}]},

    4: {color: "silver", views: [{name: "front"}, {name: "back"}, {name: "top"}, {name: "bottom"}]},

  }



const idArray = ["12", "231", "73", "4"] 

const required = ['front', 'back', 'top', 'bottom']



let newArry = idArray

        .map(colorID => ({colorID, views: blueprints[colorID].views }) )

        .filter(item => required.every(direction => item.views.some(v => v.name == direction) ))



console.log(newArry)

edited Nov 15 at 18:34

answered Nov 15 at 18:24

Mark Meyer

33k32854

add a comment |

up vote
0
down vote

You can map() over your idArray to create the array in the desired format, then use filter() to test if every() required string is in the view array and filter out the incomplete items:

const blueprints = {

    12: {color: "red", views: [{name: "front"}, {name: "back"}, {name: "top"}, {name: "bottom"}]},

    231: {color: "white", views: [{name: "front"}, {name: "back"}]},

    73: {color: "black", views: [{name: "front"}, {name: "back"}, {name: "top"}, {name: "bottom"}]},

    4: {color: "silver", views: [{name: "front"}, {name: "back"}, {name: "top"}, {name: "bottom"}]},

  }



const idArray = ["12", "231", "73", "4"] 

const required = ['front', 'back', 'top', 'bottom']



let newArry = idArray

        .map(colorID => ({colorID, views: blueprints[colorID].views }) )

        .filter(item => required.every(direction => item.views.some(v => v.name == direction) ))



console.log(newArry)

edited Nov 15 at 18:34

answered Nov 15 at 18:24

Mark Meyer

33k32854

add a comment |

up vote
0
down vote

You can map() over your idArray to create the array in the desired format, then use filter() to test if every() required string is in the view array and filter out the incomplete items:

const blueprints = {

    12: {color: "red", views: [{name: "front"}, {name: "back"}, {name: "top"}, {name: "bottom"}]},

    231: {color: "white", views: [{name: "front"}, {name: "back"}]},

    73: {color: "black", views: [{name: "front"}, {name: "back"}, {name: "top"}, {name: "bottom"}]},

    4: {color: "silver", views: [{name: "front"}, {name: "back"}, {name: "top"}, {name: "bottom"}]},

  }



const idArray = ["12", "231", "73", "4"] 

const required = ['front', 'back', 'top', 'bottom']



let newArry = idArray

        .map(colorID => ({colorID, views: blueprints[colorID].views }) )

        .filter(item => required.every(direction => item.views.some(v => v.name == direction) ))



console.log(newArry)

edited Nov 15 at 18:34

answered Nov 15 at 18:24

Mark Meyer

33k32854

You can map() over your idArray to create the array in the desired format, then use filter() to test if every() required string is in the view array and filter out the incomplete items:

const blueprints = {

    12: {color: "red", views: [{name: "front"}, {name: "back"}, {name: "top"}, {name: "bottom"}]},

    231: {color: "white", views: [{name: "front"}, {name: "back"}]},

    73: {color: "black", views: [{name: "front"}, {name: "back"}, {name: "top"}, {name: "bottom"}]},

    4: {color: "silver", views: [{name: "front"}, {name: "back"}, {name: "top"}, {name: "bottom"}]},

  }



const idArray = ["12", "231", "73", "4"] 

const required = ['front', 'back', 'top', 'bottom']



let newArry = idArray

        .map(colorID => ({colorID, views: blueprints[colorID].views }) )

        .filter(item => required.every(direction => item.views.some(v => v.name == direction) ))



console.log(newArry)

const blueprints = {

    12: {color: "red", views: [{name: "front"}, {name: "back"}, {name: "top"}, {name: "bottom"}]},

    231: {color: "white", views: [{name: "front"}, {name: "back"}]},

    73: {color: "black", views: [{name: "front"}, {name: "back"}, {name: "top"}, {name: "bottom"}]},

    4: {color: "silver", views: [{name: "front"}, {name: "back"}, {name: "top"}, {name: "bottom"}]},

  }



const idArray = ["12", "231", "73", "4"] 

const required = ['front', 'back', 'top', 'bottom']



let newArry = idArray

        .map(colorID => ({colorID, views: blueprints[colorID].views }) )

        .filter(item => required.every(direction => item.views.some(v => v.name == direction) ))



console.log(newArry)

const blueprints = {

    12: {color: "red", views: [{name: "front"}, {name: "back"}, {name: "top"}, {name: "bottom"}]},

    231: {color: "white", views: [{name: "front"}, {name: "back"}]},

    73: {color: "black", views: [{name: "front"}, {name: "back"}, {name: "top"}, {name: "bottom"}]},

    4: {color: "silver", views: [{name: "front"}, {name: "back"}, {name: "top"}, {name: "bottom"}]},

  }



const idArray = ["12", "231", "73", "4"] 

const required = ['front', 'back', 'top', 'bottom']



let newArry = idArray

        .map(colorID => ({colorID, views: blueprints[colorID].views }) )

        .filter(item => required.every(direction => item.views.some(v => v.name == direction) ))



console.log(newArry)

edited Nov 15 at 18:34

answered Nov 15 at 18:24

Mark Meyer

33k32854

edited Nov 15 at 18:34

answered Nov 15 at 18:24

Mark Meyer

33k32854

answered Nov 15 at 18:24

Mark Meyer

33k32854

answered Nov 15 at 18:24

Mark Meyer

33k32854

add a comment |

up vote
0
down vote

Object.keys(blueprints)

  .map(k => ({colorId:k, views:blueprints[k].views}))

  .filter(el =>

    ['front', 'back', 'top', 'bottom'].every(it => 

      el.views.some(s => s.name === it)

  )

)

edited Nov 16 at 6:17

answered Nov 15 at 18:39

dee zg

4,34531330

add a comment |

up vote
0
down vote

Object.keys(blueprints)

  .map(k => ({colorId:k, views:blueprints[k].views}))

  .filter(el =>

    ['front', 'back', 'top', 'bottom'].every(it => 

      el.views.some(s => s.name === it)

  )

)

edited Nov 16 at 6:17

answered Nov 15 at 18:39

dee zg

4,34531330

add a comment |

up vote
0
down vote

Object.keys(blueprints)

  .map(k => ({colorId:k, views:blueprints[k].views}))

  .filter(el =>

    ['front', 'back', 'top', 'bottom'].every(it => 

      el.views.some(s => s.name === it)

  )

)

edited Nov 16 at 6:17

answered Nov 15 at 18:39

dee zg

4,34531330

Object.keys(blueprints)

  .map(k => ({colorId:k, views:blueprints[k].views}))

  .filter(el =>

    ['front', 'back', 'top', 'bottom'].every(it => 

      el.views.some(s => s.name === it)

  )

)

edited Nov 16 at 6:17

answered Nov 15 at 18:39

dee zg

4,34531330

edited Nov 16 at 6:17

answered Nov 15 at 18:39

dee zg

4,34531330

answered Nov 15 at 18:39

dee zg

4,34531330

answered Nov 15 at 18:39

dee zg

4,34531330

add a comment |

draft saved

draft discarded

Thanks for contributing an answer to Stack Overflow!

Please be sure to answer the question. Provide details and share your research!

But avoid …

Asking for help, clarification, or responding to other answers.

Making statements based on opinion; back them up with references or personal experience.

To learn more, see our tips on writing great answers.

Some of your past answers have not been well-received, and you're in danger of being blocked from answering.

Please pay close attention to the following guidance:

Please be sure to answer the question. Provide details and share your research!

But avoid …

Asking for help, clarification, or responding to other answers.

Making statements based on opinion; back them up with references or personal experience.

To learn more, see our tips on writing great answers.

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

Post as a guest

Name

Required, but never shown

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

Name

Required, but never shown

Name

Required, but never shown

This page is only for reference, If you need detailed information, please check here

搜尋此網誌

Cfrgtkky