trying to make nr restriction 1-10, but prompt allows users to pick 13
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so i am trying to make a tick system, where i a user can pick a number from 1-10 and there will start a countdown from the number that the user picks to 0. The problem is that when a user picks number 12 in prompt, it doesnt allow him/her to continue, but when they try again it allows them.
html:
<button onclick=start()>Countdown</button>
<p id="p">
scirpt:
<script>
var ticks;
var mytimer;
function start() {
ticks = Number(prompt("where do you want to start a countdown, min 1 max 10")) + 1;
if (ticks > 11) {
ticks = Number(prompt("Skriv inn på nytt, tallet må være mellom 3-10")) + 1;
}
else if (ticks < 1) {
ticks = Number(prompt("Try again, The number must be between 1-10")) + 1;
}
mytimer = setInterval(tick, 1000)
}
function tick() {
ticks--;
document.getElementById("p").innerHTML = "Tick nr. " + ticks + "<br />";
if (ticks === 1) {
clearInterval(mytimer);
document.write("Done!");
}
}
</script>
javascript if-statement
asked Nov 22 '18 at 7:37
Raahim KhanRaahim Khan
164
add a comment |
so i am trying to make a tick system, where i a user can pick a number from 1-10 and there will start a countdown from the number that the user picks to 0. The problem is that when a user picks number 12 in prompt, it doesnt allow him/her to continue, but when they try again it allows them.
html:
<button onclick=start()>Countdown</button>
<p id="p">
scirpt:
<script>
var ticks;
var mytimer;
function start() {
ticks = Number(prompt("where do you want to start a countdown, min 1 max 10")) + 1;
if (ticks > 11) {
ticks = Number(prompt("Skriv inn på nytt, tallet må være mellom 3-10")) + 1;
}
else if (ticks < 1) {
ticks = Number(prompt("Try again, The number must be between 1-10")) + 1;
}
mytimer = setInterval(tick, 1000)
}
function tick() {
ticks--;
document.getElementById("p").innerHTML = "Tick nr. " + ticks + "<br />";
if (ticks === 1) {
clearInterval(mytimer);
document.write("Done!");
}
}
</script>
javascript if-statement
asked Nov 22 '18 at 7:37
Raahim KhanRaahim Khan
164
add a comment |
so i am trying to make a tick system, where i a user can pick a number from 1-10 and there will start a countdown from the number that the user picks to 0. The problem is that when a user picks number 12 in prompt, it doesnt allow him/her to continue, but when they try again it allows them.
html:
<button onclick=start()>Countdown</button>
<p id="p">
scirpt:
<script>
var ticks;
var mytimer;
function start() {
ticks = Number(prompt("where do you want to start a countdown, min 1 max 10")) + 1;
if (ticks > 11) {
ticks = Number(prompt("Skriv inn på nytt, tallet må være mellom 3-10")) + 1;
}
else if (ticks < 1) {
ticks = Number(prompt("Try again, The number must be between 1-10")) + 1;
}
mytimer = setInterval(tick, 1000)
}
function tick() {
ticks--;
document.getElementById("p").innerHTML = "Tick nr. " + ticks + "<br />";
if (ticks === 1) {
clearInterval(mytimer);
document.write("Done!");
}
}
</script>
javascript if-statement
asked Nov 22 '18 at 7:37
Raahim KhanRaahim Khan
164
so i am trying to make a tick system, where i a user can pick a number from 1-10 and there will start a countdown from the number that the user picks to 0. The problem is that when a user picks number 12 in prompt, it doesnt allow him/her to continue, but when they try again it allows them.
html:
<button onclick=start()>Countdown</button>
<p id="p">
scirpt:
<script>
var ticks;
var mytimer;
function start() {
ticks = Number(prompt("where do you want to start a countdown, min 1 max 10")) + 1;
if (ticks > 11) {
ticks = Number(prompt("Skriv inn på nytt, tallet må være mellom 3-10")) + 1;
}
else if (ticks < 1) {
ticks = Number(prompt("Try again, The number must be between 1-10")) + 1;
}
mytimer = setInterval(tick, 1000)
}
function tick() {
ticks--;
document.getElementById("p").innerHTML = "Tick nr. " + ticks + "<br />";
if (ticks === 1) {
clearInterval(mytimer);
document.write("Done!");
}
}
</script>
javascript if-statement
javascript if-statement
asked Nov 22 '18 at 7:37
Raahim KhanRaahim Khan
164
asked Nov 22 '18 at 7:37
Raahim KhanRaahim Khan
164
asked Nov 22 '18 at 7:37
Raahim KhanRaahim Khan
164
asked Nov 22 '18 at 7:37
Raahim KhanRaahim Khan
164
asked Nov 22 '18 at 7:37
Raahim KhanRaahim Khan
164
164
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
Untill you are not getting any number between 1 - 10, you may try calling same method again as follow:
<script>
var ticks;
var mytimer;
function start() {
ticks = Number(prompt("where do you want to start a countdown, min 1 max 10")) + 1;
if (ticks > 10) {
start();
}
else if (ticks < 1) {
start();
}
mytimer = setInterval(tick, 1000)
}
function tick() {
ticks--;
document.getElementById("p").innerHTML = "Tick nr. " + ticks + "<br />";
if (ticks === 1) {
clearInterval(mytimer);
document.write("Done!");
}
}
</script>
answered Nov 22 '18 at 7:44
Bishal GautamBishal Gautam
855519
add a comment |
The solution is simple. When you first time enter a value bigger than 11 if (ticks > 11) { statement catches it and then shows new prompt. If you again enter a value bigger than 11, there is no if statement to catch it. If you run the start function again when a faulty value entered, you can fix this problem.
answered Nov 22 '18 at 7:57
Rıdvan SumsetRıdvan Sumset
1257
ah yes, that makes sense, but do you have any examples on how i can recall the function, is it possible to keep on calling the function in an if statement.
– Raahim Khan
Nov 22 '18 at 9:49
You can do it like in @Bishal's answer ---->if (ticks > 11) { start();}
– Rıdvan Sumset
Nov 22 '18 at 10:21
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
Untill you are not getting any number between 1 - 10, you may try calling same method again as follow:
<script>
var ticks;
var mytimer;
function start() {
ticks = Number(prompt("where do you want to start a countdown, min 1 max 10")) + 1;
if (ticks > 10) {
start();
}
else if (ticks < 1) {
start();
}
mytimer = setInterval(tick, 1000)
}
function tick() {
ticks--;
document.getElementById("p").innerHTML = "Tick nr. " + ticks + "<br />";
if (ticks === 1) {
clearInterval(mytimer);
document.write("Done!");
}
}
</script>
answered Nov 22 '18 at 7:44
Bishal GautamBishal Gautam
855519
add a comment |
Untill you are not getting any number between 1 - 10, you may try calling same method again as follow:
<script>
var ticks;
var mytimer;
function start() {
ticks = Number(prompt("where do you want to start a countdown, min 1 max 10")) + 1;
if (ticks > 10) {
start();
}
else if (ticks < 1) {
start();
}
mytimer = setInterval(tick, 1000)
}
function tick() {
ticks--;
document.getElementById("p").innerHTML = "Tick nr. " + ticks + "<br />";
if (ticks === 1) {
clearInterval(mytimer);
document.write("Done!");
}
}
</script>
answered Nov 22 '18 at 7:44
Bishal GautamBishal Gautam
855519
add a comment |
Untill you are not getting any number between 1 - 10, you may try calling same method again as follow:
<script>
var ticks;
var mytimer;
function start() {
ticks = Number(prompt("where do you want to start a countdown, min 1 max 10")) + 1;
if (ticks > 10) {
start();
}
else if (ticks < 1) {
start();
}
mytimer = setInterval(tick, 1000)
}
function tick() {
ticks--;
document.getElementById("p").innerHTML = "Tick nr. " + ticks + "<br />";
if (ticks === 1) {
clearInterval(mytimer);
document.write("Done!");
}
}
</script>
answered Nov 22 '18 at 7:44
Bishal GautamBishal Gautam
855519
Untill you are not getting any number between 1 - 10, you may try calling same method again as follow:
<script>
var ticks;
var mytimer;
function start() {
ticks = Number(prompt("where do you want to start a countdown, min 1 max 10")) + 1;
if (ticks > 10) {
start();
}
else if (ticks < 1) {
start();
}
mytimer = setInterval(tick, 1000)
}
function tick() {
ticks--;
document.getElementById("p").innerHTML = "Tick nr. " + ticks + "<br />";
if (ticks === 1) {
clearInterval(mytimer);
document.write("Done!");
}
}
</script>
answered Nov 22 '18 at 7:44
Bishal GautamBishal Gautam
855519
answered Nov 22 '18 at 7:44
Bishal GautamBishal Gautam
855519
answered Nov 22 '18 at 7:44
Bishal GautamBishal Gautam
855519
answered Nov 22 '18 at 7:44
Bishal GautamBishal Gautam
855519
855519
add a comment |
add a comment |
The solution is simple. When you first time enter a value bigger than 11 if (ticks > 11) { statement catches it and then shows new prompt. If you again enter a value bigger than 11, there is no if statement to catch it. If you run the start function again when a faulty value entered, you can fix this problem.
answered Nov 22 '18 at 7:57
Rıdvan SumsetRıdvan Sumset
1257
ah yes, that makes sense, but do you have any examples on how i can recall the function, is it possible to keep on calling the function in an if statement.
– Raahim Khan
Nov 22 '18 at 9:49
You can do it like in @Bishal's answer ---->if (ticks > 11) { start();}
– Rıdvan Sumset
Nov 22 '18 at 10:21
add a comment |
The solution is simple. When you first time enter a value bigger than 11 if (ticks > 11) { statement catches it and then shows new prompt. If you again enter a value bigger than 11, there is no if statement to catch it. If you run the start function again when a faulty value entered, you can fix this problem.
answered Nov 22 '18 at 7:57
Rıdvan SumsetRıdvan Sumset
1257
ah yes, that makes sense, but do you have any examples on how i can recall the function, is it possible to keep on calling the function in an if statement.
– Raahim Khan
Nov 22 '18 at 9:49
You can do it like in @Bishal's answer ---->if (ticks > 11) { start();}
– Rıdvan Sumset
Nov 22 '18 at 10:21
add a comment |
The solution is simple. When you first time enter a value bigger than 11 if (ticks > 11) { statement catches it and then shows new prompt. If you again enter a value bigger than 11, there is no if statement to catch it. If you run the start function again when a faulty value entered, you can fix this problem.
answered Nov 22 '18 at 7:57
Rıdvan SumsetRıdvan Sumset
1257
The solution is simple. When you first time enter a value bigger than 11 if (ticks > 11) { statement catches it and then shows new prompt. If you again enter a value bigger than 11, there is no if statement to catch it. If you run the start function again when a faulty value entered, you can fix this problem.
answered Nov 22 '18 at 7:57
Rıdvan SumsetRıdvan Sumset
1257
answered Nov 22 '18 at 7:57
Rıdvan SumsetRıdvan Sumset
1257
answered Nov 22 '18 at 7:57
Rıdvan SumsetRıdvan Sumset
1257
answered Nov 22 '18 at 7:57
Rıdvan SumsetRıdvan Sumset
1257
1257
ah yes, that makes sense, but do you have any examples on how i can recall the function, is it possible to keep on calling the function in an if statement.
– Raahim Khan
Nov 22 '18 at 9:49
You can do it like in @Bishal's answer ---->if (ticks > 11) { start();}
– Rıdvan Sumset
Nov 22 '18 at 10:21
add a comment |
ah yes, that makes sense, but do you have any examples on how i can recall the function, is it possible to keep on calling the function in an if statement.
– Raahim Khan
Nov 22 '18 at 9:49
You can do it like in @Bishal's answer ---->if (ticks > 11) { start();}
– Rıdvan Sumset
Nov 22 '18 at 10:21
ah yes, that makes sense, but do you have any examples on how i can recall the function, is it possible to keep on calling the function in an if statement.
– Raahim Khan
Nov 22 '18 at 9:49
ah yes, that makes sense, but do you have any examples on how i can recall the function, is it possible to keep on calling the function in an if statement.
– Raahim Khan
Nov 22 '18 at 9:49
You can do it like in @Bishal's answer ---->
if (ticks > 11) { start();}– Rıdvan Sumset
Nov 22 '18 at 10:21
You can do it like in @Bishal's answer ---->
if (ticks > 11) { start();}– Rıdvan Sumset
Nov 22 '18 at 10:21
add a comment |
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