For loop pauses, but prints out the Array all at once
0
I'm having trouble Grok'ing a simple JavaScript for loop.
Here is an example:
var arr = ["Banana", "Orange", "Apple", "Mango"];
for (i = 1; i <= 5; ++i) {
setDelay(i);
}
function setDelay(i) {
setTimeout(function(){
for(const value of arr) {
document.write(value);
document.write("<br />");
}
}, i * 1000);
}What I'm trying to do is pause after it prints the first value, but instead it runs the entire For loop then pauses. I think I'm trying to do something with for loops that just won't work, but I would like to know why.
Any help would be appreciated.
Plunkr here: https://plnkr.co/edit/tnmFrIRTDJI8T294Qh4z?p=preview
The example Javascript, setTimeout loops? didn't help me figure it out. I still got the concept wrong as George Pantazes points out.
javascript
asked Nov 13 '18 at 19:03
RostasanRostasan
105
add a comment |
0
I'm having trouble Grok'ing a simple JavaScript for loop.
Here is an example:
var arr = ["Banana", "Orange", "Apple", "Mango"];
for (i = 1; i <= 5; ++i) {
setDelay(i);
}
function setDelay(i) {
setTimeout(function(){
for(const value of arr) {
document.write(value);
document.write("<br />");
}
}, i * 1000);
}What I'm trying to do is pause after it prints the first value, but instead it runs the entire For loop then pauses. I think I'm trying to do something with for loops that just won't work, but I would like to know why.
Any help would be appreciated.
Plunkr here: https://plnkr.co/edit/tnmFrIRTDJI8T294Qh4z?p=preview
The example Javascript, setTimeout loops? didn't help me figure it out. I still got the concept wrong as George Pantazes points out.
javascript
asked Nov 13 '18 at 19:03
RostasanRostasan
105
Possible duplicate of Javascript, setTimeout loops?
– Tyler Roper
Nov 13 '18 at 19:05
2
By running your code, you can see that you are printing the entire array within thesetTimeout. Instead of looping through the entirearrin thesetTimeoutforloop, if you access your array likearr[i]and only print one item instead of all of them, your code would work as you intend.
– George Pantazes
Nov 13 '18 at 19:06
add a comment |
0
0
0
I'm having trouble Grok'ing a simple JavaScript for loop.
Here is an example:
var arr = ["Banana", "Orange", "Apple", "Mango"];
for (i = 1; i <= 5; ++i) {
setDelay(i);
}
function setDelay(i) {
setTimeout(function(){
for(const value of arr) {
document.write(value);
document.write("<br />");
}
}, i * 1000);
}What I'm trying to do is pause after it prints the first value, but instead it runs the entire For loop then pauses. I think I'm trying to do something with for loops that just won't work, but I would like to know why.
Any help would be appreciated.
Plunkr here: https://plnkr.co/edit/tnmFrIRTDJI8T294Qh4z?p=preview
The example Javascript, setTimeout loops? didn't help me figure it out. I still got the concept wrong as George Pantazes points out.
javascript
asked Nov 13 '18 at 19:03
RostasanRostasan
105
I'm having trouble Grok'ing a simple JavaScript for loop.
Here is an example:
var arr = ["Banana", "Orange", "Apple", "Mango"];
for (i = 1; i <= 5; ++i) {
setDelay(i);
}
function setDelay(i) {
setTimeout(function(){
for(const value of arr) {
document.write(value);
document.write("<br />");
}
}, i * 1000);
}What I'm trying to do is pause after it prints the first value, but instead it runs the entire For loop then pauses. I think I'm trying to do something with for loops that just won't work, but I would like to know why.
Any help would be appreciated.
Plunkr here: https://plnkr.co/edit/tnmFrIRTDJI8T294Qh4z?p=preview
The example Javascript, setTimeout loops? didn't help me figure it out. I still got the concept wrong as George Pantazes points out.
var arr = ["Banana", "Orange", "Apple", "Mango"];
for (i = 1; i <= 5; ++i) {
setDelay(i);
}
function setDelay(i) {
setTimeout(function(){
for(const value of arr) {
document.write(value);
document.write("<br />");
}
}, i * 1000);
}var arr = ["Banana", "Orange", "Apple", "Mango"];
for (i = 1; i <= 5; ++i) {
setDelay(i);
}
function setDelay(i) {
setTimeout(function(){
for(const value of arr) {
document.write(value);
document.write("<br />");
}
}, i * 1000);
}javascript
javascript
asked Nov 13 '18 at 19:03
RostasanRostasan
105
asked Nov 13 '18 at 19:03
RostasanRostasan
105
edited Nov 13 '18 at 19:59
Rostasan
asked Nov 13 '18 at 19:03
RostasanRostasan
105
asked Nov 13 '18 at 19:03
RostasanRostasan
105
asked Nov 13 '18 at 19:03
RostasanRostasan
105
105
Possible duplicate of Javascript, setTimeout loops?
– Tyler Roper
Nov 13 '18 at 19:05
2
By running your code, you can see that you are printing the entire array within thesetTimeout. Instead of looping through the entirearrin thesetTimeoutforloop, if you access your array likearr[i]and only print one item instead of all of them, your code would work as you intend.
– George Pantazes
Nov 13 '18 at 19:06
add a comment |
Possible duplicate of Javascript, setTimeout loops?
– Tyler Roper
Nov 13 '18 at 19:05
2
By running your code, you can see that you are printing the entire array within thesetTimeout. Instead of looping through the entirearrin thesetTimeoutforloop, if you access your array likearr[i]and only print one item instead of all of them, your code would work as you intend.
– George Pantazes
Nov 13 '18 at 19:06
Possible duplicate of Javascript, setTimeout loops?
– Tyler Roper
Nov 13 '18 at 19:05
Possible duplicate of Javascript, setTimeout loops?
– Tyler Roper
Nov 13 '18 at 19:05
2
2
By running your code, you can see that you are printing the entire array within the
setTimeout. Instead of looping through the entire arr in the setTimeout for loop, if you access your array like arr[i] and only print one item instead of all of them, your code would work as you intend.– George Pantazes
Nov 13 '18 at 19:06
By running your code, you can see that you are printing the entire array within the
setTimeout. Instead of looping through the entire arr in the setTimeout for loop, if you access your array like arr[i] and only print one item instead of all of them, your code would work as you intend.– George Pantazes
Nov 13 '18 at 19:06
add a comment |
3 Answers
3
active
oldest
votes
1
Inside setTimeout - instead of printing entire array just print current element.
var arr = ["Banana", "Orange", "Apple", "Mango"];
for (let i = 1; i < 5; ++i) {
setDelay(i);
}
function setDelay(i) {
setTimeout(function(){
console.log(arr[i-1])
}, i * 1000);
}answered Nov 13 '18 at 19:09
Nitish NarangNitish Narang
2,9401815
Why are you using 1-i? Is it better to start with i = 1? I do understand that the array index starts at 0. Oh wait, so the array is four long, so you changed the i to equal 1 match the offset? Would it be better if I did i < arr.length? -Thanks
– Rostasan
Nov 13 '18 at 20:04
Yes, you are correct because you have 4 array elements and array index starts from 0
– Nitish Narang
Nov 13 '18 at 20:07
Its better to do i< arr.length. Also, its always preferable to start i with 0 rather than 1
– Nitish Narang
Nov 14 '18 at 6:45
add a comment |
1
There were multiple issues within your code:
- Be sure to mind that Javascript arrays are 0-based. You were starting at 1.
- Similarly, be sure to mind the ending bound of the index (in your for loop). It was going too far even for a 1-base (it was going to 5 out of 4 available items)
- Within the
setTimeout, you were printing the entire array by usingfor(const value of arr). You were probably trying to pass inias an index to index only one element.
With those comments in mind, here is the working code with those parts changed:
var arr = ["Banana", "Orange", "Apple", "Mango"];
for (i = 0; i < arr.length; ++i) {
setDelay(i);
}
function setDelay(i) {
setTimeout(function(){
document.write(arr[i]);
document.write("<br />");
}, i * 1000);
}answered Nov 13 '18 at 19:13
George PantazesGeorge Pantazes
472617
Refer to @Nitish-Narang's answer for better Javascript code style, although I think my advice about the indices is worth taking as well.
– George Pantazes
Nov 13 '18 at 19:15
wow thanks. I tried to do that by setting i = to the array (out of desperation), but this makes a lot more sense. I have access to the array and i. So that would let me use arr[i].
– Rostasan
Nov 13 '18 at 20:01
add a comment |
0
An alternative approach to creating a delay by using setTimeout in a loop is the setInterval function. setInterval will execute the function argument each time a specified duration of of milliseconds passes:
var arr = ["Banana", "Orange", "Apple", "Mango"];
var i = 0;
var intervalId = setInterval(logNext, 1000);
function logNext() {
if (i < arr.length) {
console.log(arr[i++]);
} else {
console.log('End of array reached!');
clearInterval(intervalId);
}
}answered Nov 13 '18 at 19:09
Tom O.Tom O.
2,36311325
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
1
Inside setTimeout - instead of printing entire array just print current element.
var arr = ["Banana", "Orange", "Apple", "Mango"];
for (let i = 1; i < 5; ++i) {
setDelay(i);
}
function setDelay(i) {
setTimeout(function(){
console.log(arr[i-1])
}, i * 1000);
}answered Nov 13 '18 at 19:09
Nitish NarangNitish Narang
2,9401815
Why are you using 1-i? Is it better to start with i = 1? I do understand that the array index starts at 0. Oh wait, so the array is four long, so you changed the i to equal 1 match the offset? Would it be better if I did i < arr.length? -Thanks
– Rostasan
Nov 13 '18 at 20:04
Yes, you are correct because you have 4 array elements and array index starts from 0
– Nitish Narang
Nov 13 '18 at 20:07
Its better to do i< arr.length. Also, its always preferable to start i with 0 rather than 1
– Nitish Narang
Nov 14 '18 at 6:45
add a comment |
1
Inside setTimeout - instead of printing entire array just print current element.
var arr = ["Banana", "Orange", "Apple", "Mango"];
for (let i = 1; i < 5; ++i) {
setDelay(i);
}
function setDelay(i) {
setTimeout(function(){
console.log(arr[i-1])
}, i * 1000);
}answered Nov 13 '18 at 19:09
Nitish NarangNitish Narang
2,9401815
Why are you using 1-i? Is it better to start with i = 1? I do understand that the array index starts at 0. Oh wait, so the array is four long, so you changed the i to equal 1 match the offset? Would it be better if I did i < arr.length? -Thanks
– Rostasan
Nov 13 '18 at 20:04
Yes, you are correct because you have 4 array elements and array index starts from 0
– Nitish Narang
Nov 13 '18 at 20:07
Its better to do i< arr.length. Also, its always preferable to start i with 0 rather than 1
– Nitish Narang
Nov 14 '18 at 6:45
add a comment |
1
1
1
Inside setTimeout - instead of printing entire array just print current element.
var arr = ["Banana", "Orange", "Apple", "Mango"];
for (let i = 1; i < 5; ++i) {
setDelay(i);
}
function setDelay(i) {
setTimeout(function(){
console.log(arr[i-1])
}, i * 1000);
}answered Nov 13 '18 at 19:09
Nitish NarangNitish Narang
2,9401815
Inside setTimeout - instead of printing entire array just print current element.
var arr = ["Banana", "Orange", "Apple", "Mango"];
for (let i = 1; i < 5; ++i) {
setDelay(i);
}
function setDelay(i) {
setTimeout(function(){
console.log(arr[i-1])
}, i * 1000);
}var arr = ["Banana", "Orange", "Apple", "Mango"];
for (let i = 1; i < 5; ++i) {
setDelay(i);
}
function setDelay(i) {
setTimeout(function(){
console.log(arr[i-1])
}, i * 1000);
}var arr = ["Banana", "Orange", "Apple", "Mango"];
for (let i = 1; i < 5; ++i) {
setDelay(i);
}
function setDelay(i) {
setTimeout(function(){
console.log(arr[i-1])
}, i * 1000);
}answered Nov 13 '18 at 19:09
Nitish NarangNitish Narang
2,9401815
answered Nov 13 '18 at 19:09
Nitish NarangNitish Narang
2,9401815
answered Nov 13 '18 at 19:09
Nitish NarangNitish Narang
2,9401815
answered Nov 13 '18 at 19:09
Nitish NarangNitish Narang
2,9401815
2,9401815
Why are you using 1-i? Is it better to start with i = 1? I do understand that the array index starts at 0. Oh wait, so the array is four long, so you changed the i to equal 1 match the offset? Would it be better if I did i < arr.length? -Thanks
– Rostasan
Nov 13 '18 at 20:04
Yes, you are correct because you have 4 array elements and array index starts from 0
– Nitish Narang
Nov 13 '18 at 20:07
Its better to do i< arr.length. Also, its always preferable to start i with 0 rather than 1
– Nitish Narang
Nov 14 '18 at 6:45
add a comment |
Why are you using 1-i? Is it better to start with i = 1? I do understand that the array index starts at 0. Oh wait, so the array is four long, so you changed the i to equal 1 match the offset? Would it be better if I did i < arr.length? -Thanks
– Rostasan
Nov 13 '18 at 20:04
Yes, you are correct because you have 4 array elements and array index starts from 0
– Nitish Narang
Nov 13 '18 at 20:07
Its better to do i< arr.length. Also, its always preferable to start i with 0 rather than 1
– Nitish Narang
Nov 14 '18 at 6:45
Why are you using 1-i? Is it better to start with i = 1? I do understand that the array index starts at 0. Oh wait, so the array is four long, so you changed the i to equal 1 match the offset? Would it be better if I did i < arr.length? -Thanks
– Rostasan
Nov 13 '18 at 20:04
Why are you using 1-i? Is it better to start with i = 1? I do understand that the array index starts at 0. Oh wait, so the array is four long, so you changed the i to equal 1 match the offset? Would it be better if I did i < arr.length? -Thanks
– Rostasan
Nov 13 '18 at 20:04
Yes, you are correct because you have 4 array elements and array index starts from 0
– Nitish Narang
Nov 13 '18 at 20:07
Yes, you are correct because you have 4 array elements and array index starts from 0
– Nitish Narang
Nov 13 '18 at 20:07
Its better to do i< arr.length. Also, its always preferable to start i with 0 rather than 1
– Nitish Narang
Nov 14 '18 at 6:45
Its better to do i< arr.length. Also, its always preferable to start i with 0 rather than 1
– Nitish Narang
Nov 14 '18 at 6:45
add a comment |
1
There were multiple issues within your code:
- Be sure to mind that Javascript arrays are 0-based. You were starting at 1.
- Similarly, be sure to mind the ending bound of the index (in your for loop). It was going too far even for a 1-base (it was going to 5 out of 4 available items)
- Within the
setTimeout, you were printing the entire array by usingfor(const value of arr). You were probably trying to pass inias an index to index only one element.
With those comments in mind, here is the working code with those parts changed:
var arr = ["Banana", "Orange", "Apple", "Mango"];
for (i = 0; i < arr.length; ++i) {
setDelay(i);
}
function setDelay(i) {
setTimeout(function(){
document.write(arr[i]);
document.write("<br />");
}, i * 1000);
}answered Nov 13 '18 at 19:13
George PantazesGeorge Pantazes
472617
Refer to @Nitish-Narang's answer for better Javascript code style, although I think my advice about the indices is worth taking as well.
– George Pantazes
Nov 13 '18 at 19:15
wow thanks. I tried to do that by setting i = to the array (out of desperation), but this makes a lot more sense. I have access to the array and i. So that would let me use arr[i].
– Rostasan
Nov 13 '18 at 20:01
add a comment |
1
There were multiple issues within your code:
- Be sure to mind that Javascript arrays are 0-based. You were starting at 1.
- Similarly, be sure to mind the ending bound of the index (in your for loop). It was going too far even for a 1-base (it was going to 5 out of 4 available items)
- Within the
setTimeout, you were printing the entire array by usingfor(const value of arr). You were probably trying to pass inias an index to index only one element.
With those comments in mind, here is the working code with those parts changed:
var arr = ["Banana", "Orange", "Apple", "Mango"];
for (i = 0; i < arr.length; ++i) {
setDelay(i);
}
function setDelay(i) {
setTimeout(function(){
document.write(arr[i]);
document.write("<br />");
}, i * 1000);
}answered Nov 13 '18 at 19:13
George PantazesGeorge Pantazes
472617
Refer to @Nitish-Narang's answer for better Javascript code style, although I think my advice about the indices is worth taking as well.
– George Pantazes
Nov 13 '18 at 19:15
wow thanks. I tried to do that by setting i = to the array (out of desperation), but this makes a lot more sense. I have access to the array and i. So that would let me use arr[i].
– Rostasan
Nov 13 '18 at 20:01
add a comment |
1
1
1
There were multiple issues within your code:
- Be sure to mind that Javascript arrays are 0-based. You were starting at 1.
- Similarly, be sure to mind the ending bound of the index (in your for loop). It was going too far even for a 1-base (it was going to 5 out of 4 available items)
- Within the
setTimeout, you were printing the entire array by usingfor(const value of arr). You were probably trying to pass inias an index to index only one element.
With those comments in mind, here is the working code with those parts changed:
var arr = ["Banana", "Orange", "Apple", "Mango"];
for (i = 0; i < arr.length; ++i) {
setDelay(i);
}
function setDelay(i) {
setTimeout(function(){
document.write(arr[i]);
document.write("<br />");
}, i * 1000);
}answered Nov 13 '18 at 19:13
George PantazesGeorge Pantazes
472617
There were multiple issues within your code:
- Be sure to mind that Javascript arrays are 0-based. You were starting at 1.
- Similarly, be sure to mind the ending bound of the index (in your for loop). It was going too far even for a 1-base (it was going to 5 out of 4 available items)
- Within the
setTimeout, you were printing the entire array by usingfor(const value of arr). You were probably trying to pass inias an index to index only one element.
With those comments in mind, here is the working code with those parts changed:
var arr = ["Banana", "Orange", "Apple", "Mango"];
for (i = 0; i < arr.length; ++i) {
setDelay(i);
}
function setDelay(i) {
setTimeout(function(){
document.write(arr[i]);
document.write("<br />");
}, i * 1000);
}var arr = ["Banana", "Orange", "Apple", "Mango"];
for (i = 0; i < arr.length; ++i) {
setDelay(i);
}
function setDelay(i) {
setTimeout(function(){
document.write(arr[i]);
document.write("<br />");
}, i * 1000);
}var arr = ["Banana", "Orange", "Apple", "Mango"];
for (i = 0; i < arr.length; ++i) {
setDelay(i);
}
function setDelay(i) {
setTimeout(function(){
document.write(arr[i]);
document.write("<br />");
}, i * 1000);
}answered Nov 13 '18 at 19:13
George PantazesGeorge Pantazes
472617
answered Nov 13 '18 at 19:13
George PantazesGeorge Pantazes
472617
answered Nov 13 '18 at 19:13
George PantazesGeorge Pantazes
472617
answered Nov 13 '18 at 19:13
George PantazesGeorge Pantazes
472617
472617
Refer to @Nitish-Narang's answer for better Javascript code style, although I think my advice about the indices is worth taking as well.
– George Pantazes
Nov 13 '18 at 19:15
wow thanks. I tried to do that by setting i = to the array (out of desperation), but this makes a lot more sense. I have access to the array and i. So that would let me use arr[i].
– Rostasan
Nov 13 '18 at 20:01
add a comment |
Refer to @Nitish-Narang's answer for better Javascript code style, although I think my advice about the indices is worth taking as well.
– George Pantazes
Nov 13 '18 at 19:15
wow thanks. I tried to do that by setting i = to the array (out of desperation), but this makes a lot more sense. I have access to the array and i. So that would let me use arr[i].
– Rostasan
Nov 13 '18 at 20:01
Refer to @Nitish-Narang's answer for better Javascript code style, although I think my advice about the indices is worth taking as well.
– George Pantazes
Nov 13 '18 at 19:15
Refer to @Nitish-Narang's answer for better Javascript code style, although I think my advice about the indices is worth taking as well.
– George Pantazes
Nov 13 '18 at 19:15
wow thanks. I tried to do that by setting i = to the array (out of desperation), but this makes a lot more sense. I have access to the array and i. So that would let me use arr[i].
– Rostasan
Nov 13 '18 at 20:01
wow thanks. I tried to do that by setting i = to the array (out of desperation), but this makes a lot more sense. I have access to the array and i. So that would let me use arr[i].
– Rostasan
Nov 13 '18 at 20:01
add a comment |
0
An alternative approach to creating a delay by using setTimeout in a loop is the setInterval function. setInterval will execute the function argument each time a specified duration of of milliseconds passes:
var arr = ["Banana", "Orange", "Apple", "Mango"];
var i = 0;
var intervalId = setInterval(logNext, 1000);
function logNext() {
if (i < arr.length) {
console.log(arr[i++]);
} else {
console.log('End of array reached!');
clearInterval(intervalId);
}
}answered Nov 13 '18 at 19:09
Tom O.Tom O.
2,36311325
add a comment |
0
An alternative approach to creating a delay by using setTimeout in a loop is the setInterval function. setInterval will execute the function argument each time a specified duration of of milliseconds passes:
var arr = ["Banana", "Orange", "Apple", "Mango"];
var i = 0;
var intervalId = setInterval(logNext, 1000);
function logNext() {
if (i < arr.length) {
console.log(arr[i++]);
} else {
console.log('End of array reached!');
clearInterval(intervalId);
}
}answered Nov 13 '18 at 19:09
Tom O.Tom O.
2,36311325
add a comment |
0
0
0
An alternative approach to creating a delay by using setTimeout in a loop is the setInterval function. setInterval will execute the function argument each time a specified duration of of milliseconds passes:
var arr = ["Banana", "Orange", "Apple", "Mango"];
var i = 0;
var intervalId = setInterval(logNext, 1000);
function logNext() {
if (i < arr.length) {
console.log(arr[i++]);
} else {
console.log('End of array reached!');
clearInterval(intervalId);
}
}answered Nov 13 '18 at 19:09
Tom O.Tom O.
2,36311325
An alternative approach to creating a delay by using setTimeout in a loop is the setInterval function. setInterval will execute the function argument each time a specified duration of of milliseconds passes:
var arr = ["Banana", "Orange", "Apple", "Mango"];
var i = 0;
var intervalId = setInterval(logNext, 1000);
function logNext() {
if (i < arr.length) {
console.log(arr[i++]);
} else {
console.log('End of array reached!');
clearInterval(intervalId);
}
}var arr = ["Banana", "Orange", "Apple", "Mango"];
var i = 0;
var intervalId = setInterval(logNext, 1000);
function logNext() {
if (i < arr.length) {
console.log(arr[i++]);
} else {
console.log('End of array reached!');
clearInterval(intervalId);
}
}var arr = ["Banana", "Orange", "Apple", "Mango"];
var i = 0;
var intervalId = setInterval(logNext, 1000);
function logNext() {
if (i < arr.length) {
console.log(arr[i++]);
} else {
console.log('End of array reached!');
clearInterval(intervalId);
}
}answered Nov 13 '18 at 19:09
Tom O.Tom O.
2,36311325
answered Nov 13 '18 at 19:09
Tom O.Tom O.
2,36311325
answered Nov 13 '18 at 19:09
Tom O.Tom O.
2,36311325
answered Nov 13 '18 at 19:09
Tom O.Tom O.
2,36311325
2,36311325
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Possible duplicate of Javascript, setTimeout loops?
– Tyler Roper
Nov 13 '18 at 19:05
2
By running your code, you can see that you are printing the entire array within the
setTimeout. Instead of looping through the entirearrin thesetTimeoutforloop, if you access your array likearr[i]and only print one item instead of all of them, your code would work as you intend.– George Pantazes
Nov 13 '18 at 19:06