Why is it setting I to the response lenght? [duplicate]

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JavaScript closure inside loops – simple practical example

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I'm trying to create a table from an ajax json response but my for loop sets i to the max value instantly.

$.ajax(settings).done(function (response) 
 console.log(response);
 var i = "";
 td = document.createElement('td');
 tr = document.createElement('tr');
 tableContent = document.getElementById('analysisTable');
 for (i = 0; i < response.segments.length; ++i) 
 td.innerHTML = response.segments[i].confidence;
 $(td).attr(
 'id': i,
 'class': "analysis",
 );
 tableContent.appendChild(tr).appendChild(td);
 
);

edited Nov 9 at 14:13

George

4,39711731

asked Nov 9 at 14:12

isetnt

marked as duplicate by vlaz, charlietfl html
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Nov 9 at 14:41

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

You're probably overwriting your previous appends every time.
– Shilly
Nov 9 at 14:15

2

You are creating a single <tr> and a single <td>. You will at the very least need to create a <td> for each segment element.
– Chris G
Nov 9 at 14:17

Here's the proper jQuery way to do this: jsfiddle.net/khrismuc/ycn9qspw
– Chris G
Nov 9 at 14:23

"my for loop sets i to the max value instantly." if you use the debugger to step through your code it will be trivial to see that this is not what is happening at all. The symptoms of your problem might make it look like that, but as others have said, it's because you keep overwriting the same variable every time, so all references to that variable which you add to the DOM will contain the same content (and since it's a loop, it'll be whatever the content was the last time the loop ran).
– ADyson
Nov 9 at 14:29

add a comment |

up vote
0
down vote

favorite

This question already has an answer here:

JavaScript closure inside loops – simple practical example

39 answers

I'm trying to create a table from an ajax json response but my for loop sets i to the max value instantly.

$.ajax(settings).done(function (response) 
 console.log(response);
 var i = "";
 td = document.createElement('td');
 tr = document.createElement('tr');
 tableContent = document.getElementById('analysisTable');
 for (i = 0; i < response.segments.length; ++i) 
 td.innerHTML = response.segments[i].confidence;
 $(td).attr(
 'id': i,
 'class': "analysis",
 );
 tableContent.appendChild(tr).appendChild(td);
 
);

edited Nov 9 at 14:13

George

4,39711731

asked Nov 9 at 14:12

isetnt

marked as duplicate by vlaz, charlietfl html
Users with the html badge can single-handedly close html questions as duplicates and reopen them as needed.

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Nov 9 at 14:41

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

You're probably overwriting your previous appends every time.
– Shilly
Nov 9 at 14:15

2

You are creating a single <tr> and a single <td>. You will at the very least need to create a <td> for each segment element.
– Chris G
Nov 9 at 14:17

Here's the proper jQuery way to do this: jsfiddle.net/khrismuc/ycn9qspw
– Chris G
Nov 9 at 14:23

"my for loop sets i to the max value instantly." if you use the debugger to step through your code it will be trivial to see that this is not what is happening at all. The symptoms of your problem might make it look like that, but as others have said, it's because you keep overwriting the same variable every time, so all references to that variable which you add to the DOM will contain the same content (and since it's a loop, it'll be whatever the content was the last time the loop ran).
– ADyson
Nov 9 at 14:29

add a comment |

up vote
0
down vote

favorite

This question already has an answer here:

JavaScript closure inside loops – simple practical example

39 answers

I'm trying to create a table from an ajax json response but my for loop sets i to the max value instantly.

$.ajax(settings).done(function (response) 
 console.log(response);
 var i = "";
 td = document.createElement('td');
 tr = document.createElement('tr');
 tableContent = document.getElementById('analysisTable');
 for (i = 0; i < response.segments.length; ++i) 
 td.innerHTML = response.segments[i].confidence;
 $(td).attr(
 'id': i,
 'class': "analysis",
 );
 tableContent.appendChild(tr).appendChild(td);
 
);

edited Nov 9 at 14:13

George

4,39711731

asked Nov 9 at 14:12

isetnt

This question already has an answer here:

JavaScript closure inside loops – simple practical example

39 answers

I'm trying to create a table from an ajax json response but my for loop sets i to the max value instantly.

$.ajax(settings).done(function (response) 
 console.log(response);
 var i = "";
 td = document.createElement('td');
 tr = document.createElement('tr');
 tableContent = document.getElementById('analysisTable');
 for (i = 0; i < response.segments.length; ++i) 
 td.innerHTML = response.segments[i].confidence;
 $(td).attr(
 'id': i,
 'class': "analysis",
 );
 tableContent.appendChild(tr).appendChild(td);
 
);

This question already has an answer here:

JavaScript closure inside loops – simple practical example

39 answers

javascript jquery html json ajax

edited Nov 9 at 14:13

George

4,39711731

asked Nov 9 at 14:12

isetnt

edited Nov 9 at 14:13

George

4,39711731

asked Nov 9 at 14:12

isetnt

edited Nov 9 at 14:13

George

4,39711731

edited Nov 9 at 14:13

George

4,39711731

edited Nov 9 at 14:13

George

4,39711731

asked Nov 9 at 14:12

isetnt

asked Nov 9 at 14:12

isetnt

asked Nov 9 at 14:12

isetnt

marked as duplicate by vlaz, charlietfl html
Users with the html badge can single-handedly close html questions as duplicates and reopen them as needed.

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Nov 9 at 14:41

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

marked as duplicate by vlaz, charlietfl html
Users with the html badge can single-handedly close html questions as duplicates and reopen them as needed.

StackExchange.ready(function()
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function()
$hover.showInfoMessage('',
messageElement: $msg.clone().show(),
transient: false,
position: my: 'bottom left', at: 'top center', offsetTop: -7 ,
dismissable: false,
relativeToBody: true
);
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StackExchange.helpers.removeMessages();

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Nov 9 at 14:41

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

You're probably overwriting your previous appends every time.
– Shilly
Nov 9 at 14:15

2

You are creating a single <tr> and a single <td>. You will at the very least need to create a <td> for each segment element.
– Chris G
Nov 9 at 14:17

Here's the proper jQuery way to do this: jsfiddle.net/khrismuc/ycn9qspw
– Chris G
Nov 9 at 14:23

"my for loop sets i to the max value instantly." if you use the debugger to step through your code it will be trivial to see that this is not what is happening at all. The symptoms of your problem might make it look like that, but as others have said, it's because you keep overwriting the same variable every time, so all references to that variable which you add to the DOM will contain the same content (and since it's a loop, it'll be whatever the content was the last time the loop ran).
– ADyson
Nov 9 at 14:29

add a comment |

You're probably overwriting your previous appends every time.
– Shilly
Nov 9 at 14:15

2

You are creating a single <tr> and a single <td>. You will at the very least need to create a <td> for each segment element.
– Chris G
Nov 9 at 14:17

Here's the proper jQuery way to do this: jsfiddle.net/khrismuc/ycn9qspw
– Chris G
Nov 9 at 14:23

"my for loop sets i to the max value instantly." if you use the debugger to step through your code it will be trivial to see that this is not what is happening at all. The symptoms of your problem might make it look like that, but as others have said, it's because you keep overwriting the same variable every time, so all references to that variable which you add to the DOM will contain the same content (and since it's a loop, it'll be whatever the content was the last time the loop ran).
– ADyson
Nov 9 at 14:29

You're probably overwriting your previous appends every time.
– Shilly
Nov 9 at 14:15

You are creating a single <tr> and a single <td>. You will at the very least need to create a <td> for each segment element.
– Chris G
Nov 9 at 14:17

Here's the proper jQuery way to do this: jsfiddle.net/khrismuc/ycn9qspw
– Chris G
Nov 9 at 14:23

"my for loop sets i to the max value instantly." if you use the debugger to step through your code it will be trivial to see that this is not what is happening at all. The symptoms of your problem might make it look like that, but as others have said, it's because you keep overwriting the same variable every time, so all references to that variable which you add to the DOM will contain the same content (and since it's a loop, it'll be whatever the content was the last time the loop ran).
– ADyson
Nov 9 at 14:29

add a comment |

1 Answer
1

active

oldest

votes

up vote
-2
down vote

You're creating the tr and td elements outside the loop. Re-Setting the innerHtml does not create new elements.

answered Nov 9 at 14:18

Ingo Bochmann

1023

That's a comment, not an answer.
– Chris G
Nov 9 at 14:28

add a comment |

1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

up vote
-2
down vote

You're creating the tr and td elements outside the loop. Re-Setting the innerHtml does not create new elements.

answered Nov 9 at 14:18

Ingo Bochmann

1023

That's a comment, not an answer.
– Chris G
Nov 9 at 14:28

add a comment |

up vote
-2
down vote

You're creating the tr and td elements outside the loop. Re-Setting the innerHtml does not create new elements.

answered Nov 9 at 14:18

Ingo Bochmann

1023

That's a comment, not an answer.
– Chris G
Nov 9 at 14:28

add a comment |

up vote
-2
down vote

You're creating the tr and td elements outside the loop. Re-Setting the innerHtml does not create new elements.

answered Nov 9 at 14:18

Ingo Bochmann

1023

You're creating the tr and td elements outside the loop. Re-Setting the innerHtml does not create new elements.

answered Nov 9 at 14:18

Ingo Bochmann

1023

answered Nov 9 at 14:18

Ingo Bochmann

1023

answered Nov 9 at 14:18

Ingo Bochmann

1023

answered Nov 9 at 14:18

Ingo Bochmann

1023

That's a comment, not an answer.
– Chris G
Nov 9 at 14:28

add a comment |

That's a comment, not an answer.
– Chris G
Nov 9 at 14:28

That's a comment, not an answer.
– Chris G
Nov 9 at 14:28

add a comment |

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