ajax success(data) to variable += data

Question

Here is my code:

 var myVariable = "<div>";

 for (i = 1; i < 5; ++i) {
 myVariable += '<div class="user' + i + '">'; 

 $.ajax({
      type: "POST",
      data: "uziv=" + i,
      url: "../php/ajax_objednavky.php",
      success: function(data){
      //alert(data);  work well
      //$('#content').html(data); work well
      myVariable += data; //doesn't work
      }              
    });

 myVariable += '</div>';

 } 

 myVariable += "</div>";
 $('#myDiv').html(myVariable);

PHP file looks like

<? echo "data from mysql"; ?>

I can view the data using alert, but how can I add data to a variable?

can u post the content of data and what you got when run your program in myVariable — iJade, Oct 17 '13 at 08:45
@DannyThunder that could not be an issue. check this http://stackoverflow.com/questions/2485423/javascript-is-using-var-to-declare-variables-optional — Praveen, Oct 17 '13 at 08:49

DannyThunder · Answer 1 · 2013-10-17T09:30:29.470

Edited.

It is probably because the ajax request take some time to finnish. By the time the first result gets back, the JS is allready finnished.

This code might be improved, but I hope you get what I mean.

var myVariable = "<div>";

 for (i = 1; i < 5; ++i) {
 myVariable += '<div class="user' + i + '">'; 

 $.ajax({
      type: "POST",
      data: "uziv=" + i,
      url: "../php/ajax_objednavky.php",
      success: function(data){
        myVariable += data; //doesn't work
        myVariable += '</div>';

        if(i == 4){
          myVariable += "</div>";
          $('#myDiv').html(myVariable);
        }
      }              
    });
 }

Instead of sending every i, you could send all of them in on ajax-request as json and get all result back in one request. This will need some reconstruction of your service, but It might be the better way of doing this.

score 0 · Answer 2 · answered Oct 17 '13 at 09:07

0

I'm surprised by your post :

data: "ter="+ objednavky + "&uziv=" + uzivatel,

I always used data: { ter: objednavky, uziv: uzivatel }

Easier to read, IMHO.

then, What do you mean by "doesnt work ?"

try this :

 myVariable += data;
 alert("foo");

Do you see the foo alert ? (I would expect that you messed up with "" and it is not PRINTED on the div, but it "works"). Exemple : ""

answered Oct 17 '13 at 09:07

mansuetus

782
5
17

yes, I see foo alert. I can see in alert(data) but, I can't add data to myVariable. – user2887469 Oct 17 '13 at 09:13
Assuming you posted the code you try to execute (no typo in div), try : var txt = new String(data) ; myVariable += txt; – mansuetus Oct 17 '13 at 15:05

Praveen · Answer 3 · 2013-10-17T09:12:33.183

-1

I guess that is what you need because there is no need to concatenate the data with myVaraible.

Updates:

The reason, the ajax data is returning an object. So concatenating will be not a good one. So I guess you might have data.something. Check this in such a way.

In-case if it is a object, then the concatenation output will be

<div>[object Object]

edited Oct 17 '13 at 09:12

answered Oct 17 '13 at 08:45

Praveen

55,303
33
133
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myVariable is simple variable – iJade Oct 17 '13 at 08:47
@user2887469 Agree@iJay , it won't work. Can you provide more details. What are you actually trying? Do you want the data to shown inside the div (myVariable) – Praveen Oct 17 '13 at 08:57
@user2887469 The reason, the ajax `data` is returning the object. So concatenating will be not a good one. So I guess you might have `data.something`. Check this in such a way. If it is a string it would have worked http://jsfiddle.net/VAyNX/ – Praveen Oct 17 '13 at 09:11
@user2887469 what is your `data` have? – Praveen Oct 17 '13 at 09:14

ajax success(data) to variable += data

3 Answers3