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Home/ Questions/Q 8062539
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Editorial Team
  • 0
Editorial Team
Asked: June 5, 20262026-06-05T10:42:15+00:00 2026-06-05T10:42:15+00:00

When a user clicks a button I want to return some data and iterate

  • 0

When a user clicks a button I want to return some data and iterate through the JSON so that I can append the results to a table row.

At this point I am just trying to get my loop to work, here’s my code.

My JSON is coming back as follows: {“COLUMNS”:[“username”,”password”],”DATA”:[[“foo”, “bar”]]}

$("#button").click(function(){

    $.ajax({
        url: 'http://localhost/test.php',
        type: 'get',
        success: function(data) {  
         $.each(data.items, function(item) {
            console.log(item);
            });
        },
        error: function(e) {
            console.log(e.message);
        }
    });

});

I’m getting a jQuery (line 16, a is not defined) error. What am I doing wrong?

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-05T10:42:17+00:00Added an answer on June 5, 2026 at 10:42 am

    Assuming your JSON is like this

    var item=  {
            "items": [
                      { "FirstName":"John" , "LastName":"Doe" },
                      { "FirstName":"Anna" , "LastName":"Smith" },
                      { "FirstName":"Peter" , "LastName":"Jones" }
                     ]
               }
    

    You can query it like this

    $.each(item.items, function(index,item) {        
        alert(item.FirstName+" "+item.LastName)
    });
    

    Sample : http://jsfiddle.net/4HxSr/9/

    EDIT : As per the JSON OP Posted later

    Your JSON does not have an items, so it is invalid.

    As per your JSON like this

    var item= {  "COLUMNS": [  "username", "password" ],
                 "DATA": [    [ "foo", "bar"  ] ,[ "foo2", "bar2"  ]]
              }
    

    You should query it like this

    console.debug(item.COLUMNS[0])
    console.debug(item.COLUMNS[1])
    
     $.each(item.DATA, function(index,item) {        
        console.debug(item[0])
        console.debug(item[1])
      });
    

    Working sample : http://jsfiddle.net/4HxSr/19/

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