Why doesn't javascript take my JSON array?

Question

My php file ends with:

echo json_encode($array1);
echo ";";
echo json_encode($array2);

and prints out e.g.

[1358499135965,68];[1358499140000,2]

My javascript code looks so:

function requestData() {
    $.ajax({
        url: 'livedata.php',
        success: function(point) {    
            var yenidata = point.split(";");
            alert(yenidata[0]);
            alert(yenidata[1]);
        });
    }

Why do I not get an alert?

if your entire response is [1358499135965,68];[1358499140000,2], then it is invalid JSON and may not be getting passed into the success handler. — Jim Deville
– Jim Deville, Commented Jan 18, 2013 at 9:04

Minko Gechev · Accepted Answer · 2013-01-18 11:20:22Z

2

Your JSON is not valid.

Try:

echo '[';
echo json_encode($array1);
echo ",";
echo json_encode($array2);
echo ']';

Now the PHP page will print you: [[1358499135965,68],[1358499140000,2]] Which can be parsed automatically as JSON using dataType:"json" in your ajax call.

When you include jQuery your code should look like:

function requestData() {
   $.ajax({
      url: 'livedata.php',
      dataType: 'json',
      success: function(point) {
         console.log(point[0]); //Array [1358499135965,68]
         console.log(point[1]); //Array [1358499140000,2]
      }
   });
}

edited Jan 18, 2013 at 11:20

answered Jan 18, 2013 at 9:05

Minko Gechev

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2 Comments

Matanya Over a year ago

if he adds dataType:"json" to the ajax func, he doesn't even need to parse as json

Susanne92 Over a year ago

But i have to echo ",;"; remove the ; ... echo ","; Now it works! Thanks

sachleen · Accepted Answer · 2013-01-18 09:05:35Z

2

Uncaught ReferenceError: $ is not defined means you're not including jQuery. You need to have that to use the features you're trying to use.

answered Jan 18, 2013 at 9:05

sachleen

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3 Comments

Susanne92 Over a year ago

This php file and javascript is included in a bigger file. So jQuery library is included

sachleen Over a year ago

Is it included before that JS?

sachleen Over a year ago

Try jQuery.ajax instead of $.ajax

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