json_encode is not working

Question

I am using Jquery Flot Chart for plotting charts. To pass the data to chart,I am getting data from the database, and then I need to pass that php array into javascript array. I am using json_encode($array),but it is not working.

     $row=array();
      while($res=mysql_fetch_array($exequery))
     {
        $row[]=$res[0];
     }

Now,in the same file(without using ajax), I want to pass this php array to javascript array,which then will be used to plot chart.

    <script src="jquery-1.8.3.min.js"></script>
     <script src="jquer.flot.min.js"></script>     
     <script type ="text/javascript">
     var dataArray=<?php echo json_encode($row)?>;

When I am using this, chart is getting disappeared and alert(dataArray) is also showing nothing. Please help.

How you're using `json_encode()` [is as recommended on other questions](http://stackoverflow.com/questions/23740548/how-to-pass-variables-and-data-from-php-to-javascript) and should work as expected. Have you verified that the array in `$row` and `dataArray` is structured as Flow Chart is expecting? — Jonathan Lonowski, Jul 16 '15 at 05:53
Side note: You appear to have a typo in one of the `src` paths -- lacking the `y` in `jquery` for `jquery.flot.min.js`. — Jonathan Lonowski, Jul 16 '15 at 05:55

score 0 · Answer 1 · answered Jul 16 '15 at 05:56

0

Json parse using jquery:

var dataArray=jQuery.parseJSON('<?php echo json_encode($row)?>');

Open this link:- http://api.jquery.com/jquery.parsejson/

answered Jul 16 '15 at 05:56

Manoj Kushwaha

114
12

1

The quotes and `parseJSON()` aren't really necessary when outputting JSON directly into JavaScript code. Provided it's output where an *Expression* is expected, such as to the right of an assignment operator, the string of JSON will be parsed as valid JavaScript, using the keywords, literals, and initializers that inspired JSON's syntax. – Jonathan Lonowski Jul 16 '15 at 06:03

json_encode is not working

1 Answers1