javascript - uncaught referenceerror: table is not defined - Stack Overflow

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I am getting this error message when trying to create a table using the Google Visualization API, with PHP & MySQL in the background.

• db connection is OK
• creating JSON from PHP array is OK
• JSON format is OK

This is the error I get: uncaught referenceerror: table is not defined

 <!--Load the AJAX API-->
<script type="text/javascript" src=""></script>
<script type="text/javascript">

 // Load the Visualization API and the table package.
  google.load('visualization', '1.0', {'packages':['table']});
  google.setOnLoadCallback(drawTable);

  function drawTable() {
        var tdata = new google.visualization.DataTable(
        <?php echo json_encode($data); ?>);
    var options = {
      title: 'A Station'
    };
    var table_draw = new google.visualization.Table(document.getElementById('table_div'));
    table_draw.draw(tdata, options);
  }

I am getting this error message when trying to create a table using the Google Visualization API, with PHP & MySQL in the background.

• db connection is OK
• creating JSON from PHP array is OK
• JSON format is OK

This is the error I get: uncaught referenceerror: table is not defined

 <!--Load the AJAX API-->
<script type="text/javascript" src="https://www.google./jsapi"></script>
<script type="text/javascript">

 // Load the Visualization API and the table package.
  google.load('visualization', '1.0', {'packages':['table']});
  google.setOnLoadCallback(drawTable);

  function drawTable() {
        var tdata = new google.visualization.DataTable(
        <?php echo json_encode($data); ?>);
    var options = {
      title: 'A Station'
    };
    var table_draw = new google.visualization.Table(document.getElementById('table_div'));
    table_draw.draw(tdata, options);
  }

• javascript
• jquery
• json

Share Improve this question Follow edited Oct 24, 2014 at 20:38 nazanin asked Oct 24, 2014 at 19:42 nazaninnazanin 7333 silver badges1414 bronze badges 1

• what this has anything to do with php and mysql? – itachi Commented Oct 24, 2014 at 20:07

Add a ment  | 

1 Answer 1

Sorted by: Reset to default 4

You are trying to use a variable, table, which you have never defined.

Possibly you wanted to use the variable table_draw which you have defined in the line before.

var table_draw = new google.visualization.Table(document.getElementById('table_div'));
table_draw.draw(tdata, options);

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