Google Visualization API php ajax

Question

I am using this Achieving Google Visualization chart reloads using ajax example to dynamically retrieve data from Mysql database and I get "Uncaught SyntaxError: Unexpected token <" error. This is my HTML file

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
  <head>
    <meta http-equiv="content-type" content="text/html; charset=utf-8"/>
    <script type="text/javascript" src="http://www.google.com/jsapi"></script>
    <script type="text/javascript">
      google.load("jquery", "1.6.1");
      google.load('visualization', '1', {packages: ['table']});
    </script>
    <script type="text/javascript">

    function drawVisualization(dataFromAjax) {
        var data = new google.visualization.DataTable();
        data.addColumn('number', 'InvoiceNo');
        data.addColumn('string', 'B/L No');
        data.addColumn('date', 'Date');
        data.addColumn('string', 'Customer Name');
        data.addColumn('number', 'Amount');
        data.addRows(dataFromAjax);
        var table = new google.visualization.Table(document.getElementById('table'));

        table.draw(data);
   }

    function makeAjaxCall() {
      $.ajax({url:'test.php',
              data: {},
              success: function(responseData) {

                         var arrayForGviz = eval("(" + responseData + ")");
                         drawVisualization(responseData);
                       }
        });
    }
    </script>
  </head>
  <body>
    <input type="button" onclick="makeAjaxCall();return false;" value="Click to get data"></input>
    <div id="table"></div>
  </body>
</html>

and This is my PHP file test.php


    $con = mysql_connect("localhost","userName","password");

    if (!$con)

    {

    die('Could not connect: ' . mysql_error());

    }

    mysql_select_db("DB_NAME", $con);

    $result = mysql_query("call cargosys.rpt_salesByDate('2013/03/05','2013/03/10')");
    $output = array();
    while($row = mysql_fetch_array($result)) {
        // create a temp array to hold the data
        $temp = array();

        // add the data
        $temp[] = $row['inv_no'];
        $temp[] = ''' . $row['bl_no'] . ''';
        $temp[] = ''' . $row['inv_date'] . ''';
        $temp[] = ''' . $row['cust_name'] . ''';
        $temp[] = $row['Amount'];

        // implode the temp array into a comma-separated list and add to the output array
        $output[] = '[' . implode(', ', $temp) . ']';
    }

    // implode the output into a comma-newline separated list and echo
    echo implode(",\n", $output);
    //echo json_encode($output);
    mysql_close($con);

Where the error occur? is this in PHP or Javascript? And my guess is, It is happening at eval() function of javascript. Am I right? — HILARUDEEN S ALLAUDEEN
– HILARUDEEN S ALLAUDEEN, Commented Mar 11, 2013 at 11:37
Php syntax error are there $temp[] = ''' . $row['bl_no'] . '''; — Tuhin Subhra Dey
– Tuhin Subhra Dey, Commented Mar 11, 2013 at 12:23

Hard worker · Accepted Answer · 2013-03-11 11:58:31Z

2

Inside the success function do:

console.log(responseData);

Then copy and paste responseData into the JSON validator at jsonlint.com

I am guessing it's a JSON syntax error, which the validator will point out to you.

answered Mar 11, 2013 at 11:58

Hard worker

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2 Comments

imsome1 Over a year ago

I checked it and it is a valid json and now it gives this error "Uncaught Error: Argument given to addRows must be either a number or an array"

imsome1 Over a year ago

its working now I changed var arrayForGviz = eval("[" + responseData + "]"); thanks for your idea

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