CODEWITHSUNDEEP
phpjson
Benjamin Felix
Benjamin Felix

Reputation: 165

How to pass an array from PHP to javascript using JSON to make a Google Chart?

I want to make a Google Chart and I am trying to send a JSON encoded array from a PHP script to a JavaScript. The two files are as follows:

<html>

<head><title>

</title></head>

<body>
<?php
$data[0]=array('State','Values');
$data[1]=array('Correct',6);
$data[2]=array('Incorrect',3);

echo json_encode($data);
?>
</body>
</html>

This is saved as test.php

The javascript file is:

 <html>
   <head>
     <script type="text/javascript" src="jquery-1.9.1.min.js"></script>
     <script type="text/javascript" src="https://www.google.com/jsapi"></script>
     <script type="text/javascript">
       google.load("visualization", "1", {packages:["corechart"]});
       google.setOnLoadCallback(drawChart);

       function drawChart() {


    var jsonData = $.ajax({
            url: "test.php",
            dataType: "json",
            async: false
        }).responseText;

    var data=google.visualization.arrayToDataTable(jsonData);

    var options = {
      title: 'My Chart'
    };

    var chart = new oogle.visualization.PieChart(document.getElementById('chart_div'));
    chart.draw(data, options);
  }
     </script>
   </head>

   <body>
     <div id="chart_div" style="width: 900px; height: 500px;"></div>
   </body>

 </html>

This is saved as test2.php.

When I run test2.php, nothing happens and the browser shows a blank screen. Where am I going wrong here? Please help.

Thanks.

Upvotes: 2

Views: 857

Answers (2)

Cups
Cups

Reputation: 6896

test.php does not need all those html tags, just return the json. When mixing js and PHP it is sometimes a good idea to just use hardcoded values first, then have PHP create those hardcoded values when you are sure your js is joined up and working.

Upvotes: 0

Alfo
Alfo

Reputation: 4819

You don't need any HTML tags in your test.php. Instead, just use:

<?php
$data[0]=array('State','Values');
$data[1]=array('Correct',6);
$data[2]=array('Incorrect',3);

header("Content-Type: application/json");
echo json_encode($data);
?>

HTML is not a part of JSON.

Upvotes: 3

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