If that's what it is printing, then PHP isn't working on your server.
On Tuesday, December 17, 2013 11:09:58 AM UTC-5, Thomas Crowell wrote:
>
> Okay, it's printing simply the PHP code:
> <?php $con = mysql_connect('localhost', 'root', 'password') or die('Error
> connecting to server'); mysql_select_db('myTest', $con); $query =
> mysql_query('SELECT price, date FROM myTest'); $table = array();
> $table['cols'] = array( array('label' => 'price', 'type' => 'number'),
> array('label' => 'date', 'type' => 'date') ); $rows = array(); while($r =
> mysql_fetch_assoc($query)) { $temp = array(); $temp[] = array('v' =>
> $r['price']); $temp[] = array('v' => $r['date']); $rows[] = array('c' =>
> $temp); } $table['rows'] = $rows; // encode the table as JSON $jsonTable =
> json_encode($table,JSON_NUMERIC_CHECK); header('Cache-Control: no-cache,
> must-revalidate'); header('Expires: Mon, 26 Jul 1997 05:00:00 GMT');
> header('Content-type: application/json');
> header('Access-Control-Allow-Origin: *'); // return the data echo
> $jsonTable; ?>
>
> Obviously that's not quite right. Any idea why it's doing that?
>
> On Tuesday, December 17, 2013 11:07:39 AM UTC-4, asgallant wrote:
>>
>> Add in the line console.dir(jsonData); between the AJAX call and the
>> DataTable constructor, then open the page in Chrome or Firefox and open the
>> developer's console. You should see the JSON string printed there.
>>
>> On Tuesday, December 17, 2013 9:26:56 AM UTC-5, Thomas Crowell wrote:
>>>
>>> Hi,
>>>
>>> I'm fairly new to javascript, PHP, and SQL, and have been having trouble
>>> getting my code to work. After searching for similar problems here and on
>>> stackexchange, I haven't been able to fix it and thought I would post here
>>> in hopes that someone could help.
>>>
>>> I'm trying to get some data from a MySQL database "myTest", convert it
>>> into JSON format, and return it to draw an annotated timeline. When I test
>>> my files in chrome, the console gives me the error "Uncaught
>>> SyntaxError: Unexpected token <" at the line var data = new
>>> google.visualization.DataTable(JSON.parse(jsonData));. In trying to fix
>>> the issue, I believe the problem is likely that the PHP code is either
>>> incorrect, or the return is not in JSON format, but I am unsure how to see
>>> the output of the PHP to check what it looks like.
>>>
>>> Any help would be greatly appreciated, and I can try to provide any
>>> other details that may be important in fixing the problem. My PHP code, and
>>> a snippet of my javascript code is below.
>>>
>>> Thanks
>>>
>>> Querydata.php:
>>> <?php
>>> $con = mysql_connect('localhost', 'root', 'password') or die('Error
>>> connecting to server');
>>>
>>> mysql_select_db('myTest', $con);
>>>
>>> $query = mysql_query('SELECT price, date FROM myTest');
>>>
>>> $table = array();
>>> $table['cols'] = array(
>>> array('label' => 'price', 'type' => 'number'),
>>> array('label' => 'date', 'type' => 'date')
>>> );
>>>
>>> $rows = array();
>>> while($r = mysql_fetch_assoc($query)) {
>>> $temp = array();
>>> $temp[] = array('v' => $r['price']);
>>> $temp[] = array('v' => $r['date']);
>>>
>>> $rows[] = array('c' => $temp);
>>> }
>>>
>>> $table['rows'] = $rows;
>>>
>>> // encode the table as JSON
>>> $jsonTable = json_encode($table,JSON_NUMERIC_CHECK);
>>>
>>> header('Cache-Control: no-cache, must-revalidate');
>>> header('Expires: Mon, 26 Jul 1997 05:00:00 GMT');
>>> header('Content-type: application/json');
>>> header('Access-Control-Allow-Origin: *');
>>>
>>>
>>> // return the data
>>> echo $jsonTable;
>>> ?>
>>>
>>>
>>> Test.js:
>>>
>>> function drawChart() {
>>> var jsonData = $.ajax({
>>> url: "http://localhost:8000/Desktop/Querydata.php";,
>>> dataType:"json",
>>> crossDomain : true,
>>> async: false
>>> }).responseText;
>>> var data = new google.visualization.DataTable(JSON.parse(jsonData));
>>> var chart = new
>>> google.visualization.AnnotatedTimeLine(document.getElementById('chart_div'));
>>> chart.draw(data, {displayAnnotations: true});
>>> };
>>>
>>
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