views:

199

answers:

3

Hi ,

I am using the php post request example given on http://code.google.com/apis/chart/docs/post_requests.html for generating chart.

The code: chartserver-image.php

<?php
 // Create some random text-encoded data for a line chart.
 header('content-type: image/png');
 $url = 'http://chart.apis.google.com/chart';
 $chd = 't:';
 for ($i = 0; $i < 150; ++$i) {
 $data = rand(0, 100000);
 $chd .= $data . ',';
 }
 $chd = substr($chd, 0, -1);

// Add data, chart type, chart size, and scale to params.
$chart = array(
'cht' => 'lc',
'chs' => '600x200',
'chds' => '0,100000',
'chd' => $chd);

// Send the request, and print out the returned bytes.
$context = stream_context_create(
array('http' => array(
  'method' => 'POST',
  'content' => http_build_query($chart))));
fpassthru(fopen($url, 'r', false, $context));
?>

another_page.html

<img width='600' height='200' src='chartserver-image.php'>

Right now when i access another_page.html, the image doesn't load when i click on view image it shows

The image “http://localhost/demo/chartserver-image.php” cannot be displayed, because it contains errors.

What is the issue i am unable to understand?

Please help me on this

Thanks

A: 

This works for me:

<?php

 // Create some random text-encoded data for a line chart.
 //header('content-type: image/png');
 $url = 'http://chart.apis.google.com/chart';
 $chd = 't:';
 for ($i = 0; $i < 150; ++$i) {
 $data = rand(0, 100000);
 $chd .= $data . ',';
 }
 $chd = substr($chd, 0, -1);

// Add data, chart type, chart size, and scale to params.
$chart = array(
'cht' => 'lc',
'chs' => '600x200',
'chds' => '0,100000',
'chd' => $chd);

$query = http_build_query($chart);

$fullurl = $url."?".$query;

$context = stream_context_create(

    array(
            'http' => array(
                    'method' => 'GET',
                    'header' => "Content-type: application/x-www-form-urlencoded\r\n" .
                                "Content-length: 0",
                    'proxy' => 'tcp://X.X.X.X:XXXX'
            )
    )
);

$ret = fopen(fullurl, 'r', false, $context);
fpassthru($ret);
the_void
Thanks for replying can you please clarify the need for 'proxy' => 'tcp://X.X.X.X:XXXX'. What is the value should i pass to this variable? Right now i am getting this errorWarning: fopen() [function.fopen]: php_network_getaddresses: getaddrinfo failed: No such host is known. in E:\xampp\htdocs\demo\chartserver-image.php on line 33
Pankaj Khurana
Pankaj Khurana
I tested it behind a proxy so that's why I had to have the `proxy` set. I'm no expert in `PHP`, but it worked for me.
the_void
+2  A: 

Hi,

Replacing 'content' => http_build_query($chart)))); with 'content' => http_build_query($chart,'', '&')))); resolves the issue.

I have added arg separator '&' to http_build_query() which avoid bug if the arg_separator.output parameter is modified in php.ini.

When i checked phpinfo the arg_separator.output was &amp. That was causing issue so adding '&' to http_build_query() resolves the problem.

Pankaj Khurana
+1 that seems to have solved my problem at http://stackoverflow.com/questions/3682622/anyone-having-problems-with-google-charts-in-xampp
Mawg
oops, sorry, no it didn't -something else did, sorry. But it's a good point to make anyway
Mawg
A: 

I had the identical problem, and I split

$context = stream_context_create(
array('http' => array(
  'method' => 'POST',
  'content' => http_build_query($chart))));

into two statements:

$x = array('http' => array(
  'method' => 'POST',
  'content' => http_build_query($chart)));

$context = stream_context_create($x);

and that seemed to solve it (please don't ask me why)

Mawg