Google 图表工具示例不起作用

Question

我正在使用 http://code.google 上给出的 php post 请求示例。 com/apis/chart/docs/post_requests.html 用于生成图表。

代码： 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'>

现在，当我访问 another_page.html 时，当我单击查看图像时，图像不会加载，它显示

图像“http://localhost/demo/chartserver-image.php” 无法显示，因为它包含错误。

我无法理解的问题是什么？

请帮我解决这个问题，

谢谢

Answer 1

替换“内容”=> http_build_query($chart))));与 '内容' => http_build_query($chart,'', '&'))));解决问题。

我添加了 arg 分隔符 '&'到 http_build_query() ，如果以下情况可以避免错误
arg_separator.output参数在php.ini中修改。

当我检查 phpinfo 时，arg_separator.output 是 &。这导致了问题，所以添加“&” http_build_query() 解决了该问题。

Answer 2

这绝对可以工作...我认为上面的例子中存在一些错误..它们在下面得到解决。

<?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",

            )
    )
);

$ret = fopen($fullurl, 'r', false, $context);
fpassthru($ret);
?>

Answer 3

这对我有用：

<?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);

Answer 4

我有同样的问题，我分为

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

两个陈述：

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

$context = stream_context_create($x);

这似乎解决了它（请不要问我为什么）