谷歌圖表工具示例不工作

Question

我正在使用在http://code.google.com/apis/chart/docs/post_requests.html上給出的PHP發佈請求示例來生成圖表。

代碼： 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'> 

現在，當我訪問another_page.html，圖像不會當我在視圖中單擊加載它顯示的圖像

圖像「http://localhost/demo/chartserver-image.php」無法顯示，因爲它包含錯誤。

我無法理解的問題是什麼？

請幫我在這

感謝

Answer 1

更換 '內容'=> http_build_query（$圖））））; 'content'=> http_build_query（$ chart，''，'&'））））;解決了這個問題。

我已經添加精氨酸分離器「&」到http_build_query（），它避免錯誤如果 arg_separator.output參數在php.ini被修改。

當我檢查phpinfo的arg_separator.output是& amp。這是一個問題，因此在http_build_query（）中添加'&'可以解決問題。

Answer 2

這個工作對我來說：

<?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 3

我有相同的問題，我劃分

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

成兩個語句：

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

$context = stream_context_create($x); 

而且似乎解決它（請不要問我爲什麼）

Answer 4

這將工作絕對好......上面的例子中的一些錯誤，我認爲......他們在下面解決。

<?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); 
?>