Google Closure Compiler和multipart/form-data无效

Question

我正在向google封闭编译器API服务发出请求:

   $content = file_get_contents('file.js');

   $url = 'http://closure-compiler.appspot.com/compile';
   $post = true;
   $postData = array('output_info' => 'compiled_code', 'output_format' => 'text', 'compilation_level' => 'SIMPLE_OPTIMIZATIONS', 'js_code' => urlencode($content)));

    $ch = curl_init();
    curl_setopt($ch, CURLOPT_URL, $url);

    curl_setopt($ch, CURLOPT_HEADER, 1);
    curl_setopt($ch, CURLOPT_RETURNTRANSFER, 1);

    if ($post) {
        curl_setopt($ch, CURLOPT_POST, $post);
        curl_setopt($ch, CURLOPT_POSTFIELDS, $postData);
    }

    curl_setopt($ch, CURLOPT_HTTPHEADER, array('Content-Type: application/x-www-form-urlencoded; charset=UTF-8'));  

但请求失败,我从谷歌收到此错误消息:

   Error(18): Unknown parameter in Http request: '------------------------------0f1f2f05fb97
   Content-Disposition: form-data; name'.
   Error(13): No output information to produce, yet compilation was requested.

我查看了标题,并发送了此Content-Type标头:

  application/x-www-form-urlencoded; charset=UTF-8; boundary=----------------------------0f1f2f05fb97

不确定添加的边界是否正常？我如何防止这种情况,因为谷歌似乎不喜欢它？

谢谢你,韦斯利

Answer 1

您必须http_build_query()在将POST数据(数组)发送到cURL之前使用.

string http_build_query ( mixed $query_data [, string $numeric_prefix [, string $arg_separator [, int $enc_type = PHP_QUERY_RFC1738 ]]] )

所以你$postData应该看起来像这样:

$postData = http_build_query(
                                array(
                                'output_info' => 'compiled_code', 
                                'output_format' => 'text', 
                                'compilation_level' => 'SIMPLE_OPTIMIZATIONS', 
                                'js_code' => urlencode($content)
                                )
                            );

Answer 2

看起来Google的API不支持多部分/表单数据数据.这对我来说似乎有点蹩脚......

根据curl_setopt()的PHP文档:

将数组传递给CURLOPT_POSTFIELDS会将数据编码为multipart/form-data,而传递URL编码的字符串会将数据编码为application/x-www-form-urlencoded.

因此,如果您将代码的第4行更改为以下内容,它应该可以工作:

$postData = 'output_info=compiled_code&output_format=text&compilation_level=SIMPLE_OPTIMIZATIONS&js_code=' . urlencode($content);

换句话说,您必须自己进行URL编码 - 您显然不能依赖cURL来获取数组并为您编码.