Question

我将图像资源存储在变量中，我需要使用其http API和PHP将其发送到服务器。我必须发送内容类型为multipart / form-data的请求。因此，我需要发出类似的请求，因为当文件输入的表单和enctype = multipart / form-data属性被发送时。我试过这个：

<?php
$url = 'here_is_url_for_web_API';
$input = fopen('delfin.jpg','r');       
$header = array('Content-Type: multipart/form-data');
$resource = curl_init();
curl_setopt($resource, CURLOPT_URL, $url);
curl_setopt($resource, CURLOPT_USERPWD, "user:password");
curl_setopt($resource, CURLOPT_HTTPAUTH, CURLAUTH_ANYSAFE);
curl_setopt($resource, CURLOPT_HTTPHEADER, $header);
curl_setopt($resource, CURLOPT_RETURNTRANSFER, 1);
curl_setopt($resource, CURLOPT_BINARYTRANSFER, true );
curl_setopt($resource, CURLOPT_INFILESIZE, 61631);
curl_setopt($resource, CURLOPT_INFILE, $input);
$result = curl_exec($resource);
curl_close($resource);
var_dump($result);
?>

我不知道响应应该如何，但这会返回： http状态405 和错误报告是：所请求的资源（）不允许指定的HTTP方法。

Answer 1

如果您使用CURL，则必须：

1，将标题'Content-Type'设置为'multipart / form-data;'

2，将curl的选项'RETURNTRANSFER'设置为true（使用curl的选项方法）

3，将curl的选项'POST'设置为true（使用curl的选项方法）

4，获取你的文件来源（你在PHP中用fopen得到的）：

$tempFile = tempnam(sys_get_temp_dir(), 'File_');                
file_put_contents($tempFile, $source);
$post = array(
    "uploadedFile" => "@" . $tempFile, //"@".$tempFile.";type=image/jpeg",
);

5，使用CURL的post方法和$ post变量

中的参数

Answer 2

使用curl在POST内容中使用multipart / form-data和boundary。

$filenames = array("/tmp/1.jpg", "/tmp/2.png");;

$files = array();
foreach ($filenames as $f){
   $files[$f] = file_get_contents($f);
}



// more fields for POST request
$fields = array("f1"=>"value1", "another_field2"=>"anothervalue");


$url = "http://example.com/upload";


$curl = curl_init();

$url_data = http_build_query($data);

$boundary = uniqid();
$delimiter = '-------------' . $boundary;

$post_data = build_data_files($boundary, $fields, $files);


curl_setopt_array($curl, array(
  CURLOPT_URL => $url,
  CURLOPT_RETURNTRANSFER => 1,
  CURLOPT_MAXREDIRS => 10,
  CURLOPT_TIMEOUT => 30,
  //CURLOPT_HTTP_VERSION => CURL_HTTP_VERSION_1_1,
  CURLOPT_CUSTOMREQUEST => "POST",
  CURLOPT_POST => 1,
  CURLOPT_POSTFIELDS => $post_data,
  CURLOPT_HTTPHEADER => array(
    //"Authorization: Bearer $TOKEN",
    "Content-Type: multipart/form-data; boundary=" . $delimiter,
    "Content-Length: " . strlen($post_data)

  ),


));


//
$response = curl_exec($curl);

$info = curl_getinfo($curl);
//echo "code: ${info['http_code']}";

//print_r($info['request_header']);

var_dump($response);
$err = curl_error($curl);

echo "error";
var_dump($err);
curl_close($curl);




function build_data_files($boundary, $fields, $files){
    $data = '';
    $eol = "\r\n";

    $delimiter = '-------------' . $boundary;

    foreach ($fields as $name => $content) {
        $data .= "--" . $delimiter . $eol
            . 'Content-Disposition: form-data; name="' . $name . "\"".$eol.$eol
            . $content . $eol;
    }


    foreach ($files as $name => $content) {
        $data .= "--" . $delimiter . $eol
            . 'Content-Disposition: form-data; name="' . $name . '"; filename="' . $name . '"' . $eol
            //. 'Content-Type: image/png'.$eol
            . 'Content-Transfer-Encoding: binary'.$eol
            ;

        $data .= $eol;
        $data .= $content . $eol;
    }
    $data .= "--" . $delimiter . "--".$eol;


    return $data;
}

请参阅此处的完整示例：https://gist.github.com/maxivak/18fcac476a2f4ea02e5f80b303811d5f

在php中使用multipart / form-data请求发送文件

2 个答案: