通过 cURL 从 PHP 中的 POST 表单发送文件

问题描述 投票：0回答：8

我正在编写一个 API，我想处理从表单上传的文件 

POST
。表单的标记并不太复杂：

<form action="" method="post" enctype="multipart/form-data">
  <fieldset>
    <input type="file" name="image" id="image" />
    <input type="submit" name="upload" value="Upload" />
  </fieldset>
</form>

但是，我很难理解如何处理此服务器端并与 cURL 请求一起发送。

我熟悉使用带有数据数组的 cURL 发送 

POST
请求，并且我读过的有关上传文件的资源告诉我在文件名中添加 
@
符号作为前缀。但这些相同的资源有一个硬编码的文件名，例如

$post = array(
    'image' => '@/path/to/myfile.jpg',
    ...
);

这是哪个文件路径？我在哪里可以找到它？是否会类似于 

$_FILES['image']['tmp_name']
，在这种情况下我的 
$post
数组应该如下所示：

$post = array(
    'image' => '@' . $_FILES['image']['tmp_name'],
    ...
);

或者我的处理方式是错误的吗？任何建议将不胜感激。

编辑：如果有人能给我一个代码片段，说明我将使用以下代码片段去哪里，那么我将不胜感激。我主要关注的是我将作为 cURL 参数发送的内容，以及如何在接收脚本中使用这些参数的示例（为了论证，我们将其称为 

curl_receiver.php
）。

我有这个网络表格：

<form action="script.php" method="post" enctype="multipart/form-data">
  <fieldset>
    <input type="file" name="image />
    <input type="submit" name="upload" value="Upload" />
  </fieldset>
</form>

这将是

script.php
：

if (isset($_POST['upload'])) {
    // cURL call would go here
    // my tmp. file would be $_FILES['image']['tmp_name'], and
    // the filename would be $_FILES['image']['name']
}
php curl image-uploading
8个回答
71
投票

这是一些将文件发送到 ftp 的生产代码（可能对您来说是一个很好的解决方案）：

// This is the entire file that was uploaded to a temp location.
$localFile = $_FILES[$fileKey]['tmp_name']; 

$fp = fopen($localFile, 'r');

// Connecting to website.
$ch = curl_init();

curl_setopt($ch, CURLOPT_USERPWD, "[email protected]:password");
curl_setopt($ch, CURLOPT_URL, 'ftp://@ftp.website.net/audio/' . $strFileName);
curl_setopt($ch, CURLOPT_UPLOAD, 1);
curl_setopt($ch, CURLOPT_TIMEOUT, 86400); // 1 Day Timeout
curl_setopt($ch, CURLOPT_INFILE, $fp);
curl_setopt($ch, CURLOPT_NOPROGRESS, false);
curl_setopt($ch, CURLOPT_PROGRESSFUNCTION, 'CURL_callback');
curl_setopt($ch, CURLOPT_BUFFERSIZE, 128);
curl_setopt($ch, CURLOPT_INFILESIZE, filesize($localFile));
curl_exec ($ch);

if (curl_errno($ch)) {

    $msg = curl_error($ch);
}
else {

    $msg = 'File uploaded successfully.';
}

curl_close ($ch);

$return = array('msg' => $msg);

echo json_encode($return);
67
投票

对于找到这篇文章并使用 PHP5.5+ 的人来说，这可能会有所帮助。

我发现 netcoder 建议的方法不起作用。即这不起作用：

$tmpfile = $_FILES['image']['tmp_name'];
$filename = basename($_FILES['image']['name']);
$data = array(
    'uploaded_file' => '@'.$tmpfile.';filename='.$filename,
);
$ch = curl_init();   
curl_setopt($ch, CURLOPT_POSTFIELDS, $data);

我会在 

$_POST
var 中收到 
'uploaded_file'
字段 - 而在 
$_FILES
var 中什么也没有。

事实证明，对于 php5.5+ 有一个新的 

curl_file_create()
函数需要使用。所以上面的内容就变成了：

$data = array(
    'uploaded_file' => curl_file_create($tmpfile, $_FILES['image']['type'], $filename)
);

由于 

@
格式现已弃用。

28
投票

这应该有效：

$tmpfile = $_FILES['image']['tmp_name'];
$filename = basename($_FILES['image']['name']);

$data = array(
    'uploaded_file' => '@'.$tmpfile.';filename='.$filename,
);

$ch = curl_init();   
curl_setopt($ch, CURLOPT_POSTFIELDS, $data);
// set your other cURL options here (url, etc.)

curl_exec($ch);

在接收脚本中，您将拥有：

print_r($_FILES);
/* which would output something like
     Array (
        [uploaded_file] => Array (
            [tmp_name] => /tmp/f87453hf
            [name] => myimage.jpg
            [error] => 0
            [size] => 12345
            [type] => image/jpeg
        )
     )
*/

然后，如果你想正确处理文件上传，你可以这样做：

if (move_uploaded_file($_FILES['uploaded_file'], '/path/to/destination/file.zip')) {
   // do stuff
}
16
投票

对于我来说，@符号不起作用，所以我做了一些研究，发现这种方法对我有用，我希望这对你有帮助。

    $target_url = "http://server:port/xxxxx.php";           
    $fname = 'file.txt';   
    $cfile = new CURLFile(realpath($fname));

        $post = array (
                  'file' => $cfile
                  );    

    $ch = curl_init();
    curl_setopt($ch, CURLOPT_URL, $target_url);
    curl_setopt($ch, CURLOPT_POST, 1);
    curl_setopt($ch, CURLOPT_HEADER, 0);
    curl_setopt($ch, CURLOPT_RETURNTRANSFER, 1); 
    curl_setopt($ch, CURLOPT_USERAGENT, "Mozilla/4.0 (compatible;)");   
    curl_setopt($ch, CURLOPT_HTTPHEADER,array('Content-Type: multipart/form-data'));
    curl_setopt($ch, CURLOPT_FRESH_CONNECT, 1);   
    curl_setopt($ch, CURLOPT_FORBID_REUSE, 1);  
    curl_setopt($ch, CURLOPT_TIMEOUT, 100);
    curl_setopt($ch, CURLOPT_POSTFIELDS, $post);

    $result = curl_exec ($ch);

    if ($result === FALSE) {
        echo "Error sending" . $fname .  " " . curl_error($ch);
        curl_close ($ch);
    }else{
        curl_close ($ch);
        echo  "Result: " . $result;
    }
5
投票

当我通过 Mercadolibre 的消息系统发送附件时，它对我有用。

答案https://stackoverflow.com/a/35227055/7656744

$target_url = "http://server:port/xxxxx.php";           
$fname = 'file.txt';   
$cfile = new CURLFile(realpath($fname));

    $post = array (
              'file' => $cfile
              );    

$ch = curl_init();
curl_setopt($ch, CURLOPT_URL, $target_url);
curl_setopt($ch, CURLOPT_POST, 1);
curl_setopt($ch, CURLOPT_HEADER, 0);
curl_setopt($ch, CURLOPT_RETURNTRANSFER, 1); 
curl_setopt($ch, CURLOPT_USERAGENT, "Mozilla/4.0 (compatible;)");   
curl_setopt($ch, CURLOPT_HTTPHEADER,array('Content-Type: multipart/form-data'));
curl_setopt($ch, CURLOPT_FRESH_CONNECT, 1);   
curl_setopt($ch, CURLOPT_FORBID_REUSE, 1);  
curl_setopt($ch, CURLOPT_TIMEOUT, 100);
curl_setopt($ch, CURLOPT_POSTFIELDS, $post);

$result = curl_exec ($ch);

if ($result === FALSE) {
    echo "Error sending" . $fname .  " " . curl_error($ch);
    curl_close ($ch);
}else{
    curl_close ($ch);
    echo  "Result: " . $result;
}
1
投票

过程方法中的cURL文件对象：

$file = curl_file_create('full path/filename','extension','filename');

Oop 方法中的 cURL 文件对象：

$file = new CURLFile('full path/filename','extension','filename');

$data = array('file' => $file);

$curl = curl_init();  
//curl_setopt ... 
curl_setopt($curl, CURLOPT_POSTFIELDS, $data);
$response = curl_exec($curl);
curl_close($curl);
0
投票

我们可以通过curl请求上传图像文件，将其转换为base64字符串。因此在post中我们将发送文件字符串，然后将其隐藏在图像中。

function covertImageInBase64()
{
    var imageFile = document.getElementById("imageFile").files;
    if (imageFile.length > 0)
    {
        var imageFileUpload = imageFile[0];
        var readFile = new FileReader();
        readFile.onload = function(fileLoadedEvent) 
        {
            var base64image = document.getElementById("image");
            base64image.value = fileLoadedEvent.target.result;
        };
        readFile.readAsDataURL(imageFileUpload);
    }
}

然后在curl请求中发送它

if(isset($_POST['image'])){
    $curlUrl='localhost/curlfile.php';  
    $ch = curl_init();
    curl_setopt($ch,CURLOPT_URL, $curlUrl);
    curl_setopt($ch,CURLOPT_POST, 1);
    curl_setopt($ch,CURLOPT_POSTFIELDS, 'image='.$_POST['image']);
    $result = curl_exec($ch);
    curl_close($ch);
}

请参阅此处http://technoblogs.co.in/blog/How-to-upload-an-image-by-using-php-curl-request/118

-4
投票

这是我的解决方案，我读了很多帖子，它们真的很有帮助，最后我用 cUrl 和 Php 为小文件构建了代码，我认为它非常有用。

public function postFile()
{


        $file_url = "test.txt";  //here is the file route, in this case is on same directory but you can set URL too like "http://examplewebsite.com/test.txt"
        $eol = "\r\n"; //default line-break for mime type
        $BOUNDARY = md5(time()); //random boundaryid, is a separator for each param on my post curl function
        $BODY=""; //init my curl body
        $BODY.= '--'.$BOUNDARY. $eol; //start param header
        $BODY .= 'Content-Disposition: form-data; name="sometext"' . $eol . $eol; // last Content with 2 $eol, in this case is only 1 content.
        $BODY .= "Some Data" . $eol;//param data in this case is a simple post data and 1 $eol for the end of the data
        $BODY.= '--'.$BOUNDARY. $eol; // start 2nd param,
        $BODY.= 'Content-Disposition: form-data; name="somefile"; filename="test.txt"'. $eol ; //first Content data for post file, remember you only put 1 when you are going to add more Contents, and 2 on the last, to close the Content Instance
        $BODY.= 'Content-Type: application/octet-stream' . $eol; //Same before row
        $BODY.= 'Content-Transfer-Encoding: base64' . $eol . $eol; // we put the last Content and 2 $eol,
        $BODY.= chunk_split(base64_encode(file_get_contents($file_url))) . $eol; // we write the Base64 File Content and the $eol to finish the data,
        $BODY.= '--'.$BOUNDARY .'--' . $eol. $eol; // we close the param and the post width "--" and 2 $eol at the end of our boundary header.



        $ch = curl_init(); //init curl
        curl_setopt($ch, CURLOPT_HTTPHEADER, array(
                         'X_PARAM_TOKEN : 71e2cb8b-42b7-4bf0-b2e8-53fbd2f578f9' //custom header for my api validation you can get it from $_SERVER["HTTP_X_PARAM_TOKEN"] variable
                         ,"Content-Type: multipart/form-data; boundary=".$BOUNDARY) //setting our mime type for make it work on $_FILE variable
                    );
        curl_setopt($ch, CURLOPT_USERAGENT, 'Mozilla/1.0 (Windows NT 6.1; WOW64; rv:28.0) Gecko/20100101 Firefox/28.0'); //setting our user agent
        curl_setopt($ch, CURLOPT_URL, "api.endpoint.post"); //setting our api post url
        curl_setopt($ch, CURLOPT_COOKIEJAR, $BOUNDARY.'.txt'); //saving cookies just in case we want
        curl_setopt ($ch, CURLOPT_RETURNTRANSFER, 1); // call return content
        curl_setopt ($ch, CURLOPT_FOLLOWLOCATION, 1); navigate the endpoint
        curl_setopt($ch, CURLOPT_POST, true); //set as post
        curl_setopt($ch, CURLOPT_POSTFIELDS, $BODY); // set our $BODY 


        $response = curl_exec($ch); // start curl navigation

     print_r($response); //print response

}

有了这个，我们应该在“api.endpoint.post”上发布以下变量 您可以使用此脚本轻松进行测试，并且您应该在最后一行的函数 postFile() 上收到此调试信息

print_r($响应); //打印响应

public function getPostFile()
{

    echo "\n\n_SERVER\n";
    echo "<pre>";
    print_r($_SERVER['HTTP_X_PARAM_TOKEN']);
    echo "/<pre>";
    echo "_POST\n";
    echo "<pre>";
    print_r($_POST['sometext']);
    echo "/<pre>";
    echo "_FILES\n";
    echo "<pre>";
    print_r($_FILEST['somefile']);
    echo "/<pre>";
}

在这里，它应该工作得很好，可能是更好的解决方案，但这确实有效，并且对于理解边界和多部分/来自数据 mime 如何在 php 和curl 库上工作非常有帮助，

我最诚挚的问候，

我对我的英语表示歉意，但这不是我的母语。

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