我想使用cURL不仅在HTTP POST中发送数据参数，而且还上传具有特定表单名称的文件。我该怎么做呢?
HTTP Post参数:
Userid = 12345 filecomment =这是一个映像文件
文件上传: 文件位置= /home/user1/Desktop/test.jpg file = image的表单名称(对应PHP端$_FILES['image'])
我认为cURL命令的一部分如下所示:
curl -d "userid=1&filecomment=This is an image file" --data-binary @"/home/user1/Desktop/test.jpg" localhost/uploader.php
我遇到的问题如下:
Notice: Undefined index: image in /var/www/uploader.php
问题是我使用$_FILES['image']在PHP脚本中拾取文件。
如何相应地调整cURL命令?
当前回答
下面是如何正确转义任意文件名的上传文件与bash:
#!/bin/bash
set -eu
f="$1"
f=${f//\\/\\\\}
f=${f//\"/\\\"}
f=${f//;/\\;}
curl --silent --form "uploaded=@\"$f\"" "$2"
其他回答
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写了一些小文件的代码，我认为这真的很有用。
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($response); //print response
}
这样我们就可以得到api。endpoint。张贴“以下vars张贴。您可以很容易地使用这个脚本进行测试，并且您应该在最后一行收到关于函数postFile()的调试。
print_r($response); //print response
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>";
}
它应该工作得很好，它们可能是更好的解决方案，但这确实有助于理解边界和multipart/from-data mime如何在PHP和cURL库上工作。
你需要使用-F选项: -F/——form <name=content>指定HTTP多部分POST数据
试试这个:
curl \
-F "userid=1" \
-F "filecomment=This is an image file" \
-F "image=@/home/user1/Desktop/test.jpg" \
localhost/uploader.php
捕获用户id作为路径变量(推荐):
curl -i -X POST -H "Content-Type: multipart/form-data"
-F "data=@test.mp3" http://mysuperserver/media/1234/upload/
捕获用户id作为表单的一部分:
curl -i -X POST -H "Content-Type: multipart/form-data"
-F "data=@test.mp3;userid=1234" http://mysuperserver/media/upload/
or:
curl -i -X POST -H "Content-Type: multipart/form-data"
-F "data=@test.mp3" -F "userid=1234" http://mysuperserver/media/upload/
我用这个命令curl -F 'filename=@/home/ yourhomedirectory /file.txt' http://yourserver/upload
经过多次尝试，这个命令对我来说是有效的:
curl -v -F filename=image.jpg -F upload=@image.jpg http://localhost:8080/api/upload
