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
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。
其他回答
如果您上传的是二进制文件，如csv，请使用以下格式上传文件
curl -X POST \
'http://localhost:8080/workers' \
-H 'authorization: eyJhbGciOiJIUzI1NiIsInR5cCI6ImFjY2VzcyIsInR5cGUiOiJhY2Nlc3MifQ.eyJ1c2VySWQiOjEsImFjY291bnRJZCI6MSwiaWF0IjoxNTExMzMwMzg5LCJleHAiOjE1MTM5MjIzODksImF1ZCI6Imh0dHBzOi8veW91cmRvbWFpbi5jb20iLCJpc3MiOiJmZWF0aGVycyIsInN1YiI6ImFub255bW91cyJ9.HWk7qJ0uK6SEi8qSeeB6-TGslDlZOTpG51U6kVi8nYc' \
-H 'content-type: application/x-www-form-urlencoded' \
--data-binary '@/home/limitless/Downloads/iRoute Masters - Workers.csv'
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"
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/
保存所有已发送文件到文件夹: 主机上的PHP文件。u.php:
<?php
$uploaddir = 'C:/VALID_DIR/';
echo '<pre>';
foreach ($_FILES as $key => $file) {
if(!isset($file) || !isset($file['name'])) continue;
$uploadfile = $uploaddir . basename($file['name']);
if (move_uploaded_file($file['tmp_name'], $uploadfile)) {
echo "$key file > $uploadfile .\n";
} else {
echo " Error $key file.\n";
}
}
print_r($_FILES);
print "</pre>";?>
客户端使用:
curl -v -F filename=ff.xml -F upload=@ff.xml https://myhost.com/u.php
这对我来说很管用。
我的虚拟机崩溃了，它只有互联网连接。 我用这种方法恢复了一些文件。
导致我到这里的问题原来是一个基本的用户错误——我没有在文件的路径中包含@符号，所以curl发布的是文件的路径/名称，而不是内容。Content-Length值因此是8，而不是我期望看到的给定测试文件长度的479。
当curl读取并发布文件时，将自动计算Content-Length头。
curl -i -H "Content-Type: application/xml"——data "@test.xml" -v -X POST https://<url>/<uri/
.．. < Content-Length: 479 .．.
把这贴在这里，以帮助其他的新手在未来。
