php 使用 file_get_contents 上传文件
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/4003989/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
Upload a file using file_get_contents
提问by Shabbyrobe
I realise I can do this with CURL very easily, but I was wondering if it was possible to use file_get_contents()with the http stream context to upload a file to a remote web server, and if so, how?
我意识到我可以很容易地用 CURL 做到这一点，但我想知道是否可以使用file_get_contents()http 流上下文将文件上传到远程 Web 服务器，如果可以，如何？
回答by netcoder
First of all, the first rule of multipartContent-Type is to define a boundarythat will be used as a delimiter between each part (because as the name says, it can have multiple parts). The boundary can be any string that is not contained in the content body. I will usually use a timestamp:
首先，multipartContent-Type的第一条规则是定义一个边界，它将用作每个部分之间的分隔符（因为顾名思义，它可以有多个部分）。边界可以是内容正文中未包含的任何字符串。我通常会使用时间戳：
define('MULTIPART_BOUNDARY', '--------------------------'.microtime(true));
Once your boundary is defined, you must send it with the Content-Typeheader to tell the webserver what delimiter to expect:
定义边界后，您必须将其与Content-Type标头一起发送，以告诉网络服务器期望的分隔符：
$header = 'Content-Type: multipart/form-data; boundary='.MULTIPART_BOUNDARY;
Once that is done, you must build a proper content body that matches the HTTP specification and the header you sent. As you know, when POSTing a file from a form, you will usually have a form field name. We'll define it:
完成后，您必须构建与 HTTP 规范和您发送的标头相匹配的适当内容正文。如您所知，从表单发布文件时，您通常会有一个表单字段名称。我们将定义它：
// equivalent to <input type="file" name="uploaded_file"/>
define('FORM_FIELD', 'uploaded_file');
Then we build the content body:
然后我们构建内容主体：
$filename = "/path/to/uploaded/file.zip";
$file_contents = file_get_contents($filename);
$content = "--".MULTIPART_BOUNDARY."\r\n".
"Content-Disposition: form-data; name=\"".FORM_FIELD."\"; filename=\"".basename($filename)."\"\r\n".
"Content-Type: application/zip\r\n\r\n".
$file_contents."\r\n";
// add some POST fields to the request too: $_POST['foo'] = 'bar'
$content .= "--".MULTIPART_BOUNDARY."\r\n".
"Content-Disposition: form-data; name=\"foo\"\r\n\r\n".
"bar\r\n";
// signal end of request (note the trailing "--")
$content .= "--".MULTIPART_BOUNDARY."--\r\n";
As you can see, we're sending the Content-Dispositionheader with the form-datadisposition, along with the nameparameter (the form field name) and the filenameparameter (the original filename). It is also important to send the Content-Typeheader with the proper MIME type, if you want to correctly populate the $_FILES[]['type']thingy.
如您所见，我们正在发送Content-Disposition带有form-data处置的标头，以及name参数（表单字段名称）和filename参数（原始文件名）。Content-Type如果您想正确填充$_FILES[]['type']事物，那么使用正确的 MIME 类型发送标头也很重要。
If you had multiple files to upload, you just repeat the process with the $contentbit, with of course, a different FORM_FIELDfor each file.
如果你有多个文件要上传，你只需用$content位重复这个过程，当然，FORM_FIELD每个文件都不同。
Now, build the context:
现在，构建上下文：
$context = stream_context_create(array(
'http' => array(
'method' => 'POST',
'header' => $header,
'content' => $content,
)
));
And execute:
并执行：
file_get_contents('http://url/to/upload/handler', false, $context);
NOTE:There is no need to encode your binary file before sending it. HTTP can handle binary just fine.
注意：在发送二进制文件之前无需对其进行编码。HTTP 可以很好地处理二进制文件。
