XHR.send（文件+参数）？ [英] XHR.send(file + params)?

问题描述

我试图找出如何在同一XMLHtt prequest发送文件和paramaters。这可能吗？

I am trying to figure out how to send a file and paramaters within the same XMLHttpRequest. Is this possible?

很显然，我可以做xhr.send（文件+参数）或XHR。（文件，则params）。而且我不认为我可以设置两个不同的请求头要做到这一点......

Obviously I can do xhr.send(file+params) or xhr.(file,params). And I don't think I can set two different request headers to do this...

xhr.setRequestHead('X_FILENAME', file.name)
xhr.send(file);
xhr.setRequestHeader("Content-type", "application/x-www-form-urlencoded");
xhr.send(params);

有没有一些方法来发送PARAMS而不必使用GET或二次XHR请求？

Is there some way to send the params without having to use GET, or a secondary xhr request?

推荐答案

如果你依赖于浏览器，支持 FORMDATA ，您可以使用下面的code（JavaScript的）：

If you rely on browser which supports FormData, you can use the code below (JavaScript):

var formData = new FormData();
formData.append('param1', 'myParam');
formData.append('param2', 12345); 
formData.append('uploadDir', 'public-data');  
formData.append('myfile', file);

xhr.send(formData);

然后，在你的服务器端，您可以使用此code（PHP）访问您的变量：

Then, on your server side you can have access to your variables by using this code (PHP):

<?
  $param1 = $_POST['param1']; //myParam
  $param2 = $_POST['param2']; //12345
  $uploaddir = $_POST['uploadDir']; //public-data
  $fileName = $_FILES['myfile']['name'];
  $fileZise = $_FILES['myfile']['size'];
  $uploaddir = getcwd().DIRECTORY_SEPARATOR.$uploaddir.DIRECTORY_SEPARATOR;
  $uploadfile = $uploaddir.basename($fileName);       
  move_uploaded_file($_FILES['file']['tmp_name'], $uploadfile);
  echo $fileName.' ['.$fileZise.'] was uploaded successfully!';
?>

要获得 $ _ FILES ['myfile的'] ，使用的var_dump（$ _ FILES [MYFILE]）

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