如何在php中读取FormData对象 [英] how to read FormData object in php
问题描述
大家好我发布了一个示例代码，我在其中使用Ajax JQuery上传文件。一切正常，但我想在我的PHP代码中读取文件内容。那么阅读它的语法是什么？
Hello everyone I am posting a sample code in which I am uploading a file using Ajax JQuery. All thing works fine but I want to read the file content in my php code. So what is the syntax to read it?
<?php
?>
<!--================================html==================================================-->
<html>
<head>
<title>AJAX UPLOAD</title>
<script type="text/javascript" src="jquery-2.0.2.js"></script>
<script type="text/javascript">
$(document).ready(function(){
$('#Button_').click(function(){
alert("hi");
var name= document.getElementById('File_');
var alpha=name.files[0];
console.log(alpha.name);
var data= new FormData();
data.append('file',alpha);
$.ajax({
url:'process.php',
data:data,
processData:false,
contentType:false,
type:'POST',
success:function(msg){
alert(msg);
}
});
});
});
</script>
</head>
<body>
<input type="file" name="File" id="File_"/>
<input type="button" name="Button" id="Button_" value="UPLOAD">
</body>
</html>
现在我不知道如何读取通过Ajax发送的文件数据。所以请让我知道代码
Now I do not know how to read the file data sent via Ajax. So please let me know the code
推荐答案
传递给 FormData.append（）<的第一个参数/ code>是传递给服务器的表单元素的名称。所以 data.append（'file'，alpha）在你的情况下相当于< input type =filename =file> ; - 您将'file'传递给append（），因此'file'是您的输入名称。
The first argument you pass to FormData.append() is the name of the form element that is passed to the server. So data.append('file', alpha) in your case is equivalent to <input type="file" name="file"> - you passed 'file' to append() so 'file' is your input name.
您应该在<$中找到您的资料c $ c> $ _ POST ['file'] 和 $ _ FILES ['file'] 。下次在提交后尝试 var_dump（$ _ POST）; 查看数组。 ：）
You should find your stuff in $_POST['file'] and $_FILES['file']. Next time try var_dump($_POST); after submit to see the array. :)
在这里了解php中的文件上传： http://php.net/manual/en/features.file-upload.php
Learn about file uploads in php here: http://php.net/manual/en/features.file-upload.php
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