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使用PHP和HTML5完成FormData对象提交表单及上传图片

FormData 对象,可以把form中所有表单元素的name与value组成一个queryString,提交到后台。在使用Ajax提交时,使用FormData对象可以减少拼接queryString的工作量。
使用FormData对象

1.创建一个FormData空对象,然后使用Append方法添加Key/Value

var formdata = new FormData();
formdata.append('name','fdipzone');
formdata.append('gender','male');

2.取得Form对象,作为参数传入到FormData对象

<form name="form1" id="form1">
<input type="text" name="name" value="fdipzone">
<input type="text" name="gender" value="male">
</form>
var form = document.getElementById('form1');
var formdata = new FormData(form);

使用FormData提交表单及上传文件:

<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.0 Transitional//EN">
<html>
<head>
 <meta http-equiv="content-type" content="text/html; charset=utf-8">
 <title> FormData Demo </title>
 <script src="//code.jquery.com/jquery-1.11.0.min.js"></script>
 <script type="text/javascript">
 <!--
 function fsubmit(){
 var data = new FormData($('#form1')[0]);
$.ajax({
url: 'server.php',
type: 'POST',
data: data,
dataType: 'JSON',
cache: false,
processData: false,
contentType: false
 }).done(function(ret){
 if(ret['isSuccess']){
 var result = '';
result += 'name=' + ret['name'] + '<br>';
result += 'gender=' + ret['gender'] + '<br>';
result += '<img src="' + ret['photo'] + '" width="100">';
$('#result').html(result);
 }else{
alert('提交失敗');
 }
 });
 return false;
 }
 -->
 </script>
</head>
<body>
 <form name="form1" id="form1">
 <p>name:<input type="text" name="name" ></p>
 <p>gender:<input type="radio" name="gender" value="1">male <input type="radio" name="gender" value="2">female</p>
 <p>photo:<input type="file" name="photo" id="photo"></p>
 <p><input type="button" name="b1" value="submit" onclick="fsubmit()"></p>
 </form>
 <div id="result"></div>
</body>
</html>

server.php

<?php
$name = isset($_POST['name'])? $_POST['name'] : '';
$gender = isset($_POST['gender'])? $_POST['gender'] : '';
$filename = time().substr($_FILES['photo']['name'], strrpos($_FILES['photo']['name'],'.'));
$response = array();
if(move_uploaded_file($_FILES['photo']['tmp_name'], $filename)){
$response['isSuccess'] = true;
$response['name'] = $name;
$response['gender'] = $gender;
$response['photo'] = $filename;
}else{
$response['isSuccess'] = false;
}
echo json_encode($response);
?>

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