转自：http://blog.csdn.net/fdipzone/article/details/38910553
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
[javascript] view plain copy 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
[html] view plain copy <form name="form1" id="form1"> <input type="text" name="name" value="fdipzone"> <input type="text" name="gender" value="male"> </form> [javascript] view plain copy var form = document.getElementById('form1'); var formdata = new FormData(form); 使用FormData提交表单及上传文件：
[html] view plain copy <!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] view plain copy <?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); ?>
