PHP结合HTML5使用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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