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PHP结合HTML5使用FormData对象提交表单及上传图片

时间：2020-11-27 来源：乌哈旅游

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="'%20+%20ret['photo']%20%20+%20'" 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>
 <p id="result"></p>
 </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);
?>