使用FormData对象提交表单及上传图片

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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" οnclick="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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