使用FormData提交表单及上传图片的方法

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);

<!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测试数据/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>  
    <p id="result"></p>  
 </body>  
</html>

<?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);  
?>

以上就是使用FormData提交表单及上传图片的方法的详细内容，更多请关注Gxl网其它相关文章！

查看更多关于使用FormData提交表单及上传图片的方法的详细内容...

声明：本文来自网络，不代表【好得很程序员自学网】立场，转载请注明出处：http://haodehen.cn/did72138

更新时间：2022-10-19 阅读：40次