2018-03-21 HTML5 FormData 文件上传 学

点击跳转HTML5 FormData 方法介绍以及实现文件上传 作者：诗渊

把 form 表单作为参数传入 FormData 构造函数

var form = document.getElementById("form1");
var fd = new FormData(form);

就可以直接通过ajax 的 send() 方法将 fd 发送到后台。

HTML部分：

<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>stu-number：<input type="text" name="number" /></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>

jQuery部分：

function fsubmit() {
        var form=document.getElementById("form1");
        var fd =new FormData(form);
        $.ajax({
             url: "server.php",
             type: "POST",
             data: fd,
             processData: false,  // 告诉jQuery不要去处理发送的数据
             contentType: false,   // 告诉jQuery不要去设置Content-Type请求头
             success: function(response,status,xhr){
                console.log(xhr);
                var json=$.parseJSON(response);
                var result = '';
                result +="个人信息：<br/>name:"+json['name']+"<br/>gender:"+json['gender']+"<br/>number:"+json['number'];
                 result += '<br/>头像：<img src="' + json['photo'] + '" height="100" style="border-radius: 50%;" />';
                 $('#result').html(result);
             }
        });
        return false;
    }

PHP部分

<?php

$name = isset($_POST['name'])? $_POST['name'] : '';  
$gender = isset($_POST['gender'])? $_POST['gender'] : '';
$number = isset($_POST['number'])? $_POST['number'] : '';  
$filename = time().substr($_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['number'] = $number;  
    $response['photo'] = $filename;  
}else{  
    $response['isSuccess'] = false;  
}  
echo json_encode($response);

?>