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PHP结合HTML5使用FormData对象提交表单及上传图片

PHP中文网Original: 2017-03-30 16:59:302265browse

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(&#39;name&#39;,&#39;fdipzone&#39;);
formdata.append(&#39;gender&#39;,&#39;male&#39;);

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(&#39;form1&#39;);
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($(&#39;#form1&#39;)[0]);
        $.ajax({
            url: &#39;server.php&#39;,
            type: &#39;POST&#39;,
            data: data,
            dataType: &#39;JSON&#39;,
            cache: false,
            processData: false,
            contentType: false
        }).done(function(ret){
            if(ret[&#39;isSuccess&#39;]){
                var result = &#39;&#39;;
                result += &#39;name=&#39; + ret[&#39;name&#39;] + &#39;<br>&#39;;
                result += &#39;gender=&#39; + ret[&#39;gender&#39;] + &#39;<br>&#39;;
                result += &#39;<img src="&#39;%20+%20ret[&#39;photo&#39;]%20%20+%20&#39;" width="100">&#39;;
                $(&#39;#result&#39;).html(result);
            }else{
                alert(&#39;提交失敗&#39;);
            }
        });
        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[&#39;name&#39;])? $_POST[&#39;name&#39;] : &#39;&#39;;
$gender = isset($_POST[&#39;gender&#39;])? $_POST[&#39;gender&#39;] : &#39;&#39;;

$filename = time().substr($_FILES[&#39;photo&#39;][&#39;name&#39;], strrpos($_FILES[&#39;photo&#39;][&#39;name&#39;],&#39;.&#39;));

$response = array();

if(move_uploaded_file($_FILES[&#39;photo&#39;][&#39;tmp_name&#39;], $filename)){
    $response[&#39;isSuccess&#39;] = true;
    $response[&#39;name&#39;] = $name;
    $response[&#39;gender&#39;] = $gender;
    $response[&#39;photo&#39;] = $filename;
}else{
    $response[&#39;isSuccess&#39;] = false;
}

echo json_encode($response);
?>

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