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How to submit forms and upload images using FormData

一个新手Original: 2017-09-14 10:52:142389browse

FormData object can combine the name and value of all form elements in the form into a queryString and submit it to the background. When submitting using Ajax, using the FormData object can reduce the workload of splicing queryString.

Use FormData object

1. Create an empty FormData object, and then use the append method to add key/value

##var formdata = new FormData();
##formdata.append(
'name',' fdipzone');
'gender'
,'male');
2. Obtain the form object and pass it into the FormData object as a parameter

    <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);

Use FormData to submit forms and upload files:

<!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; + ret[&#39;photo&#39;]  + &#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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