ajax没有向php发送数据 [英] ajax is not sending data to php

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问题描述

当我点击提交时,我收到未定义索引fname的错误。

请解决或指导我。



这是代码我的html文件

 <  表格    id   =  myForm   方法    =  post    enctype   =  multipart / form-data >  
名字:< 输入 type = text name = fname id = fname / > < span class =code-keyword>< br >
姓氏:< 输入 type = text 名称 = lname id = lname / < span class =code-keyword>> < br >
电子邮件:< 输入 < span class =code-attribute> type = text name = email id = email / > < br >
图片:< input type = file name = image < span class =code-attribute> id = image / > < br >
< 按钮 type = 按钮 < span class =code-attribute> name = btnSubmit id = btn提交 > 提交< /按钮 >
< /表格 >





 <   script     language   =  javascript    src   =  js / jquery-1.9.1.min.js >  <   / script  >  
< script language = javascript src = js / jquery.form.js > < / script >
< script 类型 = text / javascript >
$(document).ready(function(){
// alert(4) ;
$(#btnSubmit)。click(function(){
var formData = new FormData($(#myForm));
alert(formData);
console.log(formData);
$ .ajax({
type:'POST',
url:'learn_form2.php',
data:formData,
success:function(data){
alert(data);
},
cache:false,
contentType:false,
processData:false
} );
});
});
< / script >







这是我的php文件的代码

< pre lang =php> <?php
echo $ _REQUEST [' fname'];
?>

解决方案

(文件).ready (function(){
// alert(4);


(#btnSubmit)。click(function(){
var formData = new FormData(


(#myForm));
alert(formData);
console.log(formData);


when i click on submit, i get error that undefined index fname.
please resolve or guide me.

this is code of my html file

<form id="myForm" method ="post" enctype="multipart/form-data">
    First Name: <input type="text" name="fname" id="fname" /> <br>
    Last Name: <input type="text" name="lname" id="lname" /> <br>
    Email:  <input type="text" name="email" id="email" /> <br>
    Image: <input type="file" name="image" id="image" /> <br>
    <button type="button" name="btnSubmit" id="btnSubmit"> Submit </button>
</form>



<script language="javascript" src="js/jquery-1.9.1.min.js"></script>
<script language="javascript" src="js/jquery.form.js"></script>
<script type="text/javascript">
$(document).ready(function(){
//alert(4);
$("#btnSubmit").click(function(){
     var formData = new FormData($("#myForm"));
     alert(formData);
     console.log(formData);
     $.ajax({
        type: 'POST',
        url: 'learn_form2.php',
        data: formData,
         success: function (data) {
           alert(data);
         },
        cache: false,
        contentType: false,
        processData: false
      });
  });
});
</script>




and this is code of my php file

<?php
echo $_REQUEST['fname'];
?>

解决方案

(document).ready(function(){ //alert(4);


("#btnSubmit").click(function(){ var formData = new FormData(


("#myForm")); alert(formData); console.log(formData);


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