在AJAX请求中传递表单数据的麻烦

问题描述：

I'm having issues passing form values to PHP via an AJAX request. As per my code below, variables can be passed back so I suspect the issue may be with data: $('#signup-form').serialize().

JS:

$.ajax
({
    type: "POST",
    url: "http://www.domain.com/includes/register.php",
    data: $('#signup-form').serialize(),
    success: function(data)
        {
            $('#signup-response').hide();
            $('#signup-response').fadeIn();
            $('#signup-response').html(data);
        },
    error: function()
    {
        alert("fail");
    }
})

Form:

<form id="signup-form" action="" method="POST">
    <input name="email" type="email" class="form-control" id="signupEmail" placeholder="Email address">
    <input name="password1" type="password" class="form-control" id="signupPassword" placeholder="Password">
    <input name="password2" type="password" class="form-control" id="signupPassword2" placeholder="Password">
    <select name="country" id="signupCountry" class="selectpicker">
        <option value="0">Country</option>
        <option>United States</option>
        <option>United Kingdom</option>
        <option>Canada</option>
    </select>
    <select name="gender" id="signupGender" class="selectpicker">
        <option value="0">Gender</option>
        <option>Female</option>
        <option>Male</option>
    </select>
    <button id="signup" class="btn btn-success btn-block signup" type="submit">Sign up</button>
</form>

<?php
$username = $_POST['signupEmail'];
echo "hello"; // works
echo $username; // doesn't work
?>

我遇到了通过AJAX请求将表单值传递给PHP的问题。根据我的下面的代码，变量可以传回，所以我怀疑问题可能出在data：$（'＃signup-form'）。serialize（） code>。 p>

JS： p>


  $。ajax 
（{
 type：“POST”，
 url：“http://www.domain.com/includes  /register.php",nn data：$（'＃signup-form'）。serialize（），
 success：function（data）
 {
 $（'＃signup-response'）。hide（）  ; 
 $（'＃signup-response'）。fadeIn（）; 
 $（'＃signup-response'）。html（data）; 
}，
 error：function（）
 {
 警告（“失败”）; 
} 
}）
  code>  pre> 
 
 表单： p> 
 
 
 ＆lt; form  id =“signup-form”action =“”method =“POST”＆gt; 
＆lt; input name =“email”type =“email”class =“form-control”id =“signupEmail”placeholder =“电子邮件地址 “＆gt; 
＆lt; input name =”password1“type =”password“class =”form-control“id =”signupPassword“placeholder =”Password“＆gt; 
＆lt; input name =”password2“type =” 密码“class =”form-control“id =”signupPassword2“placeholder =”Password“＆gt; 
＆lt  ; select name =“country”id =“signupCountry”class =“selectpicker”＆gt; 
＆lt; option value =“0”＆gt; Country＆lt; / option＆gt; 
＆lt; option＆gt; United States＆lt; / option＆gt; 
  ＆lt; option＆gt; United Kingdom＆lt; / option＆gt; 
＆lt; option＆gt; Canada＆lt; / option＆gt; 
＆lt; / select＆gt; 
＆lt; select name =“gender”id =“signupGender”class =“selectpicker”＆gt;  
＆lt; option value =“0”＆gt;性别＆lt; /选项＆gt; 
＆lt;选项＆gt;女＆lt; /选项＆gt; 
＆lt;选项＆gt;男＆lt; /选项＆gt; 
＆lt; / select＆gt; 
＆lt;  button id =“signup”class =“btn btn-success btn-block signup”type =“submit”＆gt;注册＆lt; / button＆gt; 
＆lt; / form＆gt; 
  code>  pre> 
 \  n  register.php  p> 
 
 
 ＆lt;？php 
 $ username = $ _POST ['signupEmail']; 
echo“hello”;  // works 
echo $ username;  //不起作用
？＆gt; 
  code>  pre> 
  div>

答

Try this, You need to use input name <input name="email" type="email" class="form-control" id="signupEmail" placeholder="Email address"> rather use of input field id.

 $username = $_POST['email'];

instead of

  $username = $_POST['signupEmail'];

在AJAX请求中传递表单数据的麻烦

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