dongzhang8680 2014-04-18 17:01
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在AJAX请求中传递表单数据的麻烦

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>

register.php

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

1条回答 默认 最新

  • douju2474 2014-04-18 17:03
    关注

    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'];
    
    本回答被题主选为最佳回答 , 对您是否有帮助呢?
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