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在AJAX请求中传递表单数据的麻烦

[英]Trouble with passing form data in AJAX request

I'm having issues passing form values to PHP via an AJAX request. 我在通过AJAX请求将表单值传递给PHP时遇到问题。 As per my code below, variables can be passed back so I suspect the issue may be with data: $('#signup-form').serialize() . 根据我的下面的代码,变量可以传回,所以我怀疑问题可能与data: $('#signup-form').serialize()

JS: 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: 形成:

<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 register.php

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

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. 试试这个,你需要使用输入名称 <input name="email" type="email" class="form-control" id="signupEmail" placeholder="Email address">而不是使用输入字段id。

 $username = $_POST['email'];

instead of 代替

  $username = $_POST['signupEmail'];

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