Question

即使我的process.php有效，data.success也会返回false。我一直在尝试调试它，但我找不到导致此错误的错误。

显示“OKAY！”仅当我将data.success更改为!data.success时才会显示消息，为什么？

HTML

<form id="my" action="" method="post">
    <label for="name">Name</label>
    <input type="text" id="name" name="name" value="<?php echo $name; ?>">

    <label for="surname">Surname</label>
    <input type="text" id="surname" name="surname" value="<?php echo $surname; ?>">

    <label for="username">Username</label>
    <input type="text" id="username" name="username" value="<?php echo $username; ?>">

    <label for="email">Email</label>
    <input type="text" id="email" name="email" value="<?php echo $user_email; ?>">
    <input id="save" type="submit" name="changeit" value="Save Changes">
</form>

的jQuery

$(document).ready(function () {
    $(function () {
        $('#my').submit(function (e) {
            e.preventDefault();
            var form = $(this);
            var formData = form.serialize();
            $.ajax({
                type: 'POST',
                url: '/modules/settings/process.php', // the url where we want to POST
                data: formData // our data object
            }).done(function (data) {
                if (data.success) {
                    $('#my').append('<div class="alert alert-success">' + "OKAY!." + '</div>');
                }
            })
        });
    });
});

有人可以帮忙吗？

Answer 1

从action=""标题中删除form，少即是

尝试ajax部分

的此代码

$.ajax({url: '/modules/settings/process.php', type: 'post', data: $('form').serialize(), success: function(data){
       $('#my').append('<div class="alert alert-success">' + "OKAY!." + '</div>');
    }
});

jQuery ajax data.success返回false

1 个答案: