【发布时间】:2015-10-31 01:55:35
【问题描述】:
尝试构建一个简单的 jQuery,允许用户单击表格行中的按钮,该按钮应运行单独的 PHP 脚本以从 db 行中删除项目并在一个操作中删除表格行。 (一旦工作,将迁移到准备好的语句。)
HTML
<tr id="wish_list_row2">
<td class="text-center">
<a href="#" data-toggle="tooltip" data-placement="bottom" title="Click to download this Resource" style="margin-right:15px;"><i class="fa fa-download"></i></a>
<a onclick="deleteWishList(3,1)" data-toggle="tooltip" data-placement="bottom" title="Click to remove this Resource from your Wish List"><i class="fa fa-gift"></i></a>
</td>
</tr>
<tr id="wish_list_row3">
<td class="text-center">
<a href="#" data-toggle="tooltip" data-placement="bottom" title="Click to download this Resource" style="margin-right:15px;"><i class="fa fa-download"></i></a>
<a onclick="deleteWishList(4,1)" data-toggle="tooltip" data-placement="bottom" title="Click to remove this Resource from your Wish List"><i class="fa fa-gift"></i></a>
</td>
</tr>
JS
function deleteLike(rid,uid) {
$.post("remove_this_item.php?rid="+rid+"&uid="+uid);
$('#row'+rid).remove();
return false;
}
PHP
include_once ('[INCLUDE MYSQL CONNECTION VARIABLES - removed for this demo]');
$rid = $_GET['rid'];
$uid = $_GET['uid'];
// Create connection
$conn = mysqli_connect($servername, $username, $password, $dbname);
// delete this 'like' from the db for this user
$sql = "DELETE FROM likes WHERE user_id_REF='$uid' AND resource_id_REF='$rid' LIMIT 1";
$conn->query($sql);
mysqli_close($conn);
当我单击“删除”链接时,tr 会按预期删除。但是,db 行保持不变。如果我在 url 中手动设置 PHP 页面上的 rid 和 uid,则 db 行将被删除。我已尝试更改一些内容(例如 rid 和 uid 顺序),但尚未取得任何成功。
我对 JS 还不是很熟悉,所以也许我在编写函数时搞砸了?
【问题讨论】:
标签: javascript php jquery mysql