[英]SQL Query to Display Values in Table for Logged In User
I am just starting out, very much a rookie and any help would be greatly appreciated. 我刚刚起步,非常新手,我们将不胜感激。 I have a list of URLs in a table called "url".
我在名为“ url”的表中有一个URL列表。 I have another table called "users" and in that table a column named "username" which list the users of the site.
我有另一个名为“用户”的表,该表中有一个名为“用户名”的列,其中列出了该站点的用户。 I am looking to display urls for the logged in user via a session value of $session->username but have not yet figured out how to only render URLS for the logged in user.
我希望通过$ session-> username的会话值显示已登录用户的url,但尚未弄清楚如何仅为已登录用户呈现URL。
I apologise in advance, I may have the wrong idea but wanted to show you what I was attempting to do but am not able to get to work properly without SQL syntax errors. 提前致歉,我可能有一个错误的主意,但想向您展示我正在尝试做的事情,但是如果没有SQL语法错误,我将无法正常工作。
$result = $db->query('SELECT * from urls, users INNER JOIN username ON
$session->username');
Thank you very much 非常感谢你
Assuming you have a column named user_id
which is making a relation b/w users
and urls
table. 假设您有一个名为
user_id
的列,该列建立了黑白users
和urls
表的关系。 if that column name is named anything else replace the user_id
with that column name 如果该列名已命名,则用该列名替换
user_id
$result = $db->query('SELECT * from users u JOIN urls ur
ON u.username = ur.username
WHERE u.username="'.$session->username.'"');
You need to use the index connection your urls to your user. 您需要使用网址将索引链接到您的用户。 An user_id for example.
例如,一个user_id。 This would give you an query like:
这将给您一个查询,例如:
$sQuery = "SELECT * FROM users
INNER JOIN urls ON urls.user_id = users.user_id
WHERE users.username = '".$oSession->username."'";
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