0

下面是我试图开始工作的代码:

    $y= "SELECT ('PRV_IDX')
         FROM LLS_PRIVILEGES
         WHERE `PRV_NAME` = 'Reader';";

     mysql_query($y);

$x= "SELECT ('USER_IDX')
     FROM LLS_USERS
     WHERE `USR_LOGIN` = '".$_SESSION['tool_user']."';";

      mysql_query($x);

$w= "INSERT INTO LLS_USERS_PRIVILEGES
     (USP_USR_IDX,USP_PRV_IDX)
     VALUES ($x,$y);    ";

     mysql_query($w);

我想将这些值从选择语句插入到最终表中。但是,我不确定我的语法是否正确,并且我一直无法在线找到解决方案。我不确定您是否必须每次都执行 mysql_query 才能使 select 语句真正生效并将其放在 $variable 中。

抱歉,我是 SQL 新手,但感谢您的帮助!

4

1 回答 1

-1

mysql_*函数已弃用。改用PDO

$y=$dbh->query("SELECT ('PRV_IDX')
    FROM LLS_PRIVILEGES
    WHERE `PRV_NAME` = 'Reader'");
$x=$dbh->prepare("SELECT ('USER_IDX')
     FROM LLS_USERS
     WHERE `USR_LOGIN` = ?");
$x->execute(array($_SESSION['tool_user']));
$w=$dbh->prepare("INSERT INTO LLS_USERS_PRIVILEGES
     (USP_USR_IDX,USP_PRV_IDX)
     VALUES (?,?)");
$y="SELECT ('PRV_IDX')
    FROM LLS_PRIVILEGES
    WHERE `PRV_NAME` = 'Reader'";
$x="SELECT ('USER_IDX')
     FROM LLS_USERS
     WHERE `USR_LOGIN` = '".$_SESSION['tool_user']."'");
$w->execute(array($x,$y));

更多关于PDO 的信息:http ://www.php.net/manual/en/book.pdo.php

于 2013-10-24T13:26:18.860 回答