MySQL查询在phpMyAdmin中工作,在PHP中失败

SELECT sql_CALC_FOUND_ROWS * FROM (   SELECT *    FROM tbl_substances   liMIT 0,25) AS sleft JOIN (   SELECT subID,List1,List2,List3,List4,List5   FROM tbl_substances_Lists   WHERE orgID =  '1') AS x ON s.subst_ID = x.subIDleft JOIN (   SELECT subID,info   FROM tbl_substances_info   WHERE orgID =  '1') AS y ON s.subst_ID = y.subID

这个想法是你有一个物质的主列表(tbl_substances)然后如果你在tbl_substances_Lists或tbl_substances_info中输入了关于它们的任何信息,那么也可以显示(只要你使用正确的组织ID登录)

显示所有物质非常重要,即使它们没有自定义信息,这就是我使用left JOIN的原因.

此查询在phpmyadmin中完美运行,但当我在我的数据库脚本中使用它时,我得到：

You have an error in your sql Syntax; check the manual that corresponds to your MysqL server version for the right Syntax to use near ‘) AS s left JOIN (SELECT subID,List5 FROM tbl_substances_’ at line 2

我不确定问题是否是显而易见的我错过了,或者是否与这个代码使用MysqL_query的事实有关,我知道这个问题已被弃用和老式等等.

我不是数据库专家,所以如果这个查询看起来很难看,那我就提前道歉！

编辑2

这是构建此查询的代码(它根据您要搜索的内容动态构建,但这是基本形式)

    /*     * Length     */    if ( isset( $_POST['idisplayStart'] ) && $_POST['idisplayLength'] != '-1' )    {        $slimit = "liMIT ".MysqL_real_escape_string( $_POST['idisplayStart'] ).",".            MysqL_real_escape_string( $_POST['idisplayLength'] );    }    /*     * Ordering     */    $sOrder = "";    if ( isset( $_POST['iSortCol_0'] ) )    {        $sOrder = "ORDER BY  ";        for ( $i=0 ; $i

想象一下,$sWhere和$sOrder目前没有任何内容 – $slimit由用户选择,但在这种情况下,它将是liMIT 0,25来获得前25个记录.

在这种情况下,这一切都结合起来,以回显$squery：

SELECT sql_CALC_FOUND_ROWS * FROM (  SELECT *   FROM tbl_substances liMIT 0,25) AS s left JOIN (  SELECT subID,List5   FROM tbl_substances_Lists   WHERE orgID = '1') AS x ON s.subst_ID = x.subID left JOIN (  SELECT subID,info   FROM tbl_substances_info   WHERE orgID = '1') AS y ON s.subst_ID = y.subID

SELECT sql_CALC_FOUND_ROWS * FROM tbl_substances sleft JOIN tbl_substances_Lists x ON s.subst_ID = x.subID AND x.orgID = '1'left JOIN tbl_substances_info y ON s.subst_ID = y.subID AND y.orgID = '1'liMIT 0,25