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 s
LEFT JOIN (
   SELECT subid,list1,list2,list3,list4,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

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

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 s
LEFT 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