MySQL查询在phpMyAdmin中工作,在PHP中失败

Question

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:

您的SQL语法有错误; 查看与您的MySQL服务器版本对应的手册,以便在''附近使用正确的语法.在第2行,选择subid,list1,list2,list3,list4,list5 FROM tbl_substances_'

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<intval( $_POST['iSortingCols'] ) ; $i++ )
        {
            if ( $_POST[ 'bSortable_'.intval($_POST['iSortCol_'.$i]) ] == "true" )
            {
                $iColumnIndex = array_search( $_POST['mDataProp_'.$_POST['iSortCol_'.$i]], $aColumns );
                $sOrder .= $aColumns[ $iColumnIndex ]."
                    ".mysql_real_escape_string( $_POST['sSortDir_'.$i] ) .", ";
            }
        }

        $sOrder = substr_replace( $sOrder, "", -2 );
        if ( $sOrder == "ORDER BY" )
        {
            $sOrder = "";
        }
    }

        /*
         * Table info
         */
    $sTable = "tbl_substances ".$sLimit.") AS s 
LEFT JOIN (
  SELECT subid, list1, list2, list3, list4, list5 
  FROM tbl_substances_lists 
  WHERE orgid = '".$orgid."'
) AS x 
ON s.subst_id = x.subid 
LEFT JOIN (
  SELECT subid, info 
  FROM tbl_substances_info WHERE orgid = '".$orgid."'
) AS y 
ON s.subst_id = y.subid";

        $sWhere = "";


    /*
     * SQL queries
     * Get data to display
     */

    $sQuery = "
        SELECT SQL_CALC_FOUND_ROWS * FROM (SELECT * FROM $sTable
                $sWhere
        $sOrder
        $sLimit
    ";

        $rResult = mysql_query( $sQuery ) or die(mysql_error());

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, 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

Answer 1

没有分析代码，但查询方面我没有看到语法错误。但我建议您像这样编写查询：

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

应该给出相同的结果。