php - MySQL 查询在 phpMyAdmin 中有效，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

想法是，您有一个物质主列表 (tbl_substances)，然后如果您在 tbl_substances_lists 或 tbl_substances_info 中输入了有关它们的任何信息，那么也可以显示这些信息(只要您使用正确的组织 ID 登录)

显示所有物质很重要，即使它们没有自定义信息，这就是我使用 LEFT JOIN 的原因。

这个查询在 phpMyAdmin 中完美运行，但是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, list1, list2, list3, list4, list5 FROM tbl_substances_' at line 2

我不确定问题是否是我遗漏的明显问题，或者是否与这段代码使用 mysql_query 这一事实有关，我知道它已被弃用且过时等等。

我不是数据库专家，所以如果您觉得这个查询非常难看，那么我提前道歉!

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());

假设此时 $sWhere 和 $sOrder 中没有任何内容 - $sLimit 由用户选择，但在这种情况下它将是 LIMIT 0, 25 以获取前 25 条记录。

在这种情况下，这一切结合起来产生回显 $sQuery 的结果:

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

还没有分析代码，但在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。