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

The idea is that you have a master list of substances (tbl_substances) then if you have entered any information about them in tbl_substances_lists or tbl_substances_info then that can be displayed too (as long as you are logged in with the right organisation ID)

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）

It's important to show all the substances even if they have no custom information which is why I'm using a LEFT JOIN.

显示所有物质非常重要，即使它们没有自定义信息，这也是我使用LEFT JOIN的原因。

This query works perfectly in phpMyAdmin but when I use it in my database script I get:

这个查询在phpMyAdmin中运行得很好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

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

I'm not sure whether the problem is something obvious I'm missing or is it something to do with the fact that this bit of code uses mysql_query which I know is deprecated and old fashioned etc. etc.

我不确定问题是否是显而易见的我缺少的，或者是否与这个代码使用mysql_query这个事实有关，我知道这个问题已被弃用和老式等等。

I'm not a database expert so if this query looks very ugly to you then I apologise in advance!

我不是数据库专家，所以如果这个查询看起来非常难看，那么我提前道歉！

EDIT 2

编辑2

Here's the code for building this query (it gets built dynamically depending on what you're searching for but this is the basic form)

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());

Imagine there is nothing in $sWhere and $sOrder at the moment - $sLimit is chosen by the user but in this case it will be LIMIT 0, 25 to get the first 25 records.

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。

This all combines in this case to make the result of echoing out $sQuery:

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

2 个解决方案

#1

Haven't analyzed the code, but querywise I see no syntax error. But I'd advise, that you write the query like this:

没有分析代码，但查询我没有看到语法错误。但我建议你写这样的查询：

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

Should give the same result.

应该给出相同的结果。

#2

In my case, I was working with "ISO 8859-1 Characters" (Ç,Ã,é....), doing: utf8_decode($query); solved this problem.

在我的情况下，我正在使用“ISO 8859-1 Characters”（Ç，Ã，é....），执行：utf8_decode（$ query）;解决了这个问题。

#1