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我被迫更新到 MySQL 5 和 PHP 5，因为我的服务提供商不再支持旧版本。我有一个运行良好的 MySQL 搜索工作站点。以下是放置在用户列表页面上的包含。当我取出 SQL 语句并在服务器上运行它时，它会返回正确的结果。我的猜测是它与格式化有关。任何想法为什么我可能会收到此错误？

“查询失败：您的 SQL 语法有错误；请查看与您的 MySQL 服务器版本相对应的手册，以在第 1 行的 'ORDER BY x.id_have LIMIT 0,30' 附近使用正确的语法”

使用这种语法

<table cellpadding="15" border="0">
 <tr>
  <td valign="top">
   So far, the following members have these items which match;
   <ul>
   <?  
    $peoplewhomatch = mysql_query("SELECT x.id_customer, g.title Have, c.user_name as Member, g.title gwants, gg.title Has, y.id_customer x_username 
        FROM Want x 
        JOIN Want y 
            ON (y.id_sellversion,y.id_version) = (x.id_version,x.id_sellversion) 
        inner join Version as v 
            on x.id_version = v.id_version 
        inner join Version as vv 
            on y.id_version = vv.id_version 
        inner join Game as g 
            on g.id_game = vv.id_game 
        inner join Customer as c 
            on y.id_customer = c.id_customer 
        inner join Game as gg 
            on gg.id_game = v.id_game 
        WHERE x.id_have = $hid 
        ORDER BY x.id_have 
        LIMIT 0, 30")or die("Query failed: " . mysql_error()); 
    while($row = mysql_fetch_array($peoplewhomatch))
    { ?>
    <span class='greenitalic'><?=$row['Member']?></span> has <span class='highshadow'><?=$row['Has']?></span> and wants <span class='ishadow'><?=$row['gwants']?></span><BR><?}?>
  </td>
 </tr>
</table>
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1 回答 1

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您可能需要$hid像这样用引号括起来：

 $peoplewhomatch = mysql_query("SELECT x.id_customer, g.title Have, c.user_name as Member, g.title gwants, gg.title Has, y.id_customer x_username 
        FROM Want x 
        JOIN Want y 
            ON (y.id_sellversion,y.id_version) = (x.id_version,x.id_sellversion) 
        inner join Version as v 
            on x.id_version = v.id_version 
        inner join Version as vv 
            on y.id_version = vv.id_version 
        inner join Game as g 
            on g.id_game = vv.id_game 
        inner join Customer as c 
            on y.id_customer = c.id_customer 
        inner join Game as gg 
            on gg.id_game = v.id_game 
        WHERE x.id_have = '$hid'
        ORDER BY x.id_have 
        LIMIT 0, 30")or die("Query failed: " . mysql_error());
于 2013-06-27T01:55:21.377 回答