查询失败：您的SQL语法出错;检查与您的MySQL相对应的手册

Question

由于我的服务提供商不再支持旧版本，我不得不更新到MySQL 5和PHP 5。我有一个工作网站，这个MySQL搜索工作正常。以下是放置在用户的列表页面上的一个包含。当我将SQL语句取出并在服务器上运行时，它会返回正确的结果。我的猜测是这与格式有关。任何想法，为什么我可能会得到这个错误？ “

”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，以便在第1行“ORDER BY x.id_have LIMIT 0,30'附近使用正确的语法”

有了这个语法

<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> 

Answer 1

你可能需要包装$hid引号，像这样：

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