$foot = mysql_query("SELECT count(*)
FROM tblQA
WHERE intResponseID = ''
AND cCategory = 'Football' as qcount,
(SELECT max(dPostDateTime)
FROM tblQA
WHERE intResponseID = ''
AND cCategory = 'Football') as lastq");
$football = mysql_fetch_array($foot);
echo "<td class='forum'>" . $footbll['qcount'] . "</td>";
echo "<td class='forum'>" . $footbll['lastq'] . "</td>";
这不会显示我的表中的任何内容。我没有发布整个HTML代码,我有表结构很好。我哪里错了? SQL查询
'$ football'或'$ footbll',这是正确的拼写?你确定你的查询没有产生任何错误吗? – BoltClock 2010-10-23 15:32:08
不可能告诉。复制+粘贴完整查询,在数据库GUI中运行它,看看是否有结果出现。 – 2010-10-23 15:32:35
拼写是我的错误在这里,它的代码($ football) – BigMike 2010-10-23 15:33:41