<? while($blog2 = mysql_fetch_array($blog))
{
$tag = ("select * from blogtag_ref inner join blogtag on blogtag_ref.tag_id = blogtag.tag_id where blogtag_ref.blog_id = '".$blog2['blog_id']."' ") or die(mysql_error());
?>
<table width="1040" border="0" cellspacing="0" cellpadding="10" align="center">
<tr>
<td><div class="blogtitle"><? echo $blog2['subject']; ?></div></td>
</tr>
<tr>
<td><div class="blogdate"><? echo $date; ?></div> //
<? while($tag2 = mysql_fetch_array($tag))
{ ?>
<div class="blogtag"><? echo $tag2['tag']; ?></div>/
<? } ?>
它给了我Warning: mysql_fetch_assoc() expects parameter 1 to be resource
mysql_fetch_array错误想不通为什么
我知道我的SQL是正确的,如果我呼应$tag
...我在phpMyAdmin复制粘贴......它给我的结果...
是因为我不能在一段时间内放一段时间?
啊,我忘了...的mysql_query谢谢你这么多,我看了它20分钟... – user3011784