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#2
nuciewth2009-08-30 10:47
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Warning: mysql_fetch_row(): supplied argument is not a valid MySQL result resource in
这段代码错误,
<?php
$result1=mysql_query("select * from kuzhan");
while($row=mysql_fetch_array($result1)){
?>
<dd><?php echo "$row[dm]";?></dd>//这段代码那错了?
?>
警告:mysql_fetch_array():提供的参数不是一个有效的MySQL结果资源