if($_GET['identification'] != null){ $sql = mysql_query("select * from frontpagenews where identification = ".$_GET['identification']."order by createDate desc limit 0,7"); 这里报错 while($result = mysql_fetch_array($sql)){ ?>
} }else{
$sql = mysql_query("select * from frontpagenews where identification = 1 order by createDate desc limit 0,7"); while($result = mysql_fetch_array($sql)){ ?>
} } ?>
错误信息
Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in C:\Apache\htdocs\growers\news.php on line 147
回复讨论(解决方案)
$sql = mysql_query("select * from frontpagenews where identification = ".$_GET['identification']." order by createDate desc limit 0,7");
mysql_query("select * from frontpagenews where identification = ".$_GET['identification']. "order by createDate desc limit 0,7"); 除非 $_GET['identification'] 最后的字符是空格,不然 $_GET['identification'] 的值不就和 order 连在一起了? 如果 $_GET['identification'] = 1 你的 sql指令不就了 "select * from frontpagenews where identification = 1order by createDate desc limit 0,7" 吗?
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