Warning: mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource运行后 出现了如上代码
能告诉我 大体上是什么地方出错了吗?$sql = "SELECT entries.*, categories.cat FORM entries, categories
WHERE entries.cat_id = categories.id
ORDER BY dateposted DESC
LIMIT 1;";$result = mysql_query($sql);
$row = mysql_fetch_assoc($result);这是源代码
能告诉我 大体上是什么地方出错了吗?$sql = "SELECT entries.*, categories.cat FORM entries, categories
WHERE entries.cat_id = categories.id
ORDER BY dateposted DESC
LIMIT 1;";$result = mysql_query($sql);
$row = mysql_fetch_assoc($result);这是源代码
ORDER BY dateposted DESC
LIMIT 1;这里错了
FROM 和 FORM
http://download.csdn.net/source/2691449
$result = mysql_query($sql,$conn);
$row = mysql_fetch_assoc($result);mysql_close($conn);
sql语句可以直接输出 要执行的 复制 到命令行或者其他直接操作SQL的工具中 如果可以再说其他的 这个是必须的
WHERE entries.cat_id = categories.id
ORDER BY dateposted DESC
LIMIT 1;