Question

这段PHP代码那儿错了

<?

//基本设定
$mysql_server_name = "localhost";
$mysql_username = "root";
$mysql_password = "";
$mysql_database = "counter";

//连接

$conn=mysql_connect( $mysql_server_name, $mysql_username, $mysql_password);

//SQL语法每增加一个多一个栏位

$sql="INSERT INTO `sun`.`counter` (`new` )VALUES ('1')";

//选择处理的资料库

mysql_select_db($mysql_database,$conn);

//进行查询活执行

$result = mysql_query($sql);

//结束并且释放连接

mysql_close($conn);

//..返回值代码.........................................
//连接
$conn=mysql_connect( $mysql_server_name, $mysql_username, $mysql_password);
//SQL语法
$sql="SELECT * FROM `counter`";
//进行查询选取

$result=mysql_db_query($mysql_database,$sql,$conn);

//查询后的结果 抓出一笔
$row=mysql_fetch_row($result);

//打印出所有栏位
print_r($row);

//清楚查询结果
mysql_free_result($result);

?>

错误提示为：
Warning: mysql_fetch_row(): supplied argument is not a valid MySQL result resource in C:\AppServ\www\index.php on line 39

Warning: mysql_free_result(): supplied argument is not a valid MySQL result resource in C:\AppServ\www\index.php on line 45

Answer 1

建议你一步一步调试
先var_dump($conn)
看看
再var_dump($result);
看看是不是MySQL result resourc