select count(table1.column1),table2.column2
from table1,table2
where table1.column1 = table2.column1
group by table1.column1
from table1,table2
where table1.column1 = table2.column1
group by table1.column1
调试欢乐多
$sql = "
SELECT count(table1.column1) as column1, table2.column2
FROM table1, table2
WHERE table1.column1 = table2.column1
GROUP BY table1.column1
";给count(table1.column1)赋予一个变量就不会出错了
原因是他不是有效的列????
................................................
$result = mysql_query($query,$conn) or die(mysql_error($query));
其它的:
$conn = mysql_connect($dbhost,$dbuser,$dbpass) or die("Could not connect to database!");
mysql_select_db($dbname,$conn) or die("Could not select database!");
语句已经照 风逍遥 的方式更改了,还是报同样的错误。为什么?
Warning: mysql_error(): supplied argument is not a valid MySQL-Link resource
我的环境:
apache-2.0.48
mysql-server-4.1.7
php-5.0.4
win2003serverEnt
jdk1.5.02
问题不出在上述语句,你提出的情况,我已经测试过了。谢谢!