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回复讨论(解决方案)
试试连接表
SELECT B.* FROM tb_xmpcps AS A LEFT OUVER JOIN tb_user AS B ON A.xmpcps_user=B.online_user WHERE A.xmpcps_pcid='$xmpc_id'
$query2=mssql_query("SELECT B.* FROM tb_xmpcps AS A LEFT OUTER JOIN tb_user AS B ON A.xmpcps_user=B.online_user WHERE A.xmpcps_pcid='{$xmpc_id}'");还是报错啊?为什么?说$query2有错误?劳驾再帮我看看!
. echo mssql_get_last_message();
看看都有什么我最终还是用两个嵌套查询解决了,这样就可以了,还是感谢大家的帮助!
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