MySQL PHP count(*)返回奇怪的东西

本文介绍了MySQL PHP count(*)返回奇怪的东西的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述

我正在运行以下查询，希望它返回一个INTEGER，但是它返回"Resource id#3". 我在做什么错了?

I'm running the following query, expecting it to return an INTEGER, but it returns "Resource id #3" What am I doing wrong?

$queryPlans = mysql_query("SELECT count(*) FROM infostash.rooms"); echo $queryPlans;

此表中实际上有15行，我想返回数字15. 有什么建议吗?

There are actually 15 rows in this table, and I would like to return the number 15. Any suggestions?

推荐答案

您需要:

$queryPlans = mysql_query("SELECT count(*) FROM infostash.rooms"); $row = mysql_fetch_array($queryPlans); echo $row[0];

mysql_query() 没有返回结果.它返回的资源可以循环遍历并查询行(如上所述).

mysql_query() isn't returning the result. It's returning a resource you can loop across and interrogate for rows (as above).