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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).
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MySQL PHP count(*)返回奇怪的东西
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