试图在线表格广告,它应该设置为什么? [英] Trying to ad a online table, what should it be set to?
问题描述
好的,我正在向用户表中添加一个名为online的表。
因此,当他们登录时,它将设置(在线)为Y,当他们退出时,它将被设置为N.
但是应该创建什么,它会经常更新,我需要能够在会员在线部分从我的php中读取它吗?
varchar( 50),int(10),或文本...
我如何调用它的结果而不是文本Y但总共。
比如让我们说总共有50个用户= Y所以它会打印50个。
我已经尝试了好几次但是我不断得到9empty)或者(Y)不是总数为8
我创建了这样的表:
[PHP] ALTER TABLE用户
ADD online varchar(1)default'' N''; [/ PHP]
并按如下方式调用:
[PHP] $ sql =" SELECT online FROM FROM WHERE online = $ Y" ;;
$ result = mysql_query($ sql,$ db); [/ PHP]
并称之为:
[PHP]<?php printf($ onlin E); ?> [/ PHP]
有一种特殊的方法来计算Y的值并跳过N
Ok, i am adding a table to the users table called online.
So when they login it will set (online) to Y and when they logout it will be set to N.
But what should it be created as, it will be updated often and i need to beable to read it from my php in the Members online section?
varchar(50), int(10), or Text...
And how do i call the results of it not in text Y but in total.
Like lets say there is a total 50 users=Y so it will PRINT 50.
I have tried several times but i keep getting eather 9empty) or (Y) Not the total as 8
I created the table like this:
[PHP]ALTER TABLE users
ADD online varchar(1) default ''N'';[/PHP]
And call it like this:
[PHP]$sql="SELECT online FROM users WHERE online = $Y";
$result = mysql_query($sql ,$db);[/PHP]
And call it:
[PHP]<?php printf($online); ?>[/PHP]
is there a special way to count the value of Y and skip N
推荐答案
sql =" SELECT online FROM users WHERE online =
sql="SELECT online FROM users WHERE online =
Y" ;;
Y";
result = mysql_query(
result = mysql_query(
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