关于php的判断语句

	$pubdate = $dsql->GetOne("select pubdate from article where writer ='".$writer."' order by id desc limit 1");	if((time()-$pubdate['pubdate'])<(3600*24)){		ShowMsg("对不起，一个会员每天只能发一篇文章！","-1","0",5000);		exit;	}

以上判断一个会员在一%本文@来源gao@!dai!ma.com搞$$代^@码!网搞代gaodaima码天内只能提交一篇文档，想修改成可以提交3篇，应该怎么做？

回复讨论(解决方案)

$pubdate = $dsql->GetOne(“select pubdate,writer,count(pubdate) as total from article where writer ='”.$writer.”‘ and pubdate<".strtotime("-1 day")." limit 3");
if($pubdate[‘total’])==3){
ShowMsg(“对不起，一个会员每天只能发一篇文章！”,”-1″,”0″,5000);
exit;
}

mysql查询一天内提交的文章数,如果>=3不能提交

$pubdate = $dsql->GetOne(“select pubdate,writer,count(pubdate) as total from article where writer ='”.$writer.”‘ and pubdate<".strtotime("-1 day")." limit 3");
if($pubdate[‘total’])==3){
ShowMsg(“对不起，一个会员每天只能发一篇文章！”,”-1″,”0″,5000);
exit;
}

mysql查询一天内提交的文章数,如果>=3不能提交

mysql跑不起来。跪求检查一下。
另外：if($pubdate[‘total’])==3)应该是if(($pubdate[‘total’])==3)这样吧？
数据库字段：
数据库名称（article）
会员名称（writer）
发布时间（pubdate）

原代码改为

$n = 1; //你指定的篇数$pubdate = $dsql->GetOne("select count(*) as cnt from article where FROM_UNIXTIME(pubdate,'%Y%m%d')=DATE_FORMAT(now(),'%Y%m%d') and writer ='".$writer."' order by id desc limit 1");if($pubdate['pubdate'] > $n){    ShowMsg("对不起，一个会员每天只能发 $n 篇文章！","-1","0",5000);    exit;}

还是版主给力，学习了

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