在5个月内应仅将3个用户插入数据库 [英] only 3 users shoud be inserted into the database within a timespan of 5 months

问题描述

进行一些我先前删除的更改后，重新发布该问题. 我尽力了，但我不知道该怎么实现.

Reposting the question after making some changes which i deleted earlier. I tried hard but I could not figure out how to achieve it.

我有一张桌子fw_invitations

I have a table fw_invitations

ID  OFFICEID  Consultationdate
------------------------------
1     1       16-06-2013
2     1       16-06-2013
3     1       16-06-2013
4     1       17-08-2014
5     1       17-08-2014

$consultationdate包含用户输入的日期.

$consultationdate contains the date entered by the user.

$invcount="Select count(*) as   noofinv,
           Datediff('$consultationdate',max(consultationdate)) as datediffinv 
           from fw_invitations where OFFICEID = 1";

$db->setQuery($invcount);
$invcounts=$db->loadAssoclist();
$noofinv=$invcounts[0]['noofinv'];
$datediffinv=abs($invcounts[0]['datediffinv']);

if($noofinv >3 && $datediffinv <150)
{
  $error_message="Het maximale ";
}

我也在尝试这样做，但是下面的查询总是导致0

I am also trying this but the below query always resulting in 0

    $invcount="Select count(*) as noofinv from fw_invitations
    where consultationdate between 
    DATE_SUB($consultationdate,INTERVAL 150 DAY) and
    DATE_ADD($consultationdate,INTERVAL 150 DAY) from fw_invitations where OFFICEID = 1";

我想做的是，如果datediff超过150(5个月)，则表中插入的用户数不能超过3.

what I want to do is that not more than 3 user should be inserted into the table if the datediff is more than 150(5 months).

除非输入的时间间隔过长，否则上面的查询无法正常运行.由于我正在使用max(consultationdate)，因此上表的最大日期为2014年8月17日，因此无法验证用户再次输入日期16-06-2013的情况.

My above query was working perfectly unless I entered the date with a long interval. Since I am using max(consultationdate) so max date for the above table will be 17-08-2014, it will not validate the case in which a user again enters the date 16-06-2013.

简而言之，我想做的是在任何情况下，在5个月的时间内都应仅将3个用户插入数据库.

推荐答案

在更正了它的魅力之后，我错过了$ consultationdate中的单引号，这就是为什么它每次产生0的原因.

i missed the single quotes in the $consultationdate thats ahy it was resulting 0 everytime,after correcting that its working like a charm .

 $invcount="Select count(*) as noofinv from fw_invitations
        where consultationdate between 
        DATE_SUB('$consultationdate',INTERVAL 150 DAY) and
        DATE_ADD('$consultationdate',INTERVAL 150 DAY) from fw_invitations where OFFICEID = 1";

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