Group by Laravel 5.2

-4

Group by不工作，可能是因为我在select选项中使用了数学表达式。Group by Laravel 5.2

select FLOOR(SUM(DATEDIFF(rd.checkin_date,rd.created_at))/COUNT(DATEDIFF(rd.checkin_date,rd.created_at))) AS names, 
COUNT(rd.id) as totbooking, 
SUM(DATEDIFF(rd.checkout_date,rd.checkin_date)) AS nights, 
round(SUM(CASE WHEN h.currency_id=151 THEN rd.total_cents WHEN h.currency_id!=151 THEN rd.total_cents/cur.exchange_rate ELSE 0 END)/100) AS revenue, 
round((round(SUM(CASE WHEN h.currency_id=151 THEN rd.total_cents WHEN h.currency_id!=151 THEN rd.total_cents/cur.exchange_rate ELSE 0 END)/100))/(SUM(DATEDIFF(rd.checkout_date,rd.checkin_date)))) as adr, FLOOR((FLOOR(SUM(CASE WHEN h.currency_id=151 THEN rd.total_cents WHEN h.currency_id!=151 THEN rd.total_cents/cur.exchange_rate ELSE 0 END)/100))/COUNT(rd.id)) as atp, FLOOR(SUM(DATEDIFF(rd.checkin_date,rd.created_at))/COUNT(DATEDIFF(rd.checkin_date,rd.created_at))) as leadtime, 
SUM(DATEDIFF(rd.checkout_date,rd.checkin_date))/(COUNT(rd.id)) as alos, (SUM(DATEDIFF(rd.checkout_date,rd.checkin_date))/(h.total_rooms * 31))*100 as share from `reservation_details` as `rd` 
left join `hotels` as `h` on `rd`.`hotel_id` = `h`.`id` 
left join `currencies` as `cur` on `cur`.`id` = `h`.`currency_id` 
left join `chains` as `c` on `c`.`id` = `h`.`chain_id` 
where `rd`.`status` >= 50 and date(rd.created_at) >= "2017-04-30" and date(rd.created_at) <= "2017-05-30" GROUP BY names

来源

2017-05-30 Rakesh Tripathi

你的格式可以吗？无法阅读 – nacho

没有显示错误？那我称之为工作。 – Chay22

@nacho请立即检查 –

为什么不要你试试这个办法？如果你让我们看到一些数据会有帮助

select names,sum(totbooking),sum(nights),sum(revenue),sum(adr),leadtime,sum(alos),sum(share) 
from (
    select FLOOR(SUM(DATEDIFF(rd.checkin_date,rd.created_at))/COUNT(DATEDIFF(rd.checkin_date,rd.created_at))) AS names, COUNT(rd.id) as totbooking, 
    SUM(DATEDIFF(rd.checkout_date,rd.checkin_date)) AS nights,round(SUM(CASE WHEN h.currency_id=151 THEN rd.total_cents WHEN h.currency_id!=151 THEN rd.total_cents/cur.exchange_rate ELSE 0 END)/100) AS revenue, 
    round((round(SUM(CASE WHEN h.currency_id=151 THEN rd.total_cents WHEN h.currency_id!=151 THEN rd.total_cents/cur.exchange_rate ELSE 0 END)/100))/(SUM(DATEDIFF(rd.checkout_date,rd.checkin_date)))) as adr, FLOOR((FLOOR(SUM(CASE WHEN h.currency_id=151 THEN rd.total_cents WHEN h.currency_id!=151 THEN rd.total_cents/cur.exchange_rate ELSE 0 END)/100))/COUNT(rd.id)) as atp, FLOOR(SUM(DATEDIFF(rd.checkin_date,rd.created_at))/COUNT(DATEDIFF(rd.checkin_date,rd.created_at))) as leadtime, 
    SUM(DATEDIFF(rd.checkout_date,rd.checkin_date))/(COUNT(rd.id)) as alos, (SUM(DATEDIFF(rd.checkout_date,rd.checkin_date))/(h.total_rooms * 31))*100 as share 
    from `reservation_details` as `rd` 
    left join `hotels` as `h` on `rd`.`hotel_id` = `h`.`id` 
    left join `currencies` as `cur` on `cur`.`id` = `h`.`currency_id` 
    left join `chains` as `c` on `c`.`id` = `h`.`chain_id` 
    where `rd`.`status` >= 50 and date(rd.created_at) >= "2017-04-30" and date(rd.created_at) <= "2017-05-30") 
group by names

来源

2017-05-31 07:18:00 nacho

它返回单行，但我想要列表@nacho –

@Rakesh Tripathi如果您上传一些带有数据的表以查看您需要的内容，这将有很大帮助 – nacho

您正在使用的别名，所以GROUP BY子句中你应该使用它太我supose names属于hotels，所以你需要做这种方式

GROUP BY h.names

来源

2017-05-31 07:02:40 nacho

他的“名字”的意思是'FLOOR（SUM（DATEDIFF（rd.checkin_date，rd.created_at））/ COUNT（DATEDIFF（rd.checkin_date，rd.created_at）））AS名称' –

对不起，我没有看到它 – nacho

正确@HoàngĐăng –

Group by Laravel 5.2

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