我在MySQL中有一个查询。
完整查询
SELECT
tc.expense AS expense,
tc.tour_sub_code,
tc.login_id
FROM
tc_wallet tc
WHERE tc.login_id = 'vinod.kumbala'
AND tc.expense = 'Daily Allowance'
AND tc.delete_flag = 'F'
AND tc.status != 'reject'
结果
Expense Tour_sub_code login_id
DAILY ALLOWANCE MOS-EUROPE100119 vinod.kumbala
DAILY ALLOWANCE Test vinod.kumbala
最初,我是从tc_wallet
表中获取数据。
现在,我的要求是我需要找到特定tour_sub_code
的出勤总数。可以从attendance_master
表中找到特定于tour_sub_code
的 的出席人数。
因此,我包含了一个嵌套的选择查询
嵌套选择
(SELECT
COUNT(*)
FROM
(SELECT
*
FROM
`attendance_master`
WHERE `delete_flag` = 'F'
AND login_id = 'vinod.kumbala'
AND `tour_sub_code` = tc.`tour_sub_code`
GROUP BY `device_date`) t1) AS newNoOfdays
完整查询
SELECT
tc.expense AS expense,
tc.tour_sub_code,
tc.login_id,
(SELECT
COUNT(*)
FROM
(SELECT
*
FROM
`attendance_master`
WHERE `delete_flag` = 'F'
AND login_id = 'vinod.kumbala'
AND `tour_sub_code` = tc.`tour_sub_code`
GROUP BY `device_date`) t1) AS newNoOfdays
FROM
tc_wallet tc
WHERE tc.login_id = 'vinod.kumbala'
AND tc.expense = 'Daily Allowance'
AND tc.delete_flag = 'F'
AND tc.status != 'reject'
现在此查询给我错误为
“ where子句”中的未知列“ tc.tour_sub_code”
预期结果
Expense Tour_sub_code login_id Count
DAILY ALLOWANCE MOS-EUROPE100119 vinod.kumbala 20
DAILY ALLOWANCE Test vinod.kumbala 44
我能知道我要怎么做吗?
另外还有其他方法可以通过使用JOINS获得结果吗?
答案 0 :(得分:1)
尝试一下---
SELECT
tc.expense AS expense,
tc.tour_sub_code,
tc.login_id,
(SELECT
COUNT(am.*)
FROM `attendance_master` as am
WHERE am.delete_flag = 'F'
AND am.login_id = 'vinod.kumbala'
AND am.tour_sub_code = tc.tour_sub_code
GROUP BY am.device_date) AS count
FROM
tc_wallet tc
WHERE tc.login_id = 'vinod.kumbala'
AND tc.expense = 'Daily Allowance'
AND tc.delete_flag = 'F'
AND tc.status != 'reject'
答案 1 :(得分:1)
在最里面的子查询中,您引用的是最外面的查询中的一列。这对于FROM子句(派生表)中的子查询是不允许的。但是-您不需要该子查询。您需要的是COUNT(DISTINCT device_date)
重写
(SELECT
COUNT(*)
FROM
(SELECT
*
FROM
`attendance_master`
WHERE `delete_flag` = 'F'
AND login_id = 'vinod.kumbala'
AND `tour_sub_code` = tc.`tour_sub_code`
GROUP BY `device_date`) t1) AS newNoOfdays
到
(SELECT
COUNT(DISTINCT device_date)
FROM `attendance_master`
WHERE `delete_flag` = 'F'
AND login_id = 'vinod.kumbala'
AND `tour_sub_code` = tc.`tour_sub_code`
) AS newNoOfdays
您还可以将完整查询重写为LEFT JOIN查询:
SELECT
tc.expense AS expense,
tc.tour_sub_code,
tc.login_id,
COUNT(DISTINCT device_date) AS newNoOfdays
FROM tc_wallet tc
LEFT JOIN attendance_master am
ON am.tour_sub_code = tc.tour_sub_code
AND am.delete_flag = 'F'
AND am.login_id = 'vinod.kumbala'
WHERE tc.login_id = 'vinod.kumbala'
AND tc.expense = 'Daily Allowance'
AND tc.delete_flag = 'F'
AND tc.status != 'reject'
答案 2 :(得分:0)
尝试一下:
SELECT
tc.expense AS expense,
tc.tour_sub_code,
tc.login_id,
count(*) as count
FROM
tc_wallet tc
INNER JOIN
attendance_master am
ON
tc.tour_sub_code = am.tour_sub_code AND
tc.login_id = am.login_id
WHERE tc.login_id = 'vinod.kumbala'
AND tc.expense = 'Daily Allowance'
AND tc.delete_flag = 'F'
AND tc.status != 'reject'
GROUP BY
tc.expense AS expense,
tc.tour_sub_code,
tc.login_id
答案 3 :(得分:0)
您能尝试一下吗?
select tc.expense AS expense,
tc.tour_sub_code,
tc.login_id,
newNoOfdays.countVal
from tc_wallet tc
inner join
( select tour_sub_code, count('A') 'countval'
from attendance_master ascs (nolock)
group by tour_sub_code) as newNoOfdays
on newNoOfdays.tour_sub_code = tc.tour_sub_code
where tc.login_id = 'vinod.kumbala'
AND tc.expense = 'Daily Allowance'
AND tc.delete_flag = 'F'
AND tc.status != 'reject'