Question

我有以下数据的两张表：

1）Trans Table

TICKET  STATUS
56       0
56       0

2）Resa Table

R_TICKET    R_STATUS
56            0

如果表格中的票证数与之匹配，那么我想要这样的结果，然后打印0其他1。

我已经尝试了以下查询，但它不起作用

select case when count(a.ticket) = count(b.r_ticket) 
            then 0 
            else 1 
       end as result 
from trans a, resa b

它会显示0，但它应为1，因为56中的2点数在trans中为1而在resa中为TICKET Count(Ticket) r_Ticket Count(r_Ticket) result 56 2 56 1 1 58 1 58 1 0 }}

有些问题有所改变：我想要这样的结果：

expression: term opt_add
opt_add: '+' term opt_add
       | /* empty */

term:  factor opt_mul
opt_mul: '*' factor opt_mul
       | /* emtpty */

factor: number
      | '(' expression ')

Answer 1

请勿在{{1}}子句中使用逗号。无论如何，在这种情况下这是错误的。。。如果其中一个表为空，那么您将在聚合之前获得一个空的中间结果集。

相反，使用子查询来获取计数：

consec = []
    for key, group in groupby(enumerate(a), lambda i: i[0] - i[1]):
        consec.append(list(map(itemgetter(1), group)))
    consecHead = consec[0]
    consecTail = consec[-1]
    if 1 in consecHead:
        consecHeadcheck = consecHead
    else:
        consecHeadcheck = []
    if 21 in consecTail:
        consecTailcheck = consecTail
    else:
        consecTailcheck = []
    b_prior = [x for x in polPos if x not in consecTailcheck]
    b = [x for x in b_prior if x not in consecHeadcheck]
    print(b)
    c = []
    for i in range(len(b)): 
                   if int(int(b[i+1]) - int(b[i])) !=1:
                         c.append(int(b[i+1]) - int(b[i])
    print(c)

Answer 2

SELECT COALESCE(t1.ticket, t2.ticket),
       CASE WHEN t1.count = t2.count THEN 0 ELSE 1 END AS match
FROM
(
    SELECT TICKET AS ticket, COUNT(*) AS tcount
    FROM Trans
    GROUP BY TICKET
) t1
FULL OUTER JOIN
(
    SELECT R_TICKET AS ticket, COUNT(*) AS tcount
    FROM Resa
    GROUP BY R_TICKET
) t2
    ON t1.ticket = t2.ticket

来自不同表

2 个答案: