Question

我有一个具体的问题。我想获得账单数据。数据将用于图表，因此需要总和。以下是解释：

Sub1获得本月支付的账单金额。

Sub2获得本月未支付的账单金额。

Sub3获得了本月计算的账单金额总和，但实际上并未支付。如果其他公司没有支付5年或更长时间的账单而且该账单“支付”（法院程序在此之后），那就是这种情况。

问题是如果我没有sub1的记录，则sub2或sub3无法连接，我想在任何情况下显示它们。如果sub2没有记录且sub3有，则不显示sub3，这也是问题所在。即使存在NULL值（例如sub1）和NOT NULL for sub 2和sub3，我如何显示一行中的所有记录，反之亦然。任何帮助将不胜感激，因为我几乎整个下午都在使用这个查询解决方案。：/

SELECT sub1.y,
       sub1.m,
       sub1.val1,
       sub2.val2,
       sub2.y,
       sub2.m,
       sub3.y,
       sub3.m,
       sub3.val3
FROM
  (SELECT YEAR(curdate()) AS y,
          MONTH(curdate()) AS m,
          SUM(totwdisc) AS val1
   FROM pro_partial_inv
   WHERE paidd IS NOT NULL
     AND bringbar=0
     AND YEAR(curdate())=YEAR(paidd)
     AND MONTH(curdate())=MONTH(paidd)
   GROUP BY y,
            m) sub1
INNER JOIN
  (SELECT YEAR(curdate()) AS y,
          MONTH(curdate()) AS m,
          SUM(netto) AS val2
   FROM pro_partial_inv
   WHERE paidd IS NULL
     AND bringbar=0
     AND YEAR(curdate())=YEAR(pdate2)
     AND MONTH(curdate())=MONTH(pdate2)
   GROUP BY y,
            m) sub2
LEFT JOIN
  (SELECT YEAR(curdate()) AS y,
          MONTH(curdate()) AS m,
          SUM(totwdisc) AS val3
   FROM pro_partial_inv
   WHERE paidd IS NOT NULL
     AND bringbar=1
     AND YEAR(curdate())=YEAR(pdate2)
     AND MONTH(curdate())=MONTH(pdate2)
   GROUP BY y,
            m) sub3 
ON sub1.y=sub2.y
AND sub1.m=sub2.m
AND sub2.y=sub3.y
AND sub2.m=sub3.m

Answer 1

有一个初始选择只是获得年/月，然后离开加入其他人反对它？

SELECT sub0.y,
       sub0.m,
       sub1.val1,
       sub2.val2,
       sub2.y,
       sub2.m,
       sub3.y,
       sub3.m,
       sub3.val3
FROM  (SELECT YEAR(curdate()) AS y, MONTH(curdate()) AS m) Sub0
LEFT JOIN(SELECT YEAR(curdate()) AS y,
          MONTH(curdate()) AS m,
          SUM(totwdisc) AS val1
   FROM pro_partial_inv
   WHERE paidd IS NOT NULL
     AND bringbar=0
     AND YEAR(curdate())=YEAR(paidd)
     AND MONTH(curdate())=MONTH(paidd)
   GROUP BY y,
            m) sub1
ON sub0.y=sub1.y
AND sub0.m=sub1.m
LEFT JOIN
  (SELECT YEAR(curdate()) AS y,
          MONTH(curdate()) AS m,
          SUM(netto) AS val2
   FROM pro_partial_inv
   WHERE paidd IS NULL
     AND bringbar=0
     AND YEAR(curdate())=YEAR(pdate2)
     AND MONTH(curdate())=MONTH(pdate2)
   GROUP BY y,
            m) sub2
ON sub0.y=sub2.y
AND sub0.m=sub2.m
LEFT JOIN
  (SELECT YEAR(curdate()) AS y,
          MONTH(curdate()) AS m,
          SUM(totwdisc) AS val3
   FROM pro_partial_inv
   WHERE paidd IS NOT NULL
     AND bringbar=1
     AND YEAR(curdate())=YEAR(pdate2)
     AND MONTH(curdate())=MONTH(pdate2)
   GROUP BY y,
            m) sub3 
ON sub0.y=sub3.y
AND sub0.m=sub3.m

或简化触摸： -

SELECT sub0.y,
       sub0.m,
       sub1.val1,
       sub2.val2,
       sub3.val3
FROM  (SELECT YEAR(curdate()) AS y, MONTH(curdate()) AS m) Sub0
LEFT JOIN(SELECT YEAR(paidd) AS y,
          MONTH(paidd) AS m,
          SUM(totwdisc) AS val1
   FROM pro_partial_inv
   WHERE paidd IS NOT NULL
     AND bringbar=0
   GROUP BY y,
            m) sub1
ON sub0.y=sub1.y
AND sub0.m=sub1.m
LEFT JOIN
  (SELECT YEAR(pdate2) AS y,
          MONTH(pdate2) AS m,
          SUM(netto) AS val2
   FROM pro_partial_inv
   WHERE paidd IS NULL
     AND bringbar=0
   GROUP BY y,
            m) sub2
ON sub0.y=sub2.y
AND sub0.m=sub2.m
LEFT JOIN
  (SELECT YEAR(pdate2) AS y,
          MONTH(pdate2) AS m,
          SUM(totwdisc) AS val3
   FROM pro_partial_inv
   WHERE paidd IS NOT NULL
     AND bringbar=1
   GROUP BY y,
            m) sub3 
ON sub0.y=sub3.y
AND sub0.m=sub3.m

甚至更简化

SELECT YEAR(curdate()) AS y, 
        MONTH(curdate()) AS m,
        SUM(IF(paidd IS NOT NULL AND bringbar=0, totwdisc, 0)),
        sub2.val2,
        sub2.val3
FROM  pro_partial_inv
LEFT JOIN
  (SELECT YEAR(pdate2) AS y,
          MONTH(pdate2) AS m,   
          SUM(IF(paidd IS NULL AND bringbar=0, netto, 0)) AS val2,
          SUM(IF(paidd IS NOT NULL AND bringbar=0, totwdisc, 0)) AS val3
   FROM pro_partial_inv
   GROUP BY y,
            m) sub2
ON sub0.y=sub2.y
AND sub0.m=sub2.m
GROUP BY y, m

Answer 2

您不需要为此使用联接。您可以运行3个单独的查询来获得所需的结果。

SELECT 
    (SELECT SUM(netto) FROM ...) as sub1,
    (...) as sub2,
    (...) as sub3

您只需在每个子查询中添加所有适当的WHERE子句。

Answer 3

MySQL中没有FULL JOIN，因此您需要在一组适当的年月对上保持联接。在你的情况下，只有一个，所以：

select ...
from (select year(curdate()) as y, month(curdate()) as m) cur
left join (...) as sub1 on cur.y = sub1.y and cur.m = sub1.m
left join (...) as sub2 on cur.y = sub2.y and cur.m = sub2.m
left join (...) as sub3 on cur.y = sub3.y and cur.m = sub3.m;

JOIN为NULL - mysql

3 个答案: