回答編集履歴
7
推敲
answer
CHANGED
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@@ -75,7 +75,7 @@
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75
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group by rrh.rh_pk, rrd.c_cd, rrh.o_no, rrd.item_no
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) rrd_sum
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on rd_sum.rh_pk = rrd_sum.rh_pk
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-
and rd_sum.c_cd = rrd_sum.c_cd
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+
and rd_sum.c_cd = rrd_sum.c_cd
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and rd_sum.o_no = rrd_sum.o_no
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and rd_sum.item_no = rrd_sum.item_no
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where rd_sum.c_cd='01' and rd_sum.o_no='A100'
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6
推敲
answer
CHANGED
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@@ -31,26 +31,26 @@
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31
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from (
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select rd.rh_pk, rd.rd_pk, rd.item_no, sum(sd.quantity) as quantity
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from tbl1 rd
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-
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inner join tbl2 sd
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-
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on rd.rh_pk = sd.rh_pk
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-
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and rd.rd_pk = sd.rd_pk
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-
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and rd.c_cd = sd.c_cd
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where rd.c_cd = '01'
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and rd.o_no = 'A100'
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group by rd.rh_pk, rd.rd_pk, rd.item_no
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) rd_sum
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inner join (
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select rrh.rh_pk, rrd.item_no, sum(rrd.diff) as diff
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-
from
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+
from tbl3 rrh
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-
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inner join tbl4 rrd
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-
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on rrh.r_pk = rrd.r_pk
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-
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and rrh.c_cd = rrd.c_cd
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where rrd.c_cd = '01'
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and rrh.o_no = 'A100'
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group by rrh.rh_pk, rrd.item_no
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) rrd_sum
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on rd_sum.rh_pk = rrd_sum.rh_pk
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-
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+
and rd_sum.item_no = rrd_sum.item_no
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order by item_no
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55
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```
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追記２
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@@ -61,18 +61,17 @@
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from (
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select rd.rh_pk, rd.rd_pk, rd.c_cd, rd.o_no, rd.item_no, sum(sd.quantity) as quantity
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from tbl1 rd
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-
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+
inner join tbl2 sd
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-
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on rd.rh_pk = sd.rh_pk
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-
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+
and rd.rd_pk = sd.rd_pk
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-
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+
and rd.c_cd = sd.c_cd
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group by rd.rh_pk, rd.rd_pk, rd.c_cd, rd.o_no, rd.item_no
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-
) rd_sum
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-
inner join (
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+
) rd_sum inner join (
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select rrh.rh_pk, rrd.c_cd, rrh.o_no, rrd.item_no, sum(rrd.diff) as diff
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-
from
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+
from tbl3 rrh
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-
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inner join tbl4 rrd
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-
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on rrh.r_pk = rrd.r_pk
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-
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+
and rrh.c_cd = rrd.c_cd
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group by rrh.rh_pk, rrd.c_cd, rrh.o_no, rrd.item_no
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) rrd_sum
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on rd_sum.rh_pk = rrd_sum.rh_pk
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5
追記
answer
CHANGED
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@@ -52,4 +52,33 @@
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on rd_sum.rh_pk = rrd_sum.rh_pk
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and rd_sum.item_no = rrd_sum.item_no
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order by item_no
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+
```
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+
追記２
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+
---
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58
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+
データ件数が少ないならさほどレスポンスは落ちないでしょうから、条件部分を収斂したパターンも記述しておきます。
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59
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+
```SQL
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60
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+
select rd_sum.rh_pk, rd_sum.rd_pk, rd_sum.item_no, rd_sum.quantity, rrd_sum.diff
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61
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+
from (
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+
select rd.rh_pk, rd.rd_pk, rd.c_cd, rd.o_no, rd.item_no, sum(sd.quantity) as quantity
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63
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+
from tbl1 rd
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64
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+
inner join tbl2 sd
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65
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on rd.rh_pk = sd.rh_pk
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66
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+
and rd.rd_pk = sd.rd_pk
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67
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+
and rd.c_cd = sd.c_cd
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68
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+
group by rd.rh_pk, rd.rd_pk, rd.c_cd, rd.o_no, rd.item_no
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69
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+
) rd_sum
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+
inner join (
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71
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select rrh.rh_pk, rrd.c_cd, rrh.o_no, rrd.item_no, sum(rrd.diff) as diff
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72
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+
from tbl3 rrh
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73
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+
inner join tbl4 rrd
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74
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on rrh.r_pk = rrd.r_pk
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75
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+
and rrh.c_cd = rrd.c_cd
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76
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+
group by rrh.rh_pk, rrd.c_cd, rrh.o_no, rrd.item_no
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77
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+
) rrd_sum
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78
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+
on rd_sum.rh_pk = rrd_sum.rh_pk
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79
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+
and rd_sum.c_cd = rrd_sum.c_cd a
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80
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+
and rd_sum.o_no = rrd_sum.o_no
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81
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+
and rd_sum.item_no = rrd_sum.item_no
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82
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+
where rd_sum.c_cd='01' and rd_sum.o_no='A100'
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83
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+
order by item_no
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55
84
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```
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4
修正
answer
CHANGED
|
@@ -36,7 +36,7 @@
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36
36
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and rd.rd_pk = sd.rd_pk
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37
37
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and rd.c_cd = sd.c_cd
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38
38
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where rd.c_cd = '01'
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39
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-
and rd.o_no = '
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39
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+
and rd.o_no = 'A100'
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40
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group by rd.rh_pk, rd.rd_pk, rd.item_no
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41
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) rd_sum
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42
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inner join (
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@@ -46,7 +46,7 @@
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46
46
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on rrh.r_pk = rrd.r_pk
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47
47
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and rrh.c_cd = rrd.c_cd
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48
48
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where rrd.c_cd = '01'
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49
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-
and rrh.o_no = '
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49
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+
and rrh.o_no = 'A100'
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50
50
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group by rrh.rh_pk, rrd.item_no
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51
51
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) rrd_sum
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52
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on rd_sum.rh_pk = rrd_sum.rh_pk
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3
追記
answer
CHANGED
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@@ -24,6 +24,7 @@
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24
24
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追記
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25
25
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---
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26
26
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質問のSQLでは動作しないので、推測での記述ですけど、以下でquantityとdiffは取り出せていると思います。
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27
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+
こういった場合、そのぞれの集計結果をどの項目で結びつけるかを明らかにすると迷わなくてすみます。
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27
28
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※diffの複数行については良く分からないので、質問にデータイメージを追記して下さい。
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28
29
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```SQL
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select rd_sum.rh_pk, rd_sum.rd_pk, rd_sum.item_no, rd_sum.quantity, rrd_sum.diff
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2
追記
answer
CHANGED
|
@@ -19,4 +19,36 @@
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19
19
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> ※QuantityとDiffはレコード増える仕様
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20
20
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> この例ではDiffは１つですが複数行ある場合を仮定して頂きたいです。
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21
21
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22
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-
sum(diff1), sum(diff2)のように増えるとしても、集計の単位が同じなら、sum()の項目を増やせば良いだけ
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22
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+
sum(diff1), sum(diff2)のように増えるとしても、集計の単位が同じなら、sum()の項目を増やせば良いだけ
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23
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+
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24
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+
追記
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25
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+
---
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26
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+
質問のSQLでは動作しないので、推測での記述ですけど、以下でquantityとdiffは取り出せていると思います。
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27
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+
※diffの複数行については良く分からないので、質問にデータイメージを追記して下さい。
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28
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+
```SQL
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29
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+
select rd_sum.rh_pk, rd_sum.rd_pk, rd_sum.item_no, rd_sum.quantity, rrd_sum.diff
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30
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+
from (
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31
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+
select rd.rh_pk, rd.rd_pk, rd.item_no, sum(sd.quantity) as quantity
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32
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+
from tbl1 rd
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33
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+
inner join tbl2 sd
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34
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+
on rd.rh_pk = sd.rh_pk
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35
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+
and rd.rd_pk = sd.rd_pk
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36
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+
and rd.c_cd = sd.c_cd
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37
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+
where rd.c_cd = '01'
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38
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+
and rd.o_no = 'a100'
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39
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+
group by rd.rh_pk, rd.rd_pk, rd.item_no
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40
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+
) rd_sum
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41
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+
inner join (
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42
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+
select rrh.rh_pk, rrd.item_no, sum(rrd.diff) as diff
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43
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+
from tbl3 rrh
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44
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+
inner join tbl4 rrd
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45
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+
on rrh.r_pk = rrd.r_pk
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46
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+
and rrh.c_cd = rrd.c_cd
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47
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+
where rrd.c_cd = '01'
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48
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+
and rrh.o_no = 'a100'
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|
49
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+
group by rrh.rh_pk, rrd.item_no
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50
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+
) rrd_sum
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|
51
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+
on rd_sum.rh_pk = rrd_sum.rh_pk
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52
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+
and rd_sum.item_no = rrd_sum.item_no
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53
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+
order by item_no
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54
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+
```
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1
追記
answer
CHANGED
|
@@ -15,4 +15,8 @@
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15
15
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WHERE RD.C_Cd = '01'
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16
16
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AND RD.O_No = 'A100'
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17
17
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GROUP BY RD.RH_PK, RD.RD_PK, RD.Item_no
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18
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-
```
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18
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+
```
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19
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+
> ※QuantityとDiffはレコード増える仕様
|
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20
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+
> この例ではDiffは１つですが複数行ある場合を仮定して頂きたいです。
|
|
21
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+
|
|
22
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+
sum(diff1), sum(diff2)のように増えるとしても、集計の単位が同じなら、sum()の項目を増やせば良いだけ
|