回答編集履歴
3
typo
test
CHANGED
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@@ -94,7 +94,7 @@
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94
94
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95
95
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case
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96
96
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97
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-
when (select
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97
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+
when (select @d:=date_format(d,'%e') as dummy) between 1 and 7 then '1週目'
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98
98
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99
99
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when @d between 8 and 15 then '2週目'
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100
100
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2
訂正
test
CHANGED
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@@ -81,3 +81,43 @@
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81
81
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82
82
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83
83
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```
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84
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+
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85
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+
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86
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+
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87
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+
# 一部題意にあわせて修正
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88
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+
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89
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+
一部推奨されない表現が含まれていますが、たぶんこれで大丈夫です
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90
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+
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91
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+
```SQL
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92
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+
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93
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+
select
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94
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+
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95
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+
case
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96
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+
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97
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+
when (select (select @d:=date_format(d,'%e') as dummy)) between 1 and 7 then '1週目'
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98
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+
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99
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+
when @d between 8 and 15 then '2週目'
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100
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+
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101
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+
when @d between 15 and 21 then '3週目'
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102
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+
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103
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+
when @d between 22 and 28 then '4週目'
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104
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+
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105
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else '5週目'
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106
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+
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107
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end as d2
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108
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+
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109
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+
,d
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110
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+
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111
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+
,sum(tenpo)
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112
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+
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113
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+
,sum(tuhan)
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114
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+
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115
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from tbl
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116
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+
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117
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+
where d between '2017-07-01' and '2017-07-31'
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118
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+
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119
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+
group by d2 asc,d asc with rollup
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120
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+
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121
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+
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122
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+
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123
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```
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1
追記
test
CHANGED
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@@ -7,3 +7,77 @@
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7
7
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- 7/18はデータ0,0ですがこの場合もデータは存在するのでしょうか？
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8
8
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9
9
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それともデータはないけど7/18のカレンダーを元に0,0を表示するのでしょうか？
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10
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+
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11
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+
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12
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+
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13
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+
# sample
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14
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+
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15
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+
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16
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+
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17
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+
```SQL
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18
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+
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19
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+
create table tbl(d date unique,tenpo int,tuhan int);
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20
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+
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21
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insert into tbl value
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22
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+
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23
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+
('2017-07-01',120800,109000),
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24
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+
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25
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+
('2017-07-02',95896,371140),
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26
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+
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27
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+
('2017-07-03',44200,24000),
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28
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+
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29
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+
('2017-07-04',51600,14000),
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30
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+
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31
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+
('2017-07-05',77634,0),
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32
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+
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33
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+
('2017-07-06',87000,18303),
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34
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+
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35
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+
('2017-07-07',159240,25300),
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36
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+
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37
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('2017-07-08',0,74000),
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38
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+
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39
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+
('2017-07-09',0,79705);
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40
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+
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41
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+
```
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42
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+
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43
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+
集計単位は日付から1引いた数を7で割ったあまりを日付から引いた日
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44
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+
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45
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+
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46
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+
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47
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+
```SQL
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48
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+
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49
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+
select d
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50
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+
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51
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+
,d - interval (date_format(d,'%e')-1)%7 day as d2
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52
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+
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53
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+
,sum(tenpo)
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54
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+
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55
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+
,sum(tuhan)
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56
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+
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57
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from tbl
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58
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+
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59
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group by d
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60
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+
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61
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```
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62
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+
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63
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+
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64
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+
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65
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+
よってこれを利用して集計するとこう
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66
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+
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67
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+
```ここに言語を入力
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68
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+
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69
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+
select d
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70
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+
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71
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+
,sum(tenpo)
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72
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+
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73
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+
,sum(tuhan)
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74
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+
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75
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+
from tbl
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76
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+
|
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77
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+
group by d - interval (date_format(d,'%e')-1)%7 day
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78
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+
|
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79
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+
,d with rollup
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80
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+
|
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81
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+
|
|
82
|
+
|
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83
|
+
```
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