小编典典

Sql 不同的行组

javascript

在sql中,我想获得不同的行集:特征和值的相同组只有一次:

列特性的范围可以从 1 到 10

Table :

Name Characteristic Value
Mary eyes Blu
Mary hair blonde
Mary Sex Female
Jhon eyes Black
Jhon Hair Black
Jhon Sex Male
Jhon Nation Franch
Bill eyes Blu
Bill Hair Blond
Bill Sex Male
Will eyes Green
Will Hair Blond
Will Sex Male
Will Nation Spain
Lilly eyes Blu
Lilly Hair Blonde
Lilly Sex Female
mark eyes Black
mark Hair Black
mark Sex Male
mark Nation Franch
Anna eyes Blu
Anna Hair Blonde
Anna Sex Female
Antonio eyes Black
Antonio Hair Black
Antonio Sex Male
Antonio Nation Franch

The result that i want to achieve :

Group Characteristic Value
1 eyes Blu
1 Hair Blonde
1 Sex Female
2 eyes Black
2 Hair Black
2 Sex Male
2 Nation Franch
3 eyes Blu
3 Hair Blond
3 Sex Male
4 eyes Green
4 Hair Blode
4 Sex Male
4 Nation Spain

and finally if it’s possible :

Name Characteristic Value Group
Mary eyes Blu 1
Mary Hair Blonde 1
Mary Sex Female 1
Jhon eyes Black 2
Jhon Hair Black 2
Jhon Sex Male 2
Jhon Nation Franch 2
Bill eyes Blu 3
Bill Hair Blond 3
Bill Sex Male 3
Will eyes Green 4
Will Hair Blond 4
Will Sex Male 4
Will Nation Spain 4
Lilly eyes Blu 1
Lilly Hair Blonde 1
Lilly Sex Female 1
mark eyes Black 2
mark Hair Black 2
mark Sex Male 2
mark Nation Franch 2
Anna eyes Blu 1
Anna Hair Blonde 1
Anna Sex Female 1
Antonio eyes Black 2
Antonio Hair Black 2
Antonio Sex Male 2
Antonio Nation Franch 2

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2022-07-21

共1个答案

小编典典

您可以使用STRING_AGG将所有特征连接在一起,然后使用ROW_NUMBERDENSE_RANK计算它们。然后你将它重新加入到基表中。

对于您的第一个查询,您可以这样做。

SELECT
  Groups.GroupId,
  t.Characteristic,
  t.Value
FROM YourTable t
JOIN (
    SELECT
      t.Name,
      t.GroupDefinition,
      GroupId = DENSE_RANK() OVER (ORDER BY t.GroupDefinition),
      RowId = ROW_NUMBER() OVER (PARTITION BY t.GroupDefinition ORDER BY t.Name)
    FROM (
        SELECT
          t.Name,
          GroupDefinition = STRING_AGG(Characteristic + ':' + Value, '|')
              WITHIN GROUP (ORDER BY t.Characteristic)
        FROM YourTable t
        GROUP BY
          t.Name
    ) t
) Groups ON Groups.Name = t.Name
WHERE Groups.RowId = 1;

第二个查询如下。

SELECT
  Groups.GroupId,
  t.*
FROM YourTable t
JOIN (
    SELECT
      t.Name,
      t.GroupDefinition,
      GroupId = DENSE_RANK() OVER (ORDER BY t.GroupDefinition),
      RowId = ROW_NUMBER() OVER (PARTITION BY t.GroupDefinition ORDER BY t.Name)
    FROM (
        SELECT
          t.Name,
          GroupDefinition = STRING_AGG(Characteristic + ':' + Value, '|')
              WITHIN GROUP (ORDER BY t.Characteristic)
        FROM YourTable t
        GROUP BY
          t.Name
    ) t
) Groups ON Groups.Name = t.Name;

db<>小提琴

另一种选择是将其聚合为 JSON 或 XML 格式,然后在不重新加入基表的情况下将其切碎。

2022-07-21