将原始Sql语句按列分组，并使用相同名称的不同字符串

对于创建一个更复杂的sql语句来说，这是一个全新的概念，我正在尝试使用一个名称来进行分组，其中的名称可以有不同的形式。例如，名称可以是“凯恩，帕特里克”，“P.凯恩，帕特里克”，“凯恩，帕特里克”*

到目前为止，我在下面查询了大约7000个结果：

SELECT 
SUM(games_played) as games_played,
SUM(goals) as goals,
SUM(points) as points,
player_name
FROM player_stats
GROUP BY player_name;

生成json的示例

[
{games_played: 123, goals: 12, points: 40, player_name: "Kane, Patrick"},
{games_played: 123, goals: 12, points: 40, player_name: "P. Kane, Patrick"},
{games_played: 123, goals: 12, points: 40, player_name: "Kane, Patrick*"},
{games_played: 123, goals: 12, points: 40, player_name: "Nylander, Alex"},
{games_played: 123, goals: 12, points: 40, player_name: "A. Nylander, Alex"},
{games_played: 123, goals: 12, points: 40, player_name: "Nylander, Alex*"},
{games_played: 123, goals: 12, points: 40, player_name: "Lemieux, Mario"},
{games_played: 123, goals: 12, points: 40, player_name: "Gretzky, Wayne"},
]

问题是如何获得由相似参与者分组的每个列的总和，以便结果看起来更像如下所示：

[
{games_played: 369, goals: 36, points: 120, player_name: "Kane, Patrick"},
{games_played: 369, goals: 36, points: 120, player_name: "Nylander, Alex"},
{games_played: 123, goals: 12, points: 40, player_name: "Lemieux, Mario"},
{games_played: 123, goals: 12, points: 40, player_name: "Gretzky, Wayne"},
]

even better if i can get a knex.js query but i have no problem using a raw query here. DB is postgresSQL. 

thanks in advance

---

您需要做一些事情来将名称转换为一致的形式，可以是字符串替换、按句点拆分和只取第二个值、删除特殊字符等。没有任何人工智能可以“哦；p kane，Patrick显然与PatrickKane*相同-你必须自己进行操作。你甚至可以有一个表，它有两列，每个名称都有不同的变体，映射到一个一致的名称，然后对不同的名称进行连接，对一致的名称进行分组

我想我的第一步是整理数据：

UPDATE player_stats 
SET player_name = REPLACE(player_name, '*', '')

UPDATE player_stats 
SET player_name = SUBSTRING(player_name from 3)
WHERE player_name LIKE '_.%'

您可以停在这里，继续永远重新运行它，以不断删除表中的垃圾，随着更多变体的出现，添加更多规则

但是你应该为玩家制作一张新的桌子：

SELECT uuid_generate_v4() as player_id, player_name 
INTO players
FROM (SELECT distinct player_name FROM player_stats)x

ALTER TABLE players ADD PRIMARY KEY (player_id);

然后将列添加到stats以获取id：

ALTER TABLE player_stats ADD player_id UUID;

将数据复制到：

UPDATE player_stats d
SET d.player_id = s.player_id
FROM players s
WHERE s.player_name = d.player_name

设置外键：

ALTER TABLE player_stats
ADD CONSTRAINT fk_playersstats_playerid__players_playerid FOREIGN KEY player_id REFERENCES players(player_id)

最后转储名称列：

ALTER TABLE player_stats DROP player_name

---

然后首先修复用各种各样的垃圾填充表格的程序：）

你需要做一些事情来将名称转换为一致的形式，可以是字符串替换、按句点拆分和只取第二个值、删除特殊字符等。没有任何人工智能可以做到“哦；p kane，Patrick显然与PatrickKane*相同-你必须自己进行操作。你甚至可以有一个表，它有两列，每个名称都有不同的变体，映射到一个一致的名称，然后对不同的名称进行连接，对一致的名称进行分组

我想我的第一步是整理数据：

UPDATE player_stats 
SET player_name = REPLACE(player_name, '*', '')

UPDATE player_stats 
SET player_name = SUBSTRING(player_name from 3)
WHERE player_name LIKE '_.%'

您可以停在这里，继续永远重新运行它，以不断删除表中的垃圾，随着更多变体的出现，添加更多规则

但是你应该为玩家制作一张新的桌子：

SELECT uuid_generate_v4() as player_id, player_name 
INTO players
FROM (SELECT distinct player_name FROM player_stats)x

ALTER TABLE players ADD PRIMARY KEY (player_id);

然后将列添加到stats以获取id：

ALTER TABLE player_stats ADD player_id UUID;

将数据复制到：

UPDATE player_stats d
SET d.player_id = s.player_id
FROM players s
WHERE s.player_name = d.player_name

设置外键：

ALTER TABLE player_stats
ADD CONSTRAINT fk_playersstats_playerid__players_playerid FOREIGN KEY player_id REFERENCES players(player_id)

最后转储名称列：

ALTER TABLE player_stats DROP player_name

---

然后首先修复用各种垃圾填充表格的程序：）

如果必须这样做，您可以尝试以下方法：

SELECT 
SUM(games_played) as games_played,
SUM(goals) as goals,
SUM(points) as points,
player_name
FROM player_stats
GROUP BY
 CASE 
      when player_name like '%Patr%' then 'Kane, Patrick'
      when player_name like '%Alex%' then 'Nylander, Alex'
      when player_name like '%Mar%' then 'Lemieux, Mario'
      when player_name like '%Wayn%' then 'Gretzky, Wayne'
 ELSE NULL
 END

---

但是你应该接受Caius Jard的建议…

如果你必须这样做，你可以尝试以下方法：

SELECT 
SUM(games_played) as games_played,
SUM(goals) as goals,
SUM(points) as points,
player_name
FROM player_stats
GROUP BY
 CASE 
      when player_name like '%Patr%' then 'Kane, Patrick'
      when player_name like '%Alex%' then 'Nylander, Alex'
      when player_name like '%Mar%' then 'Lemieux, Mario'
      when player_name like '%Wayn%' then 'Gretzky, Wayne'
 ELSE NULL
 END

---

但是你应该接受Caius Jard的建议…

你应该用

唯一ID

创建一个新表，并用这些名称填充。同意Georgy；在我的答案中添加了更多细节，以说明如何使用

唯一ID创建一个新表并用这些名称填充。同意Georgy；在我的答案中添加了更多细节为了说明我如何打电话给我们建议，这对我的场景不起作用，因为有7000多条记录，所以我会写一大堆案例，谢谢：）我要打电话给我们建议，这对我的场景不起作用，因为有7000多条记录，所以我会写一大堆案例，谢谢：）让我知道如果你遇到了语法错误；它写在手机上，我无法运行任何程序来测试它让我知道你是否遇到语法错误；它写在手机上，我无法运行任何程序来测试它