SQL Server：如何在查询中使用子字符串？

我有这样的争吵：

我该怎么做

如何在“选择”操作中使用此选项

更新： 我想找到第一个和最后一个“[”””，只捕捉里面的东西

例如：

声音：英文版[钴爪]

返回

钴爪

致以最良好的祝愿，

---

Valter Henrique。

听起来您需要一个正则表达式，以便从源字符串中获取所需的数据

select substring(field, charindex('[', field) + 1, charindex(']', field) - charindex('[', field) - 1)

---

获取第一个

[

和下一个

]

之间的文本

-- cte for test data
;with actor_character(character) AS
(
  select 'voice: English version) [Cobalt Claw]' union all
  select 'voice: English version) [Cobalt Claw' union all
  select 'voice: English version)  Cobalt Claw]' union all
  select 'voice: English version) ]Cobalt Claw[' union all
  select 'voice: English version)  Cobalt Claw'
)
select *,
  case
    -- Test for not valid positions
    when Start.Pos = 1 or Stop.Pos = 0
    then character
    else substring(character, Start.Pos, Stop.Pos-Start.Pos)
  end
from actor_character
  cross apply (select charindex('[', character)+1) as Start(Pos)
  cross apply (select charindex(']', character, Start.Pos)) as Stop(Pos)

-- cte for test data
;with actor_character(character) AS
(
  select 'voice: English version) [Cobalt Claw]' union all
  select 'voice: English version) [Cobalt Claw' union all
  select 'voice: English version)  Cobalt Claw]' union all
  select 'voice: English version) ]Cobalt Claw[' union all
  select 'voice: English version) [Cobalt]Claw]' union all
  select 'voice: English version)  Cobalt Claw'
)
select *,
  case
    -- Test for not valid positions
    when Start.Pos = 0 or Stop.Pos = 0 or Start.Pos > len(character)-Stop.Pos
    then character
    else substring(character, Start.Pos+1, len(character)-Stop.Pos-Start.Pos)
  end

from actor_character
  cross apply (select charindex('[', character)) as Start(Pos)
  cross apply (select charindex(']', reverse(character))) as Stop(Pos)

---

获取第一个

[

和最后一个

]

之间的文本

-- cte for test data
;with actor_character(character) AS
(
  select 'voice: English version) [Cobalt Claw]' union all
  select 'voice: English version) [Cobalt Claw' union all
  select 'voice: English version)  Cobalt Claw]' union all
  select 'voice: English version) ]Cobalt Claw[' union all
  select 'voice: English version)  Cobalt Claw'
)
select *,
  case
    -- Test for not valid positions
    when Start.Pos = 1 or Stop.Pos = 0
    then character
    else substring(character, Start.Pos, Stop.Pos-Start.Pos)
  end
from actor_character
  cross apply (select charindex('[', character)+1) as Start(Pos)
  cross apply (select charindex(']', character, Start.Pos)) as Stop(Pos)

-- cte for test data
;with actor_character(character) AS
(
  select 'voice: English version) [Cobalt Claw]' union all
  select 'voice: English version) [Cobalt Claw' union all
  select 'voice: English version)  Cobalt Claw]' union all
  select 'voice: English version) ]Cobalt Claw[' union all
  select 'voice: English version) [Cobalt]Claw]' union all
  select 'voice: English version)  Cobalt Claw'
)
select *,
  case
    -- Test for not valid positions
    when Start.Pos = 0 or Stop.Pos = 0 or Start.Pos > len(character)-Stop.Pos
    then character
    else substring(character, Start.Pos+1, len(character)-Stop.Pos-Start.Pos)
  end

from actor_character
  cross apply (select charindex('[', character)) as Start(Pos)
  cross apply (select charindex(']', reverse(character))) as Stop(Pos)

---

@squawknull这是与行一起出现的，数据库不是我的。我明白了。所以它是字符串的一部分。我将修改我的查询。我尝试您推荐的方法，但现在我的情况是：选择前（50）个子字符串（character，charrendex（'['，character）+1，charrendex（']'，character）-charrendex（'['，character）-1）FROM actor_characterreturns:传递给LEFT或SUBSTRING函数的长度参数无效。严重性16状态2是否每一行都有“[”和“]”字符？很好：如何更新此参数以找到最后的“]”（请参阅问题中的更新）？@Jon–添加了使用last

]

的版本。我是否能够在查询中获取一列，并在同一select语句中获取该列的子字符串？