我一直在尝试设置一个 SQL 函数来构建带有“标签”的描述。例如，我想从描述开始：
"This is [length] ft. long and [height] ft. high"
并使用相关表中的数据修改描述，最终得到：
"This is 75 ft. long and 20 ft. high"
如果我们有一定数量的标签，我可以使用REPLACE 函数轻松地做到这一点，但我希望这些标签是用户定义的，并且每个描述中可能有也可能没有特定的标签。除了使用光标为每个可用标签遍历字符串一次之外，还有什么更好的方法来获得它？ SQL 是否有任何内置功能可以进行多次替换？类似：
Replace(description,(select tag, replacement from tags))
【问题讨论】：
标签： sql sql-server
我实际上建议在应用程序代码中这样做。但是，您可以使用递归 CTE：
with t as (
select t.*, row_number() over (order by t.tag) as seqnum
from tags t
),
cte as (
select replace(@description, t.tag, t.replacement) as d, t.seqnum
from t
where seqnum = 1
union all
select replace(d, t.tag, t.replacement), t.seqnum
from cte join
t
on t.seqnum = cte.seqnum + 1
)
select top 1 cte.*
from cte
order by seqnum desc;
【讨论】：
The statement terminated. The maximum recursion 100 has been exhausted before statement completion. 但您可以通过在查询末尾添加 option (maxrecursion 1000) 来修复它。
试试下面的查询：
SELECT REPLACE(DESCRIPTION,'[length]',( SELECT replacement FROM tags WHERE tag
= '[length]') )
【讨论】：
REPLACE 函数轻松做到这一点，但我想要这些标签由用户定义，并且每个描述中可能有也可能没有特定标签。
我同意 Gordon 的观点，最好在您的应用程序代码中处理这一点。
如果由于某种原因该选项不可用，并且您不想按照 Gordon 的回答使用递归，则可以使用计数表方法来交换您的值。
您需要测试 for xml 对每个值执行的性能...
假设您有一个Tag 替换值表：
create table TagReplacementTable(Tag nvarchar(50), Replacement nvarchar(50));
insert into TagReplacementTable values('[test]',999)
,('[length]',75)
,('[height]',20)
,('[other length]',40)
,('[other height]',50);
您可以创建一个内联表函数，该函数将通过您的 Descriptions 工作，并使用 TagReplacementTable 作为参考替换必要的部分：
create function dbo.Tag_Replace(@str nvarchar(4000)
,@tagstart nvarchar(1)
,@tagend nvarchar(1)
)
returns table
as
return
(
with n(n) as (select n from (values(1),(1),(1),(1),(1),(1),(1),(1),(1),(1)) n(n))
-- Select the same number of rows as characters in @str as incremental row numbers.
-- Cross joins increase exponentially to a max possible 10,000 rows to cover largest @str length.
,t(t) as (select top (select len(@str) a) row_number() over (order by (select null)) from n n1,n n2,n n3,n n4)
-- Return the position of every value that starts or ends a part of the description.
-- This will be the first character (t='f'), the start of any tag (t='s') and the end of any tag (t='e').
,s(s,t) as (select 1, 'f'
union all select t+1, 's' from t where substring(@str,t,1) = @tagstart
union all select t+1, 'e' from t where substring(@str,t,1) = @tagend
)
-- Return the start and length of every value, to use in the SUBSTRING function.
-- ISNULL/NULLIF combo handles the last value where there is no delimiter at the end of the string.
-- Using the t value we can determine which CHARINDEX to look for.
,l(t,s,l) as (select t,s,isnull(nullif(charindex(case t when 'f' then @tagstart when 's' then @tagend when 'e' then @tagstart end,@str,s),0)-s,4000) from s)
-- Each element of the string is returned in an ordered list along with its t value.
-- Where this t value is 's' this means the value is a tag, so append the start and end identifiers and join to the TagReplacementTable.
-- Where no replacement is found, simply return the part of the Description.
-- Finally, concatenate into one string value.
select (select isnull(r.Replacement,k.Item)
from(select row_number() over(order by s) as ItemNumber
,case when l.t = 's' then '[' else '' end
+ substring(@str,s,l)
+ case when l.t = 's' then ']' else '' end as Item
,t
from l
) k
left join TagReplacementTable r
on(k.Item = r.Tag)
order by k.ItemNumber
for xml path('')
) as NewString
);
然后outer apply 对函数的结果进行替换，以替换您的所有Description 值：
declare @t table (Descr nvarchar(100));
insert into @t values('This is [length] ft. long and [height] ft. high'),('[test] This is [other length] ft. long and [other height] ft. high');
select *
from @t t
outer apply dbo.Tag_Replace(t.Descr,'[',']') r;
输出：
+--------------------------------------------------------------------+-----------------------------------------+
| Descr | NewString |
+--------------------------------------------------------------------+-----------------------------------------+
| This is [length] ft. long and [height] ft. high | This is 75 ft. long and 20 ft. high |
| [test] This is [other length] ft. long and [other height] ft. high | 999 This is 40 ft. long and 50 ft. high |
+--------------------------------------------------------------------+-----------------------------------------+
【讨论】：
我不会遍历单个字符串，而是在整个字符串列上运行更新。我不确定这是否是您的意图，但这比一次一个字符串要快得多。
测试数据：
Create TABLE #strs ( mystr VARCHAR(MAX) )
Create TABLE #rpls (i INT IDENTITY(1,1) NOT NULL, src VARCHAR(MAX) , Trg VARCHAR(MAX) )
INSERT INTO #strs
( mystr )
SELECT 'hello ##color## world'
UNION ALL SELECT 'see jack ##verboftheday##! ##verboftheday## Jack, ##verboftheday##!'
UNION ALL SELECT 'on ##Date##, the ##color## StockMarket was ##MarketDirection##!'
INSERT INTO #rpls ( src ,Trg )
SELECT '##Color##', 'Blue'
UNION SELECT ALL '##verboftheday##' , 'run'
UNION SELECT ALL '##Date##' , CONVERT(VARCHAR(MAX), GETDATE(), 9)
UNION SELECT ALL '##MarketDirection##' , 'UP'
然后是这样的循环：
DECLARE @i INTEGER = 0
DECLARE @count INTEGER
SELECT @count = COUNT(*)
FROM #rpls R
WHILE @i < @count
BEGIN
SELECT @i += 1
UPDATE #strs
SET mystr = REPLACE(mystr, ( SELECT R.src
FROM #rpls R
WHERE i = @i ), ( SELECT R.Trg
FROM #rpls R
WHERE i = @i ))
END
SELECT *
FROM #strs S
产生以下内容
hello Blue world
see jack run! run Jack, run!
on May 19 2017 9:48:02:390AM, the Blue StockMarket was UP!
【讨论】：
我发现有人想做类似here 的事情，但有一定数量的选项：
SELECT @target = REPLACE(@target, invalidChar, '-')
FROM (VALUES ('~'),(''''),('!'),('@'),('#')) AS T(invalidChar)
我可以这样修改：
declare @target as varchar(max) = 'This is [length] ft. long and [height] ft. high'
select @target = REPLACE(@target,'[' + tag + ']',replacement)
from tags
然后它对 select 语句中返回的每条记录运行一次替换。
（我最初已将此添加到我的问题中，但听起来将其添加为答案是更好的协议。）
【讨论】：