在字符之间提取数据SQL

SELECT StringValue,
       Parsed = PARSENAME(REPLACE(StringValue,'-','.'), 2) 
       -- parsename starts right
  FROM dbo.RandomData;

2016+可以使用OPENJSON：

SELECT r.StringValue, Parsed = j.value
  FROM dbo.RandomData AS r
  CROSS APPLY OPENJSON ('["' 
    + REPLACE(r.StringValue, '-', '","') + '"]') AS j
  WHERE [key] = 2; -- keys are 0-based

在任何版本上，您都可以使用有序拆分函数(有几十个示例)，这里是我从this article和this earlier answer中抓取的一个：

CREATE FUNCTION dbo.SplitOrdered
(
    @list    nvarchar(max), 
    @delim   nvarchar(10)
)
RETURNS TABLE 
WITH SCHEMABINDING 
AS 
RETURN
(
  WITH w(n) AS (SELECT 0 FROM (VALUES (0),(0),(0),(0)) w(n)),
       k(n) AS (SELECT 0 FROM w a, w b),
       r(n) AS (SELECT 0 FROM k a, k b, k c, k d, k e, k f, k g, k h),
       p(n) AS (SELECT TOP (COALESCE(LEN(@list), 0)) 
                ROW_NUMBER() OVER (ORDER BY @@SPID) -1 FROM r),
       spots(p) AS 
       (
         SELECT n FROM p 
         WHERE (SUBSTRING(@list, n, LEN(@delim + 'x') - 1) LIKE @delim OR n = 0)
       ),
       parts(p,val) AS 
       (
         SELECT p, SUBSTRING(@list, p + LEN(@delim + 'x') - 1, 
           LEAD(p, 1, 2147483647) OVER (ORDER BY p) - p - LEN(@delim)) 
         FROM spots AS s
       )
       SELECT listpos = ROW_NUMBER() OVER (ORDER BY p), 
              Item    = LTRIM(RTRIM(val))
         FROM parts
);

则查询为：

SELECT r.StringValue, Parsed = s.Item
  FROM dbo.RandomData AS r
  CROSS APPLY dbo.SplitOrdered(r.StringValue, '-') AS s
  WHERE s.listpos = 3; -- listpos is 1-based

此db<>fiddle中的所有三个示例。

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