如果性能很重要,那么我建议delimitedSplit8k_Lead。您可以只使用管道作为分隔符来拆分字符串,然后排除不以 . 结尾的项目(标记)。
DECLARE @table TABLE (COL1 VARCHAR(10), COL2 VARCHAR(1000));
INSERT @table
VALUES
('ID1','1234567890<abc>|4312314124<abc>|1232345133<def>|4131234131<abc>|41234134132<def>'),
('ID2','2662314129<abc>|7868845133<abc>|6831234131<abc>|41234139999<xxx>|1234567999<abc>')
SELECT t.COL1, ds.item
FROM @table t
CROSS APPLY dbo.DelimitedSplit8K_LEAD(t.COL2,'|') ds
WHERE ds.Item LIKE '%<abc>';
退货
COL1 item
---------- -----------------
ID1 1234567890<abc>
ID1 4312314124<abc>
ID1 4131234131<abc>
ID2 2662314129<abc>
ID2 7868845133<abc>
ID2 6831234131<abc>
ID2 1234567999<abc>
然后您使用 XML PATH 进行连接,如下所示:
DECLARE @table TABLE (COL1 VARCHAR(10), COL2 VARCHAR(1000));
INSERT @table
VALUES
('ID1','1234567890<abc>|4312314124<abc>|1232345133<def>|4131234131<abc>|41234134132<def>'),
('ID2','2662314129<abc>|7868845133<abc>|6831234131<abc>|41234139999<xxx>|1234567999<abc>')
SELECT t.COL1, stripBadNumbers.newString
FROM @table t
CROSS APPLY
(VALUES((
SELECT ds.item
FROM dbo.DelimitedSplit8K_LEAD(t.COL2,'|') ds
WHERE ds.Item LIKE '%<abc>'
FOR XML PATH(''), TYPE
).value('.', 'varchar(1000)'))) stripBadNumbers(newString);
返回:
COL1 newString
---------- -------------------------------------------------------------------
ID1 1234567890<abc>4312314124<abc>4131234131<abc>
ID2 2662314129<abc>7868845133<abc>6831234131<abc>1234567999<abc>