SQL 计算字符串中的数字

SQL Count digits in a string

我有table以下信息。

description: adjkfa34kj34kj33j4k3jk4
description: adfkjjkdf34_3434kjkjkj3
description: akjdfkjadfkjadkjfkjadfj
description: 34394394093040930949039

我如何编辑下面的 SQL 查询来计算字符串中的位数(即 [0-9])?

select description as num_count
from posts;

一种方法是去掉所有其他字符:

select length(regexp_replace(description, '[^0-9]', '', 'g'))

试试这个,

 CREATE FUNCTION dbo.udf_GetNumeric
    (
        @strAlphaNumeric VARCHAR(256)
    )
    RETURNS VARCHAR(MAX)
    AS
    BEGIN
        DECLARE @intAlpha INT
        SET @intAlpha = PATINDEX('%[^0-9]%', @strAlphaNumeric)

        BEGIN
            WHILE @intAlpha > 0
            BEGIN
                    SET @strAlphaNumeric = STUFF(@strAlphaNumeric, @intAlpha, 1, '' )
                    SET @intAlpha = PATINDEX('%[^0-9]%', @strAlphaNumeric )
            END
        END

        RETURN @strAlphaNumeric
    END
    GO

/*For Example,*/
SELECT LEN(dbo.udf_GetNumeric('') )
SELECT LEN(dbo.udf_GetNumeric('asdf1234a1s2d3f4@@@') )
SELECT LEN(dbo.udf_GetNumeric('123ghdfhdhh456'))
SELECT LEN(dbo.udf_GetNumeric(NULL) )

SELECT LEN(dbo.udf_GetNumeric(描述)) AS num_count 来自帖子;

我在 presto 中使用了以下内容:

SELECT CARDINALITY(REGEXP_EXTRACT_ALL('a123', '\d')) -- returns 3