SQL Split函数，用于处理文本限定符之间出现delimeter的字符串？

有几个SQL拆分函数，从循环驱动到使用xml命令，甚至使用数字表。我还没有找到一个支持文本限定符的

使用下面的示例字符串，我希望在、、上拆分，但不希望在双引号或单引号之间拆分

示例数据：

jsmith@anywhere.com, "Sally \"Heat\" Jones" <sally@anywhere.com>, "Mark Jones" <mjones@anywhere.com>, "Stone, Ron" <rstone@anywhere.com>

应返回一个表：

jsmith@anywhere.com
"Sally \"Heat\" Jones" <sally@anywhere.com>
"Mark Jones" <mjones@anywhere.com>
"Stone, Ron" <rstone@anywhere.com>

---

我知道这是一个复杂的查询/函数，但任何建议或指导都将不胜感激。

这是一个快速解决方案，它不够完美，没有堆栈，因此它将把引号内的逗号作为分隔符

alter function fnSplit
(
    @Delim char(1),
    @List nvarchar(4000)
)
returns table as
return
    with 
    Strings(PosIdx) as 
    (
        select 1 
        union all 
        select PosIdx + 1 from Strings where PosIdx < 4000
    )
    select
        ltrim(rtrim(substring(@List, PosIdx, charindex(@Delim, @List + @Delim, PosIdx) - PosIdx))) as value
    from   
        Strings
    where  
        PosIdx <= convert(int, len(@List))
    and substring(@Delim + @List, PosIdx, 1) = @Delim 
go
select * from fnSplit(',', 'jsmith@anywhere.com, "Sally \"Heat\" Jones" <sally@anywhere.com>, "Mark Jones" <mjones@anywhere.com>, "Stone, Ron" <rstone@anywhere.com>') 
    option (maxrecursion 0)

---

这是一个快速的解决方案，它不够完美，没有堆栈，因此它将把引号内的逗号作为分隔符

alter function fnSplit
(
    @Delim char(1),
    @List nvarchar(4000)
)
returns table as
return
    with 
    Strings(PosIdx) as 
    (
        select 1 
        union all 
        select PosIdx + 1 from Strings where PosIdx < 4000
    )
    select
        ltrim(rtrim(substring(@List, PosIdx, charindex(@Delim, @List + @Delim, PosIdx) - PosIdx))) as value
    from   
        Strings
    where  
        PosIdx <= convert(int, len(@List))
    and substring(@Delim + @List, PosIdx, 1) = @Delim 
go
select * from fnSplit(',', 'jsmith@anywhere.com, "Sally \"Heat\" Jones" <sally@anywhere.com>, "Mark Jones" <mjones@anywhere.com>, "Stone, Ron" <rstone@anywhere.com>') 
    option (maxrecursion 0)

以下是我的解决方案：

CREATE FUNCTION fnSplitString
(
    @input nvarchar(MAX) 
)
RETURNS @emails TABLE
(
    email nvarchar(MAX) 
)
AS
BEGIN

DECLARE @len int = LEN(@input)
DECLARE @pos int = 1;
DECLARE @start int = 1;
DECLARE @ignore bit = 0;
WHILE(@pos<=@len)
BEGIN

    DECLARE @ch nchar(1) = SUBSTRING(@input, @pos, 1);

    IF ( @ch = '"' or @ch = '''')
    BEGIN
        SET @ignore = 1 - @ignore;
    END

    IF (@ch = ',' AND @ignore = 0)
    BEGIN
        INSERT @emails VALUES (SUBSTRING(@input, @start, @pos-@start));
        SET @start = @pos+1;
    END

    SET @pos = @pos + 1;
END

IF (@start<>@pos)
BEGIN
    INSERT @emails VALUES (SUBSTRING(@input, @start, @pos-@start));
END

RETURN
END
GO

DECLARE @input nvarchar(max) = 'jsmith@anywhere.com, "Sally \"Heat\" Jones" <sally@anywhere.com>, "Mark Jones" <mjones@anywhere.com>, "Stone, Ron" <rstone@anywhere.com>';

select * from fnSplitString(@input)

以下是我的解决方案：

CREATE FUNCTION fnSplitString
(
    @input nvarchar(MAX) 
)
RETURNS @emails TABLE
(
    email nvarchar(MAX) 
)
AS
BEGIN

DECLARE @len int = LEN(@input)
DECLARE @pos int = 1;
DECLARE @start int = 1;
DECLARE @ignore bit = 0;
WHILE(@pos<=@len)
BEGIN

    DECLARE @ch nchar(1) = SUBSTRING(@input, @pos, 1);

    IF ( @ch = '"' or @ch = '''')
    BEGIN
        SET @ignore = 1 - @ignore;
    END

    IF (@ch = ',' AND @ignore = 0)
    BEGIN
        INSERT @emails VALUES (SUBSTRING(@input, @start, @pos-@start));
        SET @start = @pos+1;
    END

    SET @pos = @pos + 1;
END

IF (@start<>@pos)
BEGIN
    INSERT @emails VALUES (SUBSTRING(@input, @start, @pos-@start));
END

RETURN
END
GO

DECLARE @input nvarchar(max) = 'jsmith@anywhere.com, "Sally \"Heat\" Jones" <sally@anywhere.com>, "Mark Jones" <mjones@anywhere.com>, "Stone, Ron" <rstone@anywhere.com>';

select * from fnSplitString(@input)

---

这显然不是解决原来问题的办法。我想在、、上拆分，但不是在双引号或单引号之间拆分。这显然不是原始问题的解决方案。我希望在、、上拆分，但不希望在双引号或单引号之间拆分。