如何将sql中的字符串拆分为多列

我有一张桌子

身份证件 名称 命名路径 1. 产品1 无效的 2. 产品2 产品1 3. 产品3 产品1 |产品2 4. 产品4 产品1 |产品2 |产品3

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这是我想到的最好的方法：

select 
    case 
        when name_path is null then name
        when CHARINDEX ('|', name_path) = 0 then name_path
        else  LEFT(name_path, CHARINDEX ('|', name_path)-1) --CAST( CHARINDEX ('|', name_path) as nvarchar)
        --else LEFT(name_path, CHARINDEX ('|', name_path))
        --else SUBSTRING(name_path, LEN(LEFT(name_path, CHARINDEX ('|', name_path))) + 1, LEN(name_path) - LEN(LEFT(name_path, 
            --  CHARINDEX ('|', name_path))) - LEN(RIGHT(name_path, LEN(name_path) - CHARINDEX ('.', name_path))) - 1)
    end as producttype,
    case
        when LEN(name_path) - LEN(REPLACE(name_path, '|', '')) = 0 then name
        else REPLACE( PARSENAME(REPLACE(replace(name_path,'.','---') , '|' , '.'),LEN(name_path) - LEN(REPLACE(name_path, '|', ''))),'---','.')
    end as product,
    case
        when LEN(name_path) - LEN(REPLACE(name_path, '|', '')) = 1 then name
        else REPLACE( PARSENAME(REPLACE(replace(name_path,'.','---') , '|' , '.'),LEN(name_path) - LEN(REPLACE(name_path, '|', ''))-1),'---','.')
    end as productflavor1,
    case
        when LEN(name_path) - LEN(REPLACE(name_path, '|', '')) = 2 then name
        else REPLACE( PARSENAME(REPLACE(replace(name_path,'.','---') , '|' , '.'),LEN(name_path) - LEN(REPLACE(name_path, '|', ''))-2),'---','.')   end as productflavor2,
    case
        when LEN(name_path) - LEN(REPLACE(name_path, '|', '')) = 3 then name
        else REPLACE( PARSENAME(REPLACE(replace(name_path,'.','---') , '|' , '.'),LEN(name_path) - LEN(REPLACE(name_path, '|', ''))-3),'---','.')
    end as productflavor3,
    case
        when LEN(name_path) - LEN(REPLACE(name_path, '|', '')) = 4 then name
        else REPLACE( PARSENAME(REPLACE(replace(name_path,'.','---') , '|' , '.'),LEN(name_path) - LEN(REPLACE(name_path, '|', ''))-4),'---','.')
    end as productflavor4,
    case
        when LEN(name_path) - LEN(REPLACE(name_path, '|', '')) = 5 then name
        else REPLACE( PARSENAME(REPLACE(replace(name_path,'.','---') , '|' , '.'),LEN(name_path) - LEN(REPLACE(name_path, '|', ''))-5),'---','.')
    end as productflavor5,
    case
        when LEN(name_path) - LEN(REPLACE(name_path, '|', '')) = 6 then name
        else REPLACE( PARSENAME(REPLACE(replace(name_path,'.','---') , '|' , '.'),LEN(name_path) - LEN(REPLACE(name_path, '|', ''))-6),'---','.')
    end as productflavor6,
    case
        when LEN(name_path) - LEN(REPLACE(name_path, '|', '')) = 7 then name
        else REPLACE( PARSENAME(REPLACE(replace(name_path,'.','---') , '|' , '.'),LEN(name_path) - LEN(REPLACE(name_path, '|', ''))-7),'---','.')
    end as productflavor7,
    case
        when LEN(name_path) - LEN(REPLACE(name_path, '|', '')) = 8 then name
        else REPLACE( PARSENAME(REPLACE(replace(name_path,'.','---') , '|' , '.'),LEN(name_path) - LEN(REPLACE(name_path, '|', ''))-8),'---','.')
    end as productflavor8,
    table.*
from table


     


 

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哪个数据库管理系统？在SQL Server中，有一个

STRING\u SPLIT

我曾对此进行过研究，但未能使其发挥作用，以获得所需的结果。