SQL 区分大小写的比较

问题描述

我喜欢用 CASE 语句知道两个字段是否相同。

我设法通过以下查询做到了，但我想知道是否有更好的方法？ 我只希望 '=' 进行区分大小写的比较。我尝试使用 COLLATE 但后来我需要使用一个 WHERE 子句来过滤我的结果，我想要所有的行。使用 HASHBYTES 似乎有点矫枉过正，尤其是当我需要将它与 ISNULL 结合使用时

DECLARE @myTable AS TABLE (old varchar(255),new varchar(255));
INSERT INTO @myTable VALUES
    ('test','test'),('test','TEST'),(null,null)


SELECT old,new,CASE WHEN HASHBYTES('SHA2_512',old) = HASHBYTES('SHA2_512',new) THEN 'same' ELSE 'changed' END AS updated,ISNULL(old,'')) = HASHBYTES('SHA2_512',ISNULL(new,'')) THEN 'same' ELSE 'changed' END AS updated_isnull
FROM @myTable
--where old = new COLLATE latin1_General_CS_AS

我需要列“updated_isnull”

| old  | new  | updated | updated_isnull |
| ---- | ---- | ------- | -------------- |
| test | test | same    | same           |
| test | TEST | changed | changed        |
| NULL | NULL | changed | same           |

解决方法

正确的解决方案是修复您的列，以便您的数据存储在区分大小写的排序规则中。然后你会这样做：

select old,new,case when old = new
             or (old is null and new is null)
         then 1
        else 0 
    end as is_same
from @mytable

使用不区分大小写的排序规则，我们可以像这样解决

select old,case when old collate Latin1_General_Bin = new collate Latin1_General_Bin 
             or (old is null and new is null)
         then 1
        else 0 
    end as is_same
from @mytable

Demo on DB Fiddle：

old  | new  | is_same
:--- | :--- | ------:
test | test |       1
test | TEST |       0
null | null |       1

,

试试这个：

{
    "_id": ObjectId("5ee3e2deee404124a8ba4382"),"geoJSON": {
         "type": "Feature","geometry": {
            "type": "LineString","coordinates": [
                [
                    -85.5,31.0
                ],[
                    -85.6,31.0
                ]
            ]
        }
    }
}
{
    "_id": ObjectId("5ee3e2deee404124a8ba4383"),"coordinates": [
                [
                    -85.55,31.5
                ],31.5
                ]
            ]
        }
    }
}
{
    "_id": ObjectId("5ee3e2deee404124a8ba4384"),32.0
                ],32.0
                ]
            ]
        }
    }
}