sql - oracle sql - 选择具有多个 "case when"的语句并检查包含文本

Question

我寻找一个工作示例，我可以在其中使用 mutliple when case 语句和 check 来验证是否包含特定文本:
例如

SELECT 
  ID,
  NAME,
  (SELECT 
      (Case when Contains(Descr,"Test") Then "contains Test" 
        when Contains(Descr, "Other") Then "contains Other"
        Else "No Match" End) From DESCRIPTION
       where item_id = id
  ) as "Match"
  From Item

Answer 1

在 Oracle 中，字符串文字需要用单引号括起来。

要查找子字符串匹配，您可以使用 LIKE :

SELECT  ID,
        NAME,
        CASE WHEN Descr LIKE '%Test%'  THEN 'Contains Test'
             WHEN Descr LIKE '%Other%' THEN 'Contains Other'
             ELSE 'No Match'
        END AS Match
FROM    Item i
        LEFT OUTER JOIN
        Description d
        ON i.id = d.item_id

或 INSTR() :

SELECT  ID,
        NAME,
        CASE WHEN INSTR( Descr, 'Test' ) > 0  THEN 'Contains Test'
             WHEN INSTR( Descr, 'Other' ) > 0 THEN 'Contains Other'
             ELSE 'No Match'
        END AS Match
FROM    Item i
        LEFT OUTER JOIN
        Description d
        ON i.id = d.item_id

或 REGEXP_LIKE() :

SELECT  ID,
        NAME,
        CASE WHEN REGEXP_LIKE( Descr, 'Test' )  THEN 'Contains Test'
             WHEN REGEXP_LIKE( Descr, 'Other' ) THEN 'Contains Other'
             ELSE 'No Match'
        END AS Match
FROM    Item i
        LEFT OUTER JOIN
        Description d
        ON i.id = d.item_id