如何在mysql视图中传递动态参数

jav*_*aba 6 mysql sql mysql-select-db mysql-workbench

我在mysql中创建了这个

CREATE VIEW MYVIEW AS (
SELECT A.FNAME
     , A.LNAME
     , B.EMAIL
FROM EMPLOYEE A, EMPEMAIL B
WHERE A.EID = :empId
  AND A.EID = B.EID
AND B.EMAILTYP = :emailType)
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现在我想让"empId"和"emailType"动态化.我的意思是在选择时传递值.什么需要改变代码?提前

Dev*_*art 10

你可以使用这个功能的解决方案 -

CREATE FUNCTION func() RETURNS int(11)
  RETURN @var;

CREATE VIEW view1 AS
  SELECT * FROM table1 WHERE id = func();
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使用示例:

SET @var = 1;
SELECT * FROM view1;
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Mur*_*nik 8

只需创建没有参数的视图(即,仅处理连接):

CREATE VIEW MYVIEW AS (
SELECT A.FNAME
     , A.LNAME
     , B.EMAIL
     , A.EID AS EID             -- added to be used in the WHERE
     , B.EMAILTYP AS EMAILTYP   -- added to be used in the WHERE
FROM EMPLOYEE A, EMPEMAIL B
WHERE A.EID = B.EID)
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并在查询时应用动态参数:

SELECT FNAME, LNAME, EMAIL
FROM   my_view
WHERE  eid = 'your_empId' AND emailtyp = 'your_emailType'
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