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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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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