SELECT NAME,YEAR,
(SELECT NVL(SUM(A.Time),0) FROM tb_workA A WHERE a.id = tb_user.ID AND a.yid = tb_year.yid) AS SUMA,
(SELECT NVL(SUM(B.Time),0) FROM tb_workB B WHERE b.id = tb_user.ID AND b.yid = tb_year.yid) AS SUMb,
(SELECT NVL(SUM(C.Time),0) FROM tb_workC C WHERE c.id = tb_user.ID AND c.yid = tb_year.yid) AS SUMc
FROM tb_user,tb_year
ORDER BY TB_USER.NAME,TB_YEAR.YID;上面的表UID 是sql里面的id, varchar 修改为varchar2....
(SELECT NVL(SUM(A.Time),0) FROM tb_workA A WHERE a.id = tb_user.ID AND a.yid = tb_year.yid) AS SUMA,
(SELECT NVL(SUM(B.Time),0) FROM tb_workB B WHERE b.id = tb_user.ID AND b.yid = tb_year.yid) AS SUMb,
(SELECT NVL(SUM(C.Time),0) FROM tb_workC C WHERE c.id = tb_user.ID AND c.yid = tb_year.yid) AS SUMc
FROM tb_user,tb_year
ORDER BY TB_USER.NAME,TB_YEAR.YID;上面的表UID 是sql里面的id, varchar 修改为varchar2....
解决方案 »
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