oracle 三表关联查询
CreationTime--2018年7月4日17点52分
Author:Marydon
左连接实现三表关联
表A---------------------------------关联第一张表B-----------------------关联第二张表c
1.语法
select * from 表名A left join 表B on A.columnX=B.columnM and A.columnY=B.columnN left join 表c on 表A=表c的id
2.应用场景
四张表 GJPT_BASY、GJZY_BASY、GJPT_BASY_ERROR、GJZY_BASY_ERROR
根据四张表,要求返回:医疗机构名称,医疗机构编号,总数,合格数和问题数
3.SQL实现
SELECT TEMP1.*, TEMP2.HEGESUM, TEMP3.TROUBLESUM FROM (SELECT T1.YLNAME, T1.YLCODE, SUM(RS1) AS YLSUM--总数 FROM (SELECT COUNT(1) AS RS1, HDSD00_11_118 AS YLNAME, HDSD00_11_119 AS YLCODE FROM GJPT_BASY GROUP BY HDSD00_11_119, HDSD00_11_118 UNION ALL SELECT COUNT(1) AS RS1, HDSD00_12_133 AS YLNAME, HDSD00_12_134 AS YLCODE FROM GJZY_BASY GROUP BY HDSD00_12_133, HDSD00_12_134 UNION ALL SELECT COUNT(1) AS RS1, HDSD00_11_118 AS YLNAME, HDSD00_11_119 AS YLCODE FROM GJPT_BASY_ERROR GROUP BY HDSD00_11_119, HDSD00_11_118 UNION ALL SELECT COUNT(1) AS RS1, HDSD00_12_133 AS YLNAME, HDSD00_12_134 AS YLCODE FROM GJZY_BASY_ERROR GROUP BY HDSD00_12_133, HDSD00_12_134) T1 GROUP BY T1.YLNAME, T1.YLCODE) TEMP1 LEFT JOIN (SELECT * FROM (SELECT T2.YLNAME, T2.YLCODE, SUM(RS2) AS HEGESUM--合格数 FROM (SELECT COUNT(1) AS RS2, HDSD00_11_118 AS YLNAME, HDSD00_11_119 AS YLCODE FROM GJPT_BASY GROUP BY HDSD00_11_119, HDSD00_11_118 UNION ALL SELECT COUNT(1) AS RS2, HDSD00_12_133 AS YLNAME, HDSD00_12_134 AS YLCODE FROM GJZY_BASY GROUP BY HDSD00_12_133, HDSD00_12_134) T2 GROUP BY T2.YLNAME, T2.YLCODE)) TEMP2 ON TEMP2.YLNAME = TEMP1.YLNAME AND TEMP2.YLCODE = TEMP1.YLCODE LEFT JOIN (SELECT * FROM (SELECT T3.YLNAME, T3.YLCODE, SUM(RS3) TROUBLESUM--问题数 FROM (SELECT COUNT(1) AS RS3, HDSD00_11_118 AS YLNAME, HDSD00_11_119 AS YLCODE FROM GJPT_BASY_ERROR GROUP BY HDSD00_11_119, HDSD00_11_118 UNION ALL SELECT COUNT(1) AS RS3, HDSD00_12_133 AS YLNAME, HDSD00_12_134 AS YLCODE FROM GJZY_BASY_ERROR GROUP BY HDSD00_12_133, HDSD00_12_134) T3 GROUP BY T3.YLNAME, T3.YLCODE)) TEMP3 ON TEMP3.YLNAME = TEMP1.YLNAME AND TEMP3.YLCODE = TEMP1.YLCODE WHERE TEMP3.YLCODE='41580781841010511A1001';