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ORACLE-SQL(二)
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UpdateTime--2017年1月22日14:06:25
方法二:使用(+)
SELECT T.* FROM (SELECT S.*, C.CLAZZNAME, ROWNUM ROWNO FROM STUDENT3 S, CLAZZ3 C WHERE S.CLAZZID = C.CLAZZID(+) AND ROWNUM <= ?) T WHERE T.ROWNO >= ?
注意:
a.只有Oracle数据库可以使用(+)来代替左连接和右连接;
b."(+)"在"="右侧,表示的是:左连接,如:S.CLAZZID = C.CLAZZID(+);左表(学生表)为主表;
c."(+)"在"="左侧,表示的是:右连接,如:S.CLAZZID(+) = C.CLAZZID;右表(班级表)为主表。
方法三:from后面跟两张表
SELECT T.* FROM (SELECT S.*, C.CLAZZNAME, ROWNUM ROWNO FROM STUDENT3 S, CLAZZ3 C WHERE S.CLAZZID = C.CLAZZID AND ROWNUM <= ?) T WHERE T.ROWNO >= ?
1.2.2 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
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
情景:
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
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‘;
1.2.3 分页,分组,计数,排序
/** * 返回数据 * FORGID 医疗机构ID * FRCODE 县区编码 * FORGNAME 医疗机构名称 * FCENNAME_TEM_COUNT 名称不对称计数 * ISNULLCENCODE_COUNT 未对照计数 * TOTALCOUNT L_DIAITEM_ERROR表中共有多少条数据 */ SELECT T3.* FROM (SELECT T2.*, ROWNUM ROWNO FROM (SELECT T.FORGID, T1.FRCODE, T1.FORGNAME, SUM(NVL(T.NCCW, 0)) FCENNAME_TEM_COUNT, SUM(NVL(T.WDZ, 0)) ISNULLCENCODE_COUNT, TOTALCOUNT FROM L_DIAITEM_ERROR T, TORGANIZATION T1 WHERE T.FORGID = T1.FORGID GROUP BY T.FORGID, T1.FORGNAME, T1.FORGSEQ, T1.FRCODE ORDER BY T1.FORGSEQ) T2 WHERE ROWNUM <= V_END) T3 WHERE T3.ROWNO >= V_START;
注意:
a.TOTALCOUNT是已存在的一个变量,这里不作为关注重点;
b.被group by的数据,要想查询表中的某个字段,有且只有两种方式:
方式一:
在group by 后面加上想要查询出来的字段
方式二:
对于数字列,使用求和函数sum()实现查询
否则,会报错:不是group by 表达式
ORACLE-SQL(二)
标签:nbsp from 存在 主表 求和 表达 变量 oracle数据库 author
本文系统来源:http://www.cnblogs.com/Marydon20170307/p/6928963.html
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