我该如何在oracle中找到没有相同案件的人?

时间:2018-09-20 03:45:01

标签: sql oracle oracle10g

我有一个名为CASE_PERSON的表。DDL是

CREATE TABLE CASE_PERSON
(
  ID VARCHAR2(50) NOT NULL   PRIMARY KEY,
  p_id VARCHAR2(50) DEFAULT NULL NOT NULL,
  id_card VARCHAR2(18) DEFAULT NULL NOT NULL,
  p_name VARCHAR2(50) NOT NULL,
  case_no VARCHAR2(60) NOT NULL,
  case_name VARCHAR2(200) NOT NULL,
 )

还有一些DML

insert into CASE_PERSON(ID, P_ID, ID_CARD, P_NAME, CASE_NO, CASE_NAME)
values ('001', '123', '123-123123', 'tom', 'c-001', 'case-one');

insert into CASE_PERSON(ID, P_ID, ID_CARD, P_NAME, CASE_NO, CASE_NAME)
values ('002', '234', '123-123234', 'rose', 'c-001', 'case-one');

insert into CASE_PERSON(ID, P_ID, ID_CARD, P_NAME, CASE_NO, CASE_NAME)
values ('003', '345', '123-123345', 'jack', 'c-001', 'case-one');

insert into CASE_PERSON(ID, P_ID, ID_CARD, P_NAME, CASE_NO, CASE_NAME)
values ('005', '555', '123-123555', 'roman', 'c-002', 'case-two');

insert into CASE_PERSON(ID, P_ID, ID_CARD, P_NAME, CASE_NO, CASE_NAME)
values ('006', '666', '123-123666', 'jim', 'c-002', 'case-two');

该表是根据Person表和case表构建的。 p的前缀来自PERSON,大小写来自CASE。首先,我从PERSON获取所有身份证,并从CASE_PERSON查询以查找具有相同大小写的人匹配给定的身份证。

现在,我想得到的结果称为“相同病例的人”。输出应为以下格式:

id   p_id   id_card    case_no case_name  same_case_id_card same_case_p_name
 1    123  123-123123   c-001  case-one     123-123234           rose
 2    123  123-123123   c-001  case-one     123-123345           jack
 3    555  123-123555   c-002  case-two     123-123666           jim  

如何查询?

2 个答案:

答案 0 :(得分:2)

我认为您要查找的查询是:

SELECT * FROM case a, case b 
WHERE a.case_no = b.case_no;

答案 1 :(得分:1)

您似乎想要第一人称的案件详细信息,以及非第一人称的详细信息:

WITH cte AS (
  SELECT *, ROW_NUMBER() OVER(PARTITION BY case_no ORDER BY p_id) as rown FROM case_person
)
SELECT 
  ROW_NUMBER() OVER(ORDER BY two.p_id) as id,  
  one.p_id,
  one.id_card,
  one.case_no,
  one.case_name,
  two.id_card as same_case_id_card,
  two.case_name as same_case_p_name
FROM
  (SELECT * FROM cte WHERE rown = 1) one
  INNER JOIN
  (SELECT * FROM cte WHERE rown > 1) two
  ON one.case_no = two.case_no