【发布时间】:2011-09-14 02:26:15
【问题描述】:
谁能帮我弄清楚为什么我在cms.CRIME_ID 上收到错误:
标识符无效
select c.criminal_id, c.first, c.last, cms.CRIME_ID, cc.crime_code, cc.fine_amount
from criminals c join crimes cms on c.criminal_id = cms.criminal_id
join crime_charges cc using (crime_id)
order by c.first, c.last;
我绝对知道该列存在,除此之外,我可以引用该表中的所有其他列。
该列唯一不同的是它是该表的主键。
编辑:这是完整的错误和表创建脚本。
Error starting at line 1 in command:
select c.criminal_id, c.first, c.last, cms.CRIME_ID, cc.crime_code, cc.fine_amount
from criminals c join crimes cms on c.criminal_id = cms.criminal_id
join crime_charges cc using (crime_id)
order by c.first, c.last
Error at Command Line:1 Column:39
Error report:
SQL Error: ORA-00904: "CMS"."CRIME_ID": invalid identifier
00904. 00000 - "%s: invalid identifier"
*Cause:
*Action:
CREATE TABLE crimes
(crime_id NUMBER(9),
criminal_id NUMBER(6),
classification CHAR(1),
date_charged DATE,
status CHAR(2),
hearing_date DATE,
appeal_cut_date DATE);
ALTER TABLE crimes
MODIFY (classification DEFAULT 'U');
ALTER TABLE crimes
ADD (date_recorded DATE DEFAULT SYSDATE);
ALTER TABLE crimes
MODIFY (criminal_id NOT NULL);
ALTER TABLE crimes
ADD CONSTRAINT crimes_id_pk PRIMARY KEY (crime_id);
ALTER TABLE crimes
ADD CONSTRAINT crimes_class_ck CHECK (classification IN('F','M','O','U'));
ALTER TABLE crimes
ADD CONSTRAINT crimes_status_ck CHECK (status IN('CL','CA','IA'));
ALTER TABLE crimes
ADD CONSTRAINT crimes_criminalid_fk FOREIGN KEY (criminal_id)
REFERENCES criminals(criminal_id);
ALTER TABLE crimes
MODIFY (criminal_id NOT NULL);
EDIT2:另外,我应该提一下,当不使用连接而只使用常规选择语句时,我可以很好地访问该列,如下面的代码示例所示:
select c.criminal_id, c.first, c.last, cms.crime_id, cc.crime_code, cc.fine_amount
from criminals c, crime_charges cc, crimes cms
where c.criminal_id = cms.criminal_id
and cms.crime_id = cc.crime_id
order by c.first, c.last;
【问题讨论】:
-
请贴出涉及表的详细信息
-
您的用户是否拥有犯罪表的权限?
-
crime_charges 表的定义是什么?
-
F,M,O,U = 重罪、轻罪、条例 (?)、???
-
您是否尝试过使用完全限定名称(例如:
crimes.crime_id而不是cms.crime_id)?