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社区首页 >问答首页 >SQL错误ORA-00923:未在预期位置找到FROM关键字

SQL错误ORA-00923:未在预期位置找到FROM关键字
EN

Stack Overflow用户
提问于 2021-06-07 15:29:39
回答 1查看 53关注 0票数 0
代码语言:javascript
复制
SELECT Person_ID, CONCAT('First_name','Surname') AS "Person_Name", Next_of_kin, '~ No next of kin ~' AS Next_of_kin_name, Next_of_kin_age AS NULL  
FROM PERSON  
WHERE Next_of_kin IS NULL
UNION
SELECT Childs.Person_ID, CONCAT('Childs.First_name','Childs.Surname') AS "Person_name", Next_of_kin, CONCAT('Fathers.First_name','Fathers.Surname') AS "Next_of_kin_name",TO_CHAR(sysdate, 'YYYY') - TO_CHAR(Fathers.birth_date, 'YYYY')
(TO_CHAR(sysdate, 'YYYY')- TO_CHAR(Birth_date, 'YYYY')) AS Next_of_kin_age 
FROM Person Childs, Person Fathers
WHERE Childs.next_of_kin = Fathers.Person_ID
AND TO_CHAR(sysdate, 'YYYY') - TO_CHAR(Fathers.birth_date, 'YYYY') >= 50;
EN

回答 1

Stack Overflow用户

发布于 2021-06-07 15:36:00

  • 上一个列名作为NULL返回,但实际上您希望返回值作为Next_of_kin_age.

NULL

  • 还有,在第二个select的最后一列中,你遗漏了运算符,我假设是减运算符。

作为Next_of_kin_age的TO_CHAR(sysdate,'YYYY')- TO_CHAR(Fathers.birth_date,'YYYY') (TO_CHAR(sysdate,'YYYY') - TO_CHAR(Birth_date,'YYYY'))

尝尝这个

代码语言:javascript
复制
SELECT Person_ID, 
       CONCAT('First_name','Surname') AS "Person_Name", 
       Next_of_kin, 
       '~ No next of kin ~' AS Next_of_kin_name, 
       NULL AS Next_of_kin_age 
FROM PERSON  
WHERE Next_of_kin IS NULL
UNION
SELECT Childs.Person_ID, 
       CONCAT('Childs.First_name','Childs.Surname') AS "Person_name", 
       Next_of_kin, 
       CONCAT('Fathers.First_name','Fathers.Surname') AS "Next_of_kin_name",
       TO_CHAR(sysdate, 'YYYY') - TO_CHAR(Fathers.birth_date, 'YYYY') - (TO_CHAR(sysdate, 'YYYY') - TO_CHAR(Birth_date, 'YYYY')) AS Next_of_kin_age 
FROM Person Childs, Person Fathers
WHERE Childs.next_of_kin = Fathers.Person_ID
AND TO_CHAR(sysdate, 'YYYY') - TO_CHAR(Fathers.birth_date, 'YYYY') >= 50;
票数 0
EN
页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接:

https://stackoverflow.com/questions/67867586

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