需要用同一数据库中另一张表的数据填充一张表 [英] Need to populate one table with data from another table in same database
问题描述
我正在尝试使用 mysql 从另一个表中填充一个表.第一个表是用户,第二个表是技术人员.
I am trying to populate one table from another table with mysql. First table is users and the second table is technicians .
用户包含:userID, surname, firstname, loginid, password, accesslevel .
Users contains : userID, surname, firstname, loginid, password, accesslevel .
技术人员包含:techID、tech_surname、tech_firstname、tech_loginid、tech_password、tech_accesslevel.
Technicians contains : techID, tech_surname, tech_firstname, tech_loginid, tech_password, tech_accesslevel.
当我添加用户时,如果 accesslevel=tech 和 users.loginid 不等于 Technicals.tech_loginid,我希望填充技术人员表.
When I add a user, I want the technicians table to populate if accesslevel=tech and users.loginid is not equal to technicians.tech_loginid.
我尝试了几件事,结果要么没有添加记录,要么每次都添加了 accesslevel=tech 的用户中的所有记录,给了我几个重复的记录.
I have tried several things and the result is that either no record is added or all records in users where accesslevel=tech are added each time, giving me several duplicate records.
我试过了:
INSERT INTO technicians (techID, tech_surname, tech_firstname, tech_loginid, tech_passwrd, tech_accesslevel)
SELECT firstname, surname, loginid, accesslevel, passwrd, tech_loginid
FROM users, technicians
WHERE accesslevel='tech' AND 'loginid!=tech_loginid'
它不起作用,如果我从 WHERE 子句中删除 AND 语句,它会在每次使用 accesslevel=tech 添加新用户时推送所有记录.
It doesn't work and if I remove the AND statement from the WHERE clause it pushes all the records each time a new user is added with accesslevel=tech.
我做错了什么?我已经搜索了几个小时的答案.
What am I doing wrong? I've searched for hours for an answer.
干杯
推荐答案
我认为您正在尝试插入那些已经不存在于该表中的技术人员,因此查询将是这样的
I think you're trying to insert those technicians that already do not exist in that table so a query would be something like this
INSERT INTO technicians (techID, tech_surname, tech_firstname, tech_loginid, tech_passwrd, tech_accesslevel)
SELECT loginid, surname, firstname, tech_loginid, passwrd, accesslevel
FROM users
LEFT OUTER JOIN technicians ON loginid = tech_loginid
WHERE accesslevel='tech' and tech_loginid IS null
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