Mysql字段列表中的'nume'是模棱兩可的錯誤

Question

select nume as "Nume", adresa as "Adresa",localitate as "Localitatea" 
from info 
left join angajati 
on id_i = id_a 
where localitate like "Orhei" 
and year(curdate()) - year(data_nast) >=50 

我有2個表與字段和我得到的錯誤，請幫助我現在不現在爲什麼這段代碼沒有工作！

Answer 1

由於您加入id_i和id_a列，因此兩個nume列在連接結果中是不同的。指定您的意思是該表的nume列：

SELECT info.nume AS "Nume" ... 

或

SELECT angajati.nume AS "Nume" ... 

Answer 2

您需要定義nume字段應來自哪個表。下面是一個使用表的別名的一種方法：

select i.nume as "Nume", 
     adresa as "Adresa", 
     localitate as "Localitatea" 
from info i 
    left join angajati a on id_i = id_a 
where localitate like "Orhei" 
    and year(curdate()) - year(data_nast) >=50 

當使用joins，如果同一字段在多個表來表示，你必須定義你指的是哪個表。