有两个表:
表1:
id = 0 name=JOHN speciality =dentist
id = 1 name=ABY speciality= dentist
id = 2 name = SARA speciality= cardiologist
,...
表 2:
id = 0 name = JOHN city=paris
id = 1 name = ABY city=tokio
id = 1 name = SARA city=london
$city='pa';
$speciality='dentist';
(这两个变量来自用户输入的搜索表单)
我想说的是:
select * FROM Table 1 WHERE speciality=$speciality AND city LIKE %$city% (from Table2) ;
我该说什么? (很明显,两个表的每一行的id都是相同的,每个id代表一个人)
最佳答案
尝试这样:
select t1.* FROM Table 1 as t1,table2 as t2 WHERE t1.speciality='$speciality' AND t1.city LIKE %$city% and t1.name=t2.name;
如果两个表的行 ID 相同,则:
select t1.* FROM Table 1 as t1,table2 as t2 WHERE t1.speciality='$speciality' AND t1.city LIKE %$city% and t1.name=t2.name and t1.id=t2.id ;
关于php - 如何从两个有条件的表中进行选择,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/22134298/