我希望这两个不同的表结果出现在一个查询中。
$result=mysqli_query("SELECT * FROM table_1 WHERE keyskill LIKE'php%'");
$result=mysqli_query("SELECT * FROM table_2 WHERE parentaddress LIKE'chennai%'");
有人可以帮我吗?
这是表结构...
表 1:
ID | F. Name | L.Name | Gender | Parentaddress
___________________________________________________
12 | Peter | Son | Male | Chennai
13 | Johny | Depp | Male | Coimbatore
表 2:
S.No| Name | Title | Gender | Keyskill
___________________________________________________
13 | Johny | Student | Male | PHP
12 | Peter | Student | Male | JAVA
你的数据库不应该是这样的吗?(只是猜测!
表1(人员):
ID | F. Name | L.Name | Gender | Parentaddress | Title
__________________________________________________________
12 | Peter | Son | Male | Chennai | Student
13 | Johny | Depp | Male | Coimbatore | Student
表 2(技能):
S.No| ID | Keyskill
___________________________________________________
70 | 13 | PHP
71 | 13 | Whatever Language
72 | 12 | JAVA
表 1 定义了要为其保存技能的人员,表 2 仅按 iD 字段引用了这些人员。
使用 UNION :
select a.`ID` as ID, a.`F.Name` as first_name, a.`L.Name` as last_name, b.`Title` as tittle, a.`Gender` as gender, a.`Parentaddress` as parent_address, b.keyskill as keyskill from table_1 a inner join table_2 b on a.id = b.`S.No` where b.keyskill LIKE 'php%'
union all
select a.`S.No` as ID, b.`F.Name` as first_name, b.`L.Name` as last_name, a.`Title` as tittle, a.`Gender` as gender, b.`Parentaddress` as parent_address, a.`Keyskill` as keyskill from table_2 a inner join table_1 b where b.parentaddress LIKE 'chennai%'
或者可能通过ID加入:
select * from table_1 a left outer join table_2 b on a.id = b.`S.No` where b.keyskill LIKE 'php%' and a.parentaddress LIKE 'chennai%'