My db tables are like that:
我的db表是这样的:
- routes - table for routes, with columns:
a. destination_id
b. departure_id
- destinations - table for all the destinations, with columns:
a. destination_id
b. destination_name.
routes - 路由表,包含列:a。 destination_id b。 departure_id
destination - 所有目标的表,包含列:a。 destination_id b。 DESTINATION_NAME。
Because every destination is also departure point I use one table for both.
因为每个目的地也是出发点我两个都使用一个表。
Relation is as follows :
1. routes.destination_id = destinations.destination_id,
also
2. routes.departure_id = destinations.destination_id
关系如下:1. routes.destination_id = destinations.destination_id,也是2. routes.departure_id = destinations.destination_id
The result I need is as described:
我需要的结果如下所述:
If routes.destination_id = 3 and routes.departure_id = 5, I want to match each of them to the corresponds name in the destinations table, and to put in the result the correct column names, for departure I want to call it departure_city and same for destination. Right now I get the same name for both different ids, so if destination_id #3 is New York and destination_id #5 is Las Vegas, I will get both as New York and that is not what I need.
如果routes.destination_id = 3和routes.departure_id = 5,我想将它们中的每一个与目的地表中的对应名称相匹配,并在结果中输入正确的列名,对于出发我想称之为departure_city和相同目的地。现在我得到了两个不同ID的相同名称,所以如果destination_id#3是纽约而destination_id#5是拉斯维加斯,那么我将同时获得纽约,这不是我需要的。
So far, my query looks like that:
到目前为止,我的查询看起来像这样:
SELECT routes.*, destinations.destination_name as 'departure_city', destinations.destination_name as 'destination_city'
FROM routes, destinations
WHERE routes.route_departure_id = destinations.destination_id
Before you are telling me to use JOIN, I have done it with JOIN but couldn't change the column name! Also, I always received all the table's columns and I do not know how to get only necessary columns with JOIN!
在你告诉我使用JOIN之前,我已经使用JOIN完成了但是无法更改列名!此外,我总是收到所有表的列,我不知道如何只用JOIN获得必要的列!
Right now my issues are that I can't get the value from destinations twice, once for the route_destination and one for the route_departure and show different values.
现在我的问题是我无法从目的地获取两次值,一次是route_destination,另一次是route_departure,并显示不同的值。
1 个解决方案
#1
2
You are looking for two joins:
您正在寻找两个连接:
SELECT r.*,
dep.destination_name as departure_city,
dest.destination_name as destination_city
FROM routes r JOIN
destinations dep
ON r.departure_id = dep.destination_id JOIN
destinations dest
ON r.destination_id = dest.destination_id;
Additional advice:
- Use table aliases. They make the query easier to write and to read.
-
Never use commas in the
FROM
clause. Always use explicitJOIN
syntax. - Do not use single quotes to define column aliases. Only use single quotes for string and date constants. Use backticks if you are using a name that needs to be quoted (and then, change the name so backticks aren't necessary).
使用表别名。它们使查询更容易编写和阅读。
切勿在FROM子句中使用逗号。始终使用显式JOIN语法。
不要使用单引号来定义列别名。仅对字符串和日期常量使用单引号。如果您使用需要引用的名称,请使用反引号(然后,更改名称,以便不需要反引号)。
#1
2
You are looking for two joins:
您正在寻找两个连接:
SELECT r.*,
dep.destination_name as departure_city,
dest.destination_name as destination_city
FROM routes r JOIN
destinations dep
ON r.departure_id = dep.destination_id JOIN
destinations dest
ON r.destination_id = dest.destination_id;
Additional advice:
- Use table aliases. They make the query easier to write and to read.
-
Never use commas in the
FROM
clause. Always use explicitJOIN
syntax. - Do not use single quotes to define column aliases. Only use single quotes for string and date constants. Use backticks if you are using a name that needs to be quoted (and then, change the name so backticks aren't necessary).
使用表别名。它们使查询更容易编写和阅读。
切勿在FROM子句中使用逗号。始终使用显式JOIN语法。
不要使用单引号来定义列别名。仅对字符串和日期常量使用单引号。如果您使用需要引用的名称,请使用反引号(然后,更改名称,以便不需要反引号)。