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MYSQL select statement, 2 fields with same name in different table

Posted on 2008-06-14
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Last Modified: 2013-12-13
I have 2 tables:
Car and Car_Make

In Car I have 2 fields, car_make and car_model, both are int.
Example:
Car_Make, Car_Model
1,20

In Car_Make I have 2 fields, make_id and make_name
Example:
make_id, make_name
1,Buick
20,Skylark

I want to be able to use my select statement to select all the cars in the Car table, however I need the make and model name from the car_make table.

I don't know how to write the statement, since the make_name field has both of the fields I need.

Here is a sample select to start with, that just selects the int values:
$query1  = "SELECT car_model, car_make, car_year, car_price, car_mileage, car_avail FROM site_cars ORDER BY $sortby";
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Question by:baumli1
[X]
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4 Comments
 
LVL 3

Accepted Solution

by:
Xorlev earned 125 total points
ID: 21787275
What I recommend you do is have 3 tables
cars:
entry_id, make_id, model_id, year, price, mileage, etc.

car_models:
model_id, make_id, model_name

car_makes:
make_id, make_name

Then you use a query like so:


SELECT model_name, make_name, year, price, mileage FROM cars c LEFT JOIN car_models cmo USING (model_id) LEFT JOIN car_makes cma USING (make_id) ORDER BY etc. etc.

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Author Comment

by:baumli1
ID: 21787290
Although it doesn't answer the question, in the long run this will be easier and better for other selects.
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LVL 3

Expert Comment

by:Xorlev
ID: 21787302
When I was writing the solution for the initial question I noticed how messy (and difficult) it'd make the query to have both make AND model in the same table. This will simplify things immediately.
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LVL 143

Expert Comment

by:Guy Hengel [angelIII / a3]
ID: 21787571
I do agree with the above suggestion.

here is the answer to the original question:
$query1  = "SELECT cma.name car_model, cmo.name car_make, c.car_year, c.car_price, c.car_mileage, c.car_avail 
FROM site_cars c
JOIN site_car_makes cma
  ON cma.make_id = c.car_make
JOIN site_car_makes cmo
  ON cmo.make_id = c.car_model
ORDER BY $sortby";

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