Forum: Ruby on Rails Help with ruby query!

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Jay M. (Guest)
on 2008-12-04 05:43
I am using Rails 2.0.2 with mysql
database

I have two tables: book and author
Table book has 3 columns: id, isbn, and title
Table author has 3 columns: id, name, and age

In the models I have:
class Book < ActiveRecord::Base
belongs_to :author
end

class Author < ActiveRecord::Base
   has_many :books
end

I have author_id as the foreign key in the book's table

The user will select  'age' from a list
Then I want to display names of all authors with the selected age,
and also the 'title' from book's table.

In plain SQL, the query will be like this:

SELECT name, title
FROM author
INNER JOIN book
ON author.id = book.id
WHERE age = "the selected age"

How can I write this in Ruby?

Thanks
Cypray
Teedub (Guest)
on 2008-12-04 07:24
(Received via mailing list)
I think you can find that answer here:

http://railscasts.com/episodes/3-find-through-association
Ryan B. (Guest)
on 2008-12-04 07:27
(Received via mailing list)
Jay M. (Guest)
on 2008-12-04 16:28
Thanks for the info guys. Now I am getting this error coming from the
author's model:

"syntax error, unexpected tASSOC, expecting ')'"

This is what I have in author's class


class Author < ActiveRecord::Base
   has_many :books

def author_age
    Author.find :all, (:conditions => ["age = ?", params[:authors]])
    end

  end

What I want to do is to display all author's name base on the selected
age.

please, What is wrong with my query?

Thanks
Cypray
Frederick C. (Guest)
on 2008-12-04 16:36
(Received via mailing list)
On 4 Dec 2008, at 14:28, Jay M. wrote:
> What I want to do is to display all author's name base on the selected
> age.
>
> please, What is wrong with my query?
>
because hashes are written with {} not () (and in fact you don't even
need the {} here)

Fred
Jay M. (Guest)
on 2008-12-04 18:48
Thanks a lot Mr. Fred.

Now, I am having trouble with displaying the query result on
authors\show.html
I am having problems referencing the returned object and getting the two
columns displayed. The error is: "You have a nil object when you didn't
expect it!"


I have this in my model:
class Author < ActiveRecord::Base

has_many :books

def author_age
    @results = Author.find :all, :conditions => ["age = ?",
params[:authors]]
     end
end


I have this on view\authors\show.html

<table border="1">
 <tr>
 <td width="20%"><p align="center"><i><b>Author Name</b></i></td>
 <td width="20%"><p align="center"><i><b>Age</b></i></td>
 </tr>


 <% @results.each do |result| %>  // I am not sure if this is right!

 <tr>

 <td><%=h @result.name %></td>
 <td><%=h @result.age %></td>
 </tr>

</table>

What I want to do is to display the 'name' and 'age' that the query
returned.

Please help!
Cypray



Frederick C. wrote:
> On 4 Dec 2008, at 14:28, Jay M. wrote:
>> What I want to do is to display all author's name base on the selected
>> age.
>>
>> please, What is wrong with my query?
>>
> because hashes are written with {} not () (and in fact you don't even
> need the {} here)
>
> Fred
Thorsten M. (Guest)
on 2008-12-04 18:59
(Received via mailing list)
<% @results.each do |result| %>  // I am not sure if this is right!

 <tr>

 <td><%=h @result.name %></td>
 <td><%=h @result.age %></td>
 </tr>

should be

<% @results.each do |result| %>  // I am not sure if this is right!

 <tr>

 <td><%=h result.name %></td>
 <td><%=h result.age %></td>
 </tr>
<% end %>
Jay M. (Guest)
on 2008-12-04 19:52
I made those changes, but I am still getting this error: "You have a nil
object when you didn't expect it!"

It is pointing to the line with the loop:
<% @results.each do |result|

My model look like this:

class Author < ActiveRecord::Base

has_many :books

def author_age
    @results = Author.find :all, :conditions => ["age = ?",
params[:authors]]
     end
end

I have this in view\authors\show.html


 <tr>
 <td width="20%"><p align="center"><i><b>Author Name</b></i></td>
 <td width="20%"><p align="center"><i><b>Age</b></i></td>
 </tr>


 <% @results.each do |result| %>  // It is flagging this line.

 <tr>

 <td><%=h result.name %></td>
 <td><%=h result.age %></td>
 </tr>
<% end %>

Is something wrong with the private method in the model or something
else?
Please help.
Cypray


















Thorsten M. wrote:
> <% @results.each do |result| %>  // I am not sure if this is right!
>
>  <tr>
>
>  <td><%=h @result.name %></td>
>  <td><%=h @result.age %></td>
>  </tr>
>
> should be
>
> <% @results.each do |result| %>  // I am not sure if this is right!
>
>  <tr>
>
>  <td><%=h result.name %></td>
>  <td><%=h result.age %></td>
>  </tr>
> <% end %>
Thorsten M. (Guest)
on 2008-12-04 20:06
(Received via mailing list)
> It is pointing to the line with the loop:
> <% @results.each do |result|

So for one reason or another @results is nil
Most likely your query doesn't return any records.
(btw: you better name that @authors instead of @results for
readablity)

So what is in params[:authors] ?
Where does this param come from and does it contain an age?
After all, that's what you ask Rails to search for in
@results = Author.find :all, :conditions => ["age = ?", params
[:authors]]

So if for example params[:authors] would contain an authors id or
name (as the name suggests) then it wouldn't find anything.

You could easily debug this with something like:
@results = Author.find :all, :conditions => ["age = ?", params
[:authors]]
logger.info "RESULT: #{@results.size}"
Then you would find the number of found records in development.log
(Where btw you could see, which params where sent to your controller)
Jay M. (Guest)
on 2008-12-04 21:28
The problem could be the query because when I search for one record, it
works.
So, this works fine without the query and the loop.
<tr>
 <td width="20%"><p align="center"><i><b>Author Name</b></i></td>
 <td width="20%"><p align="center"><i><b>Age</b></i></td>
 </tr>



 <tr>

 <td><%=h @author.name %></td>
 <td><%=h @author.age %></td>
 </tr>
<% end %>

But this returns one 'Name' for the selected 'age'. That means it is
returning the name that is in the same row with the selected age, that
is, having same id.

In the database, there are several names with same age. So, what I want
is a query that will return all names with age value equal to the value
of the selected age and not the id of the selected age.

In SQL, the query will be:
SELECT name, age
FROM authors
WHERE age = 'the selected age'  // the value, not the id of the selected
age.

So, I need help with a query that will return all the names with the
selected age, and also the proper way to display those names and the age
on author\show.html

cypray





Thorsten M. wrote:
>> It is pointing to the line with the loop:
>> <% @results.each do |result|
>
> So for one reason or another @results is nil
> Most likely your query doesn't return any records.
> (btw: you better name that @authors instead of @results for
> readablity)
>
> So what is in params[:authors] ?
> Where does this param come from and does it contain an age?
> After all, that's what you ask Rails to search for in
> @results = Author.find :all, :conditions => ["age = ?", params
> [:authors]]
>
> So if for example params[:authors] would contain an authors id or
> name (as the name suggests) then it wouldn't find anything.
>
> You could easily debug this with something like:
> @results = Author.find :all, :conditions => ["age = ?", params
> [:authors]]
> logger.info "RESULT: #{@results.size}"
> Then you would find the number of found records in development.log
> (Where btw you could see, which params where sent to your controller)
Hassan S. (Guest)
on 2008-12-04 22:50
(Received via mailing list)
On Thu, Dec 4, 2008 at 11:28 AM, Jay M.
<removed_email_address@domain.invalid> wrote:
>
> The problem could be the query because when I search for one record, it
> works.

> In SQL, the query will be:
> SELECT name, age
> FROM authors
> WHERE age = 'the selected age'  // the value, not the id of the selected
> age.

So answer the previous question:

> Thorsten M. wrote:

>> So what is in params[:authors] ?
>> Where does this param come from and does it contain an age?

You can look in your logs (Rails and/or MySQL) to see the query
that's being constructed from your code, if nothing else.

--
Hassan S. ------------------------ removed_email_address@domain.invalid
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