Help with LinqtoSql - linq-to-sql

Im using the Repository pattern and I want to write a method that receives a role and returns an Iqueryable of the users that belong to that role. (Im not sure if the right way would be to receive the role object or the role_id... in any case, how can I do this?? I dont like the query structure, I prefer the method structure of linq.
users and roles is many to many with a users_roles join table.
private ClasesDataContext db = new ClasesDataContext();
public IQueryable GetByRole(Role role)
{
return db.Users.Where();
}

Maybe try something like:
public IQueryable<User> GetByRoleId(Role role) {
return db.UsersRoleJoinTable.Where(ur => ur.Role == role).select(ur => ur.User);
}
Where UsersRoleJoinTable is your many-to-many join table.
Hope it helps.
Update: the select(ur => ur.User) is telling linq that for every row returned by "db.UsersRoleJoinTable.Where(ur => ur.Role == role)" we want to get the user associated with the UsersRoleJoinTable object. If you wanted a list of user ids instead, you could tell linq to select only user.id by doing select(ur => ur.id). Think of linq's select as a some sort of "for every row do this and put it in the list returned instead of the original row"
There is one downside to this approach tho, I believe in this case Linq is generating the sql to get the rows from the Join table (UsersRoleJoinTable) and then for every row returned, is executing another query to look up the User. I might be wrong on this, so to check the SQL generated by Linq do:
string sql_query = db.UsersRoleJoinTable.Where(ur => ur.Role == role).select(ur => u.User).ToString();
and then print the value of sql_query or watch it in debug mode. If Linq is in fact doing multiple queries, then I think the best solution is to create a view or stored procedure in SQL Server to get the users associated with the role and then add the view or stored procedure to Visual Studio designer so that you can call the view like:
db.GetUsers(role_id) //if using a GetUsers stored procedure
or
db.UsersByRoleView.where(ur => ur.role_id == passed_role_id) //if using a UsersByRoleView view

If you have an instance of the Role object
public IQueryable<User> GetByRole(Role role) {
return db.Users.Where(u => u.Role == role);
}
would work.
If you don't but just know the Id or some other property of the role something like this might be better.
public IQueryable<User> GetByRoleId(int roleId) {
return db.Users.Where(u => u.Role.Id == roleId);
}

Related

Laravel Join Query returning empty column

I have a member table with some entries. Each member can create a user account in laravel's users table.
They each have a field called "person_id" and that's how the connection is made.
I have a search that returns a list with all of them. I have a checkbox "Search only registered" that means it returns only members that have users account, otherwise if the check doesn't check, return a mix with all of them.
The thing is, no matter if the checkbox is checked or not, the person_id must be pulled for each one.
if($reg == 'on') {
$Members = $Members->rightJoin('users', 'users.person_id', '=', 'members.person_id');
}
else {
$Members = $Members->leftJoin('users', 'users.person_id', '=', 'members.person_id');
}
I tried with leftJoin but person_id comes empty
at first look if you are using Eloquent i can tell you are missing the "->get();" at the end of each query.
Hope this helps.
Use relation in member model:
public function user()
{
return $this->belongsTo('App\User', 'person_id', 'person_id' );
}
public function getMemberWithUser()
{
return $this->select('*')->with('user')->get()->toArray();
}
and use (new Member)->getMemberWithUser(); in controller. That will return you member detail with user.
Neverming guys I found it out.
Most members don't have yet a user account, only 2. And the select wasn't specifying which table to take the person_id from. And with most members missing an user account, it was trying to get it from users.
I did this:
$Participants = Member::select(
'members.first_name',
'members.last_name',
'members.person_id',
'members.email'
);

Yii2 build relation many to many

Yii2 build relation many to many
I have 2 tables users and friends
Code query
$friends = Friends::find()
->select(['friends.user_id', 'users.name'])
->leftJoin('users','users.id = friends.friend_user')
->with('users')
->all();
In result error
Invalid Parameter – yii\base\InvalidParamException. app\models\Friends has no relation named "users".
Friends has a column called user_id and thus only belongs to one user. If you auto-generated the Friends ActiveRecord it probably has a function getUser (singular because it is only one) that will look something like this:
public function getUser() {
return $this->hasOne(User::className(), ['id' => 'user_id']);
}
So you're getting the error because no getUsers function exists (that returns a valid ActiveQuery object). Because there can only be one user per friend I think you should use the singular version. And if that still gives the same error you should implement the function above and maybe change it a bit to match your classname.
When you use with(['relation']) to load relations Yii will convert the entry to getRelation and call that function on the model to get the query that is needed to load the relation.

CakePhp mysql raw query error

I'm new to cakephp. I'm trying to search through mysql tables. I want to use nested query.
class TableController extends AppController{
.
.
public function show(){
$this->set('discouns', $this->DiscounsController->query("SELECT * FROM discoun as Discoun WHERE gcil_id = 1"));//(SELECT id FROM gcils WHERE genre = 'Shoes' AND company_name = 'Adidas')"));
}
}
Error:
Error: Call to a member function query() on a non-object
I've also tried
public function show(){
$this->DiscounsController->query("SELECT * FROM count as Count WHERE ctr_id = (SELECT id FROM ctrs WHERE genre = 'Shoes' AND company_name = 'Adidas')");
}
Error:
Error: Call to a member function query() on a non-object
File: C:\xampp\htdocs\cakephppro\myapp\Controller\CountsController.php
Please help. I've been trying this for last few hours. :/
As mentioned in the comments there are a few problems with your code.
Firstly, you are trying to call the query() method on a Controller, whereas you should be executing it on a Model, as it is models that handle database queries and the controller should simply be used to call these methods to get the data and pass them to the view.
The second thing is that you are executing a very simple SQL query raw instead of using CakePHPs built in functions <- Be sure to read this page in full.
Now for your problem, as long as you have setup your model relationships correctly and followed the correct naming conventions, this should be your code to run your SQL query from that controller:
public function show(){
$this->set('discouns', $this->Discouns->find('all', array(
'conditions' => array(
'gcil_id' => 1,
'genre' => 'shoes',
'company_name' => 'Adidas'
)
));
}
query() is not a Controller, but a Model method. That's what the error (Call to a member function on a non-object) is trying to tell you.
So the correct call would be:
$this->Discount->query()
But you are calling this in a TableController, so unless Table and Discount have some type of relationship, you won't be able to call query().
If the Table does have a relationship defined you will be able to call:
$this->Table->Discount->query()
Please not that query() is only used when performing complex SQL queries in scenarios where the standard methods (find, save, delete, etc.) are less practical.
$this->Counts->find('all',array(
'conditions' => array(
'ctrs.genre' => 'Shoes',
'ctrs.company_name' => 'Adidas'
), 'recursive' => 1
));
The above is with tables named counts and ctrs.
This is assuming you have the model set up to have some sort of relationship between the counts table and the ctrs table. It's kind of hard to tell in your code exactly what you tables are.
The CakePHP book should have all the answers you need. One of the reasons to run CakePHP over regular PHP is the FIND statement. Once you have your models set up correctly, using the find statement should be really easy.
http://book.cakephp.org/2.0/en/models.html

EF4.1 Eager loaded linked objects are returned null

Can anyone explain why a Company is returned but Company.CompanyServices is null (even though I've created one in the test)?
public List<Company> GetContactCompanies(int contactId)
{
var query = (
from directorCompany in ctx.CompanyDirectors
.Where(d => d.ContactAddress.Contact.Id == contactId)
.Include(d => d.Company.CompanyServices)
select directorCompany.Company
).OrderBy(c => c.CompanyName).Distinct();
return query.ToList();
}
Note substituting the Include for .Include("Company.CompanyServices") has no effect
Is the Company.CompanyServices property marked as virtual? Check out ScottGu's blog on entity framework, where he creates POCO classes with one to many relationships he marks the collection properties as virtual.
When I first started using EF 4, that had me stumped for quite a while.
Obviously I can't see your entity classes so this may be a moot point!
Found an answer that's not wholly inuitive, but it works which is the main thing.
Happy to see one that plays on the original query...
var query = (
from directorCompany in ctx.CompanyDirectors
.Where(d => d.ContactAddress.Contact.Id == contactId)
select directorCompany.Company
).OrderBy(c => c.CompanyName).Distinct();
return query.Include(c => c.CompanyServices).ToList();

Codeigniter Noob Question--active record and results?

I'm writing a simple app. I need to block user from a page if their credit is < 0.
I have a table "User_profiles" with a "credit" row.
How can I set up a model in conjunction with the controller to send the user to another page if the value of "credit" is 0?
This should be straightforward, but I'm new at the select->where stuff...
It has to be the row of the current user too--I don't know how to traverse arrays very well yet.
Thanks!
Well, the easiest solution would be to just load a different view...
As for the model, it would look like this:
class UserModel extends Model {
public function getUserCredit($id) {
$this->load->database();
//effectively generates: SELECT `credit` FROM `User_profiles` WHERE `id`=$id
$query = this->db->select('credit')->where('id',$id)->get('User_profiles');
//row() executes the query for a single result, returns the credit property
return $query->row()->credit;
}
}
Then in the controller:
class Users extends Controller {
//....
public function credit() {
$this->load->model('userModel','users');
// assuming the session library has been loaded
$user_id = $this->session->userdata('id');
$credit = $this->users->getUserCredit($user_id);
if ($credit == '0') $this->load->view('users/no_credit');
else $this->load->view('users/credit');
}
}
That's untested, but it should at least help you get the idea.
When you request the page /users/credit/1, CI will call the Users::credit(1) action.
It then loads UserModel as $this->users
You call $this->users->getUserCredit(1), which translates to UserModel::getUserCredit(1), to store as $credit
The model loads the database.
You tell the db to select('credit') (select the credit column), where('id',1) (where the id = 1), then get('User_profiles') (get matching rows from the User_profiles table). That returns a query, which you store as $query for readability.
getUserCredit returns the credit property of the single-row result of the query
If $credit == 0, you load the view views/users/no_credit.php
Otherwise, you load the view views/users/credit.php (it's conventional to name the views after the actions they represent and put them in a folder corresponding to the controller)