I have composite primary keys so Laravel's save() method didn't work. Therefore I need to override the save method. For example, my primary key is a composite key consisting of column_a and column_b. Can you give me example how to override the save method and where I put it?
My Primary Key: column_a, column_b
I've tried the followings:
protected function setKeysForSaveQuery(Builder $query)
{
parent::setKeysForSaveQuery($query);
$query->where('locale', '=', $this->locale);
return $query;
}
I found above code, but it causes 500 Internal Server Error even I call a method that only reads value from database. I have set $primaryKey = 'column_a' and 'locale' (inside where) to column_b. I didn't know what $this->locale refers to?
Here is another way I found, but it failed too
protected $secondaryKey = 'column_b';
function newQuery()
{
$query = parent::newQuery();
$query->where($this->secondaryKey, '=', $this->type);
return $query;
}
Again, I didn't know what $this->type refers to.
I have found the solution. In the model, add primaryKey variable and the following function
protected $primaryKey = array('column1','column2');
protected function setKeysForSaveQuery(\Illuminate\Database\Eloquent\Builder $query) {
if (is_array($this->primaryKey)) {
foreach ($this->primaryKey as $pk) {
$query->where($pk, '=', $this->original[$pk]);
}
return $query;
}else{
return parent::setKeysForSaveQuery($query);
}
}
Source: https://github.com/laravel/framework/issues/5517#issuecomment-113655441
You can override it. You can put it in your model, or you can put it in a trait and use this trate in multiple models if you want.
The original save() method is here (line 832), maybe this will help.
Related
I want to get all user's online friends, how can I call a custom model function inside the eloquent condition?
this is my code
$friends = $user->friends()->where(function (Builder $query){
$query->where('friend', 'yes');
})
->get();
and this is my function in model
public function getIsOnlineAttribute(): bool
{
// check if the user is online or not
return $this->is_online;
}
I can access is_online after eloquent by foreach, but in my case, I want to check everything in one step ( inside where condition in eloquent). how can I do that???
You can't use conditions for eloquent accessors, in this case you can use (assume 1 is database column value):
$friends = $user->friends()->where('is_online', 1)->get();
or
$friends = $user->friends()->whereIsOnline(1)->get();
or you can create eloquent scope on your model:
public function scopeIsOnline($query) {
$query->where('is_online',1);
}
and you can use this eloquent scope on your controller in this way:
$friends = $user->friends()->isOnline()->get();
this worked for me :)
$friends = $user->friends()
->simplePaginate()
->reject(function ($friend) {
return $friend->is_online === false;
});
On my User model I have the following:
public function isOnline()
{
return $this->hasMany('App\Accounting', 'userid')->select('rtype')->latest('ts');
}
The accounting table has activity records and I'd like this to return the latest value for field 'rtype' for a userid when used.
In my controller I am doing the following:
$builder = App\User::query()
->select(...fields I want...)
->with('isOnline')
->ofType($realm);
return $datatables->eloquent($builder)
->addColumn('info', function ($user) {
return $user->isOnline;
}
})
However I don't get the value of 'rtype' for the users in the table and no errors.
It looks like you're not defining your relationship correctly. Your isOnline method creates a HasMany relation but runs the select method and then the latest method on it, which will end up returning a Builder object.
The correct approach is to only return the HasMany object from your method and it will be treated as a relation.
public function accounts()
{
return $this->hasMany('App\Accounting', 'userid');
}
Then if you want an isOnline helper method in your App\User class you can add one like this:
public function isOnline()
{
// This gives you a collection of \App\Accounting objects
$usersAccounts = $this->accounts;
// Do something with the user's accounts, e.g. grab the last "account"
$lastAccount = $usersAccounts->last();
if ($lastAccount) {
// If we found an account, return the rtype column
return $lastAccount->rtype;
}
// Return something else
return false;
}
Then in your controller you can eager load the relationship:
$users = User::with('accounts')->get(['field_one', 'field_two]);
Then you can do whatever you want with each App\User object, such as calling the isOnline method.
Edit
After some further digging, it seems to be the select on your relationship that is causing the problem. I did a similar thing in one of my own projects and found that no results were returned for my relation. Adding latest seemed to work alright though.
So you should remove the select part at very least in your relation definition. When you only want to retrieve certain fields when eager loading your relation you should be able to specify them when using with like this:
// Should bring back Accounting instances ONLY with rtype field present
User::with('accounts:rtype');
This is the case for Laravel 5.5 at least, I am not sure about previous versions. See here for more information, under the heading labelled Eager Loading Specific Columns
Thanks Jonathon
USER MODEL
public function accounting()
{
return $this->hasMany('App\Accounting', 'userid', 'userid');
}
public function isOnline()
{
$rtype = $this->accounting()
->latest('ts')
->limit(1)
->pluck('rtype')
->first();
if ($rtype == 'Alive') {
return true;
}
return false;
}
CONTROLLER
$builder = App\User::with('accounting:rtype')->ofType($filterRealm);
return $datatables->eloquent($builder)
->addColumn('info', function (App\User $user) {
/*
THIS HAS BEEN SUCCINCTLY TRIMMED TO BE AS RELEVANT AS POSSIBLE.
ARRAY IS USED AS OTHER VALUES ARE ADDED, JUST NOT SHOWN HERE
*/
$info[];
if ($user->isOnline()) {
$info[] = 'Online';
} else {
$info[] = 'Offline';
}
return implode(' ', $info);
})->make();
Laravel has a built-in method called findOrfail() described here:
Not Found Exceptions
Sometimes you may wish to throw an exception if a model is not found. This is particularly useful in routes or controllers. The findOrFail and firstOrFail methods will retrieve the first result of the query. However, if no result is found, a Illuminate\Database\Eloquent\ModelNotFoundException will be thrown:
$model = App\Flight::findOrFail(1);
$model = App\Flight::where('legs', '>', 100)->firstOrFail();
But what if you want to write your own code which deals with no result found. So in my controller I want something like:
public function show($id)
{
$data = JobCardHead::where('JobCardNum', $id)->first();
If no data found then
some code
else
some code
What does Eloquent return in case no model is found for the methods that do not throw an exception? I'm guessing it's either null or false. So, you could check the return value of the first() call:
$data = JobCardHead::where('JobCardNum', $id)->first();
if ($data) {
// found it
} else {
// not found
}
Or, knowing that firstOrFail() throws an exception, you could wrap the call in a try/catch block:
use Illuminate\Database\Eloquent\ModelNotFoundException;
//..
try {
$data = JobCardHead::where('JobCardNum', $id)->first();
} catch (ModelNotFoundException $e) {
// Data not found. Here, you should make sure that the absence of $data won't break anything
}
// Here $data is an instance of the model you wanted
The exact approach you should pick actually depends on your usecase a lot.
UPD. By the way, if $id here is your primary key here, you should simply use find($id) or findOrFail($id), as described here
Laravel returns null on first() calls to your model when no records are returned.
https://github.com/laravel/framework/blob/5.2/src/Illuminate/Database/Query/Builder.php#L1548
So for example you could do this as an even cleaner solution:
public function show($id)
{
if ($job = JobCardHead::where('JobCardNum', $id)->first()) {
// Record was found!
return view('job.show', compact('job'));
}
return view('404');
}
Is it possible to return an value for an hasOne relation directly with an Model?
For example:
$List = Element::orderBy('title')->get();
The Element has a "hasOne" Relation to an column:
public function type()
{
return $this->hasOne('App\Type', 'id', 'type_id');
}
How can i now return automatically the "type" for the Model?
At the Moment i am looping through all Elements, and build my own "Array" of Objects, including the Key of "type" in this example. But ill prefer to do this only in my Model.
Ill know how to add a "normal" property, but can it be someone from an relation?
public function getTypeAttribute($value)
{
return // how ?
}
protected $appends = array('type');
Is this possible?
Edit:
A workaround could be to use DB:: to return the correct value - but ill dont thing thats a good workaround: like:
public function getTypeAttribute($value)
{
// make a query with $this->type_id and return the value of the type_name
}
protected $appends = array('type');
you need to eager load your relations when getting the Element:
$list = Element::with('type')->orderBy('title')->get();
then access the type using
foreach ($list as $item) {
echo $item->type->type_name;
}
where type_name would be the name of a column in the types table
Make a query scope and in that scope, join your attributes from other tables.
http://laravel.com/docs/5.1/eloquent#query-scopes
http://laravel.com/docs/5.1/queries#joins
I'm working on a project with a bit of a complex model that has joins in its relations and also requires a parameter. It all works pretty well, except for when I need to eager load the relationship, as I couldn't figure out if there is a way to pass a parameter/variable to it.
The Controller
$template = Template::find($request->input('id'));
$this->output = $template->zones()->with('widgets_with_selected')->get();
The Model
public function widgets_with_selected($banner_id)
{
return $this->belongsToMany('App\Models\Widget', 'zone_has_widgets')
->leftJoin('banner_has_widgets', function($join) use($banner_id) {
$join->on('widgets.id', '=', 'banner_has_widgets.widget_id')
->where('banner_has_widgets.banner_id', '=', $banner_id);
})
->select('widgets.*', 'banner_has_widgets.banner_id');
}
This is returning a Missing argument error as the variable is not being passed.
I have resolved the issue by moving the logic to the controller, but I want to know if there is a way to keep the relationship in the model and just call it with a parameter.
Looking at the laravel code I dont think this is possible as you'd like to do it. You simply cant pass parameters to a with() call.
A possible workaround is to have an attribute on your model for $banner_id.
$template = Template::find($request->input('id'));
$template->banner_id = 1;
$this->output = $template->zones()->with('widgets_with_selected')->get();
Then change your relationship
public function widgets_with_selected()
{
return $this>belongsToMany('App\Models\Widget','zone_has_widgets')
->leftJoin('banner_has_widgets', function($join) use($this->banner_id) {
$join->on('widgets.id', '=', 'banner_has_widgets.widget_id')
->where('banner_has_widgets.banner_id', '=', $banner_id);
})
->select('widgets.*', 'banner_has_widgets.banner_id');
}
You could perhaps alter it a bit by passing the banner_id through a method. Sortof like this in your model:
public function setBanner($id) {
$this->banner_id = $id;
return $this;
}
Then you can do:
$template->setBanner($banner_id)->zones()->with('widgets_with_selected')->get();
Not sure if this works, and it's not really a clean solution but a hack.