I need to delete and detach all relationship between BrandName table and stellar_category table.
public function destroy($id)
{
$brand_name = BrandName::find($id);
$brand_name->stellar_category->detach();
$brand_name->delete();
return back()->withMessage('Your Brand is Deleted');`enter code here`
}
Try to change it to
$brand_name->stellar_category()->detach();
Related
public function complaint_pending(){
$complaints = Complaint::whereNull('answear')->get();
return view('admin.complaint.panding',compact('complaints'));
}
I want to show Complaint all data which answer row is NULL
My DB pic is below
enter image description here
you have type error in function.
change answear to answer in complaint_pending method.
try the following code
public function complaint_pending(){
$complaints = Complaint::where('answear',NULL)->get();
return view('admin.complaint.panding',compact('complaints'));
}
That's not work but it's work perfectly ...
public function complaint_pending(){
$complaints = Complaint::where('answear', NULL)->get();
return view('admin.complaint.panding',compact('complaints'));
}
I have 2 tables called jobs & job_records. Its relationship as below:
JobRecords Model
public function job()
{
return $this->belongsTo(Job::class);
}
Job Model:
public function jobRecord()
{
return $this->hasOne(JobRecord::class);
}
jobs table has 2 columns that I need to display alongside my job_records table view. It's total_pges & status.
In my JobRecords Controller, I have tried the following method. It throws me an error of Call to undefined relationship.
JobRecordController:
$job_records = JobRecord::whereStatus('In Progress')
->with('jobs', 'jobs.status', 'jobs.total_pges')
->get();
return DataTables::of($job_records)
I am still beginning with Laravel and PHP. I can sense that there is something wrong with the relationship. But I couldn't figure out what it is exactly. Can anyone help me out with this matter?
In your JobRecord model change the relation ship as
public function job()
{
return $this->hasOne('App\Models\Job','foreign_key','local_key');
}
Similarly, in Job model
public function job()
{
return $this->belongsTo('App\Models\JobRecord','foreign_key','local_key');
}
Replace foreign_key and local_key with appropriate values...
I deleted my previous answer. What are you trying to do exactly? You can't use "jobs" in the "with function" without to define "jobs" as function in the model.
If you change it to "job" (instead of "jobs), then it would work, but I don't know if you want this. With your query you saying that a record have many jobs? But your model doesn't define that.
I Have three tables
#1 Table timeline which is my reference table with an Auto incremented ID which is stored in column id
#2 timeline_videos
#3 timeline_else
What happens is on post if a video is uploaded with the post
it will go into Table #2 ,anything else goes into Table #3.
#2-3 have the Auto Increment Id from the Table timeline stored in a column pid
On query of The Timeline I need to join both tables data using id=pid
so I can use the rest of the Relational Data with the post.
I have done a bit of research and can't seem to find a method for doing so.
So far the code I have
Controller
$groupposts = timeline::where([
['owner','=',$owner],['id','<',$lastid],
])
->join('timeline_videos','timeline.id','=','timeline_videos.pid')
//->join('timeline_else','timeline.id','=','timeline_else.pid')
->orderBy('id','desc')
->limit(5)
->get();
This works with no errors with the second Join commented out but I need to also grab the timeline_else data .
Update --
I have now decided to use Eloquent Relationships to join the tables,
my question now is what type of relationship do I have between the
tables?
I realize it basically needs to be able to switch between two tables to
grab data based on the fact that timeline_videos and timeline_else will not be "JOIN" but separated by type .
The tables need to Join with table #1 timeline based on a column I now have named type for clarifying where to look and matching/joining using the id = pid
You can use relationships.
it sounds like timelines has many videos and has many video failures
https://laravel.com/docs/5.5/eloquent-relationships#one-to-many
you would have a model for each table and set up the relationships
timelines model:
public function videos()
{
return $this-> hasMany('App\Videos');
}
public function videoFailures()
{
return $this-> hasMany('App\videoFailures');
}
videos model:
public function timeline()
{
return $this->belongsTo('App\Timelines');
}
videos failures model:
public function timeline()
{
return $this->belongsTo('App\Timelines');
}
You can then go:
$timeLine = Timmeline::find($id);
to find videos of the time lines you would do:
$videos = $timeLine->videos();
to find else:
$videoElse = $timeLine-> videoFailures();
By using some of what Parker Dell supplied and a bit more trial and error
My Models Looks like
timeline
class timeline extends Model
{
protected $table ='timeline';
public $timestamps = false;
public function videos()
{
return $this->hasMany('App\timeline_videos','pid','id');
}
public function else()
{
return $this->hasMany('App\timeline_ect','pid','id');
}
}
timeline_ect.php ,I changed the name of the table
class timeline_ect extends Model
{
protected $table='timeline_ect';
public $timestamps = false;
public function timeline()
{
return $this->belongsTo('App\Models\timeline','pid','id');
}
}
timeline_videos
class timeline_videos extends Model
{
protected $table='timeline_videos';
public $timestamps = false;
public function timeline()
{
return $this->belongsTo('App\timeline','id','pid');
}
}
Then Lastly my Controller
$timeline = timeline::with('videos','else')
->orderBy('id','desc')
->limit(5)
->get();
So far no Problem query is correct.
I have two table witch named users & Inbox
In the Inbox table I have a column named sender_id that have the user_id of the sender
I want to show this message in the view. I need a query to get the sender_id from the inbox table and use that to select a certain user from the users table
I need to do this with all messages and all users.
Laravel is basicly straith foward when you use eloquent. You can always customise it.
First, almost all the time, I create a model and a migration at the same time using this : php artisan make:model Something --migration
I know you already make some models and/or migrations, but I'll go step by step to help you understand it.
So, in your case, it'll be php artisan make:model User --migration and php artisan make:model Inbox --migration. Doing this, you get two model named User and Inbox and two migration named date_create_users_table.php and date_create_inboxs_table.php. Maybe you already did the default user table with php artisan make:auth. If it's the case, don't remake one.
I'm not sure about how laravel will name the Inbox model migration... Since, I think, Laravel 5.3, the plurialisation changed and don't always just add an "S" at the end.
Then, now you got your models and migrations, let's add some line into your migration files. Since you want to do a one to many relationship. You don't need to touch the user one. Only the Inbox migration. Each Inbox is related to one User and Users can have many Inboxs. Add something like this in your migration:
public function up()
{
Schema::create('inboxs', function (Blueprint $table) {
$table->increments('id');
$table->integer('user_id');
$table->foreign('user_id')->references('id')->on('users');
all other columns...
});
}
There, you can change the column's name if you need to have a sender, a recipient, etc... Do this instead :
public function up()
{
Schema::create('inboxs', function (Blueprint $table) {
$table->increments('id');
$table->integer('sender_id');
$table->foreign('sender_id')->references('id')->on('users');
$table->integer('recipient_id');
$table->foreign('recipient_id')->references('id')->on('users');
all other columns...
});
}
What we just did, it's creating the Foreign key that Laravel will use to build the query. There is one last part before the fun one. We need to create the relation in our Model. Begin with the user one:
App/User.php
public function inboxs() {
return $this->hasMany(Inbox::class);
}
And now into the App/Inbox.php model:
public function user() {
return $this->belongsTo(User::class);
}
If you need to have a Sender/Recipient/etc... go this way instead:
public function sender() {
return $this->belongsTo(User::class);
}
public function recipient() {
return $this->belongsTo(User::class);
}
Note that each of your function need to be writen in the same way it's into your migration. sender_id need a relation named sender().
Now, that our relations are done, we can simply call everything using eloquent.
$inboxs = Inbox::with('sender')->get();
This will return an array of all your Inbox into the inboxs table. You can access the sender this way: $inboxs[0]->sender();
You need the id, do this: $sender_id = $inboxs[0]->sender_id;
The sender name : $sender_name = $inboxs[0]->sender->name;
If you want to get one Inbox and you have the id, just do this $inbox = Inbox::with('sender')->find($id);
This way you don't get an array, only one result and can access the sender directly using $sender_name = $inbox->sender->name; instead of having to add [0] or using a foreach loop.
You can get all messages sended by a user using something like this:
$inboxs = Inbox::where('sender_id', $sender_id)->get();
Finally, you can pass your data to the view using:
return view('path.to.view')->with('inbox',$inbox);
Into the view you do this to show the sender's name:
//If view.blade.php
{{$inbox['sender']['name']}} //work a 100%
{{$inbox->sender->name}} //I'm not sure about this one
//If not using blade
<?php echo $inbox['sender']['name']; ?>
There is a lot of thing you can do using Eloquent and you can add as much condition you want. The only thing I suggest you to really do if you want to use Eloquent, be aware about the n+1 problem. There is a link where I explain it. Look for the EDIT section of my answer.
If you need some documentation:
Laravel 5.3 Relationships
Laravel 5.3 Migrations
Laravel 5.3 Eloquent
I think you should update your code like:
$user_messages = DB::table('messages')
->select('messages.id as msgId','messages.message as message','users.id as userId','users.user_name as user_name')
->join('messages','messages.user_id','=','users.id')
->where('messages.user_id',$user_id)
->get();
return view("view.path")
->with('messages',$user_messages);
Hope this work for you!
In Model :
namespace App;
use Illuminate\Database\Eloquent\Model;
class Messages extends Model
{
protected $table = 'table_name';
public function sender()
{
return $this->belongsTo('App\User', 'sender_id', 'id');
}
}
In Controller :
public function functionName($user_id){
$messages = Messages::where('sender_id', $user_id)->get();
return view("view.path")
->with('messages',$messages);
}
In view, you can access seder details like this $message->sender->name for name for id $message->sender->id
My understanding is that find only takes the primary key as the parameter. That works great if the value you are looking for is actually the primary key. In my case, I have a class like this:
public class Chamber
{
[Key]
public int Id {get;set;}
public string ChamberName { get; set; }
}
I want to check whether a given ChamberName exists in either my context or the database itself. How can I do that? Do I have to somehow enumerate of the context myself first, then, look it up in the database with a call like db.Chambers.where(a=>a.ChamberName.equals...?
I can see it working well if ChamberName is my primary key, but it is not.
THanks,
There is a property called Local in the DbSet. You can query that first to find entities loaded to the context.
var entity = db.Chambers.Local.Where(/**/).SingleOrDefault();
if (entity == null)
{
entity = db.Chambers.Where(/**/).SingleOrDefault();
}
You can't use the .Find() method - but how about:
public Chamber FindByChamberName(string chamberName)
{
using(MyDbContext ctx = new MyDbContext())
{
Chamber result = ctx.Chambers
.FirstOrDefault(c => string.Compare(c.ChamberName, chamberName, true));
return result;
}
}
You don't have to manually enumerate anything - just retrieve the first occurence of a chamber by that name - or none.
If you just need to know whether a given chamber (specified by its ChamberName) exists or not, you could use the .Any() method in Linq:
using(MyDbContext ctx = new MyDbContext())
{
return ctx.Chambers.Any(c => string.Compare(c.ChamberName, chamberName, true));
}