Symfony 1.4 improve doctrine save() method - performance

I have 500 entries in my db. In my backend I have action. For example:
public function executeMyAction(sfWebRequest $request) {
// Get some data from table
$templates = Doctrine_Core::getTable('SeoTemplates')->findOneByEntity('training');
//Get data from other table(500 items)
$trainings = Doctrine::getTable('Training')->getTraining();
// Make some operations with data
foreach ($trainings as $training) {
$training->setSomeValue1('some_data');
$training->setSomeValue2('some_data');
$training->setSomeValue2('some_data');
}
// Problem part (try to save)
$trainings->save();
}
save() performed for a long time. How to solve this problem? Is it possible?
In my problem part I have all known error Fatal error: Maximum execution time of 30 seconds exceeded in

Save each record instead of a collection
$templates = Doctrine_Core::getTable('SeoTemplates')->findOneByEntity('training');
$trainings = Doctrine::getTable('Training')->getTraining();
foreach ($trainings as $training) {
$training->setSomeValue1('some_data');
$training->setSomeValue2('some_data');
$training->setSomeValue2('some_data');
$training->save();
}
or use Doctrine to update the records using a query
$q = Doctrine_Query::create()
->update('TABLE')
->set($val1, '?', $val1)
->set($val2, '?', $val2)
->set($val3, '?', $val3)
->where('id = ?', $id)
->execute();

Related

Cache eloquent model with all of it's relations then convert it back to model with relations?

I am trying to optimize a project that is working pretty slow using caching. I'm facing a problem that I don't quite understand how to cache full eloquent models with their relationships and later on covert them back to a model with all relations intact. Here's a fragment of my code
if (Cache::has($website->id.'_main_page')) {
$properties = (array) json_decode(Cache::get($website->id.'_main_page'));
$page = Page::hydrate($properties);
}else{
$expiresAt = now()->addMinutes(60);
$page = Page::with(['sections', 'pageSections.sectionObjects'])->where('website_id', $website->id)->where('main', 1)->first();
Cache::put($website->id.'_main_page', $page->toJson(), $expiresAt);
}
Problem is, hydrate seems to be casting this data as a collection when in fact it's suppose to be a single model. And thus later on I am unable to access any of it's properties without getting errors that they don't exists. $properties variable looks perfect and I would use that but I need laravel to understand it as a Page model instead of stdClass, I also need all of the relationships to be cast into their appropriate models. Is this even possible? Any help is much appreciated
Is there any reason you can't use the cache like this
$page = Cache::remember($website->id.'_main_page', 60 * 60, function () use ($website) {
return Page::with(['sections', 'pageSections.sectionObjects'])
->where('website_id', $website->id)
->where('main', 1)
->first();
});
Conditional
$query = Page::query();
$key = $website->id.'_main_page';
if (true) {
$query = $query->where('condition', true);
$key = $key . '_condition_true';
}
$query::with(['sections', 'pageSections.sectionObjects'])
->where('main', 1)
$page = Cache::remember($key, 60 * 60, function () use ($query) {
return $query->first();
});

laravel more than one result from single query

I am trying to get all rows and distinct column from single query. but paginate method is only giving result but not pagination option like total prev next etc..
$offers = Offer::whereHas('users', function ($q) use ($authUser) {
$q->where('user_id', $authUser->parent_id);
$q->where('publisher_id', '=', $authUser->id);
});
and distinct column
$websites = $offers->distinct()->get(['website']);
with pivot columns (just wanted to show my full query)
$offers->orderBy($sortBy, $orderBy)->paginate($perPage)->map(function ($offer) {
if (!empty($offer->users)) {
$manager = $publisher = '';
foreach ($offer->users as $user) {
$manager = $user->pivot->user_id;
$publisher = $user->pivot->publisher_id;
}
$offer->manager = $manager;
$offer->publisher = $publisher;
}
return $offer;
});
Return
return response()->json([
'offers' => $offers,
'websites' => $websites
], 200);
hope my question will make sense.
Thanks.
You should run getCollection() before mapping to get the paginator's underlying collection.
(https://laravel.com/api/7.x/Illuminate/Pagination/LengthAwarePaginator.html#method_getCollection)
$offers->orderBy($sortBy, $orderBy)->paginate($perPage)
->getCollection()
->map(function ($offer) {
// ...
return $offer;
});
I'm answering based on it being $offers:
Your usage of map() is copying the modified results of your paginate() call to a new collection and that collection does not include the pagination information. That's why you no longer have pagination information.
Since there result of paginate() is already a usable collection, you can use each() instead of map() which will alter the objects in-place.

Any short way for handling multiple queries in laravel?

I need to get multiple results from different queries on one table.
For example I need to get Count, Sum, Average of one table. Should I do like this or is there a shorter way?
public function index()
{
$count = Patient::all()->count();
$dateCount = Patient::where('date', date("Y-m-d"))->count();
$loanAmount = DB::table('patients')->sum('loan_amount');
$payAmount = DB::table('patients')->sum('pay_amount');
return view('index', compact('count','dateCount','loanAmount' ,'payAmount'));
}
If you see All queries are for one table to get specific results, So basically is there a short way to get these results not by single queries for each?
You can do this by DB query as below :
$data=\DB::table('patients')
->selectRaw('count(id) as count,sum(loan_amount) as loanAmount,sum(pay_amount) as payAmount,sum(case when date = "'.date("Y-m-d").'" then 1 else 0 end) AS dateCount')
->first();
You can also do this using eloquent but it will return you collection.
$data=Patient::selectRaw('count(id) as count,sum(loan_amount) as loanAmount,sum(pay_amount) as payAmount,sum(case when date = "'.date("Y-m-d").'" then 1 else 0 end) AS dateCount')
->first();
You can do something like this.
public function index()
{
$patient = Patient::all();
$count = $patient->count();
$dateCount = Patient::today()->count();
$loanAmount = $patient->sum('loan_amount');
$payAmount = $patient->sum('pay_amount');
return view('index', compact('count','dateCount','loanAmount' ,'payAmount'));
}
Also you can create scope in your patient model:
public function scopeToday($query) {
return $query->where('date', date("Y-m-d"));
}

Laravel simplePaginate() for Grouped Data

I have the following query.
$projects = Project::orderBy('created_at', 'desc');
$data['sorted'] = $projects->groupBy(function ($project) {
return Carbon::parse($project->created_at)->format('Y-m-d');
})->simplePaginate(5);
When I try to paginate with the simplePaginate() method I get this error.
stripos() expects parameter 1 to be string, object given
How can I paginate grouped data in this case?
The created_at attribute is already casted as a Carbon Object (by default in laravel models). that's why you are getting that error. Try this:
$projects = Project::orderBy('created_at', 'desc')->get();
$data['sorted'] = $projects->groupBy(function ($project) {
return $project->created_at->format('Y-m-d');
})->simplePaginate(5);
this answer is just for the error you're getting. now if you want help with the QueryBuilder, can you provide an example of the results you're expecting to have and an example of the database structure ?
The pagination methods should be called on queries instead of collection.
You could try:
$projects = Project::orderBy('created_at', 'desc');
$data['sorted'] = $projects->groupBy('created_at');
The problem was solved. I was create custom paginator via this example:
https://stackoverflow.com/a/30014621/6405083
$page = $request->has('page') ? $request->input('page') : 1; // Use ?page=x if given, otherwise start at 1
$numPerPage = 15; // Number of results per page
$count = Project::count(); // Get the total number of entries you'll be paging through
// Get the actual items
$projects = Project::orderBy('created_at', 'desc')
->take($numPerPage)->offset(($page-1)*$numPerPage)->get()->groupBy(function($project) {
return $project->created_at->format('Y-m-d');
});
$data['sorted'] = new Paginator($projects, $count, $numPerPage, $page, ['path' => $request->url(), 'query' => $request->query()]);
simplePaginate Method is exist in the path below:
Illuminate\Database\Eloquent\Builder.php::simplePaginate()

Laravel: How to update pivot tables withou using first

Im new in Laravel. I want to update my leaves pivot table. I am trying with below code but it only updates the single row i have multiple rows in db with same leave_id and i want to update all this where leave_id = xyz
I have following function in my model Leave:
public function relLeave(){
return $this->belongsToMany(User::class)->withPivot('days');
}
LeaveController:
public function saveUpdate(Request $request)
{
$leave = Leave::find($request->id);
$msg = $leave->relLeave()->Where('leave_id', $request->id)->get()->first();
$msg->pivot->days = $request->days;
$msg->pivot->save();
}
I followed #option's instruction and it works for me i removed the first();
below is my updated code.
$msg = $leave->relLeave()->Where('leave_id', $request->id)->get();
foreach($msg as $msgs)
{
$msgs->pivot->days = $request->days;
$msgs->pivot->save();
}
you can update extra fields in pivot table when updating relationship
$leave->relLeave()->sync([$request['id'] => ['days' => $request['days']]]);
You can use Query Builder for that if it's an option:
DB::table('leave_user')->where('leave_id', $request->id)->update(['days' => $request->days]);
This is just one DB query and it's pretty simple one.
If you want Eloquent solution, use updateExistingPivot() in a loop:
$leave = Leave::find($request->id);
$usersIds = $leave->relLeave()->pluck('id')->toArray();
foreach ($usersIds as $userId) {
$leave->relLeave()->updateExistingPivot($userId, ['days' => $request->days]);
}

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