I have a table Quotes
that has a one-to-many relationship with the QuoteDetails
table.
The table structure looks like this:
return $this->quotes->with('quotedetails')
->orderBy('created_at', 'DESC')->paginate(10);
Results in:
{
"current_page":1,
"data":[
{
"id":2,
"subject":"demo",
"quotedetails":[
{
"id":2,
"quotes_id":2,
"description":"testing data",
"total":1080
},
{
"id":3,
"quotes_id":2,
"description":"testing",
"total":180
}
]
},
{
"id":1,
"subject":"demo",
"quotedetails":[
{
"id":1,
"quotes_id":1,
"description":"data",
"total":360
}
]
}
],
"per_page":10,
"prev_page_url":null,
"to":2,
"total":2
}
I want to include the sum of quotedetails.total
rather than the full list of QuoteDetails
rows. I am not sure how this should be done.
Here is how I am querying in my model ( Quotes
)
public function quotedetails(){
return $this->hasMany('App\Models\QuotesDetail', 'quotes_id', 'id');
}
How can i get the total
sum.
Here is how i expect the output
{
"current_page":1,
"data":[
{
"id":2,
"subject":"demo link data",
"quotesum": 1260
},
{
"id":1,
"subject":"demo",
"quotesum": 360
}
],
"per_page":10,
"prev_page_url":null,
"to":2,
"total":2
}
$query->withCount([
'quotedetail AS quotesum' => function ($query) {
$query->select(DB::raw("SUM(total) as quotesum"));
}
]);
I understand that you want to do it by query, But taking this without even query you'll be able to transform the paginated data this way, giving this example:
$quotes = Quotes::whereNotNull('id')->with('quotedetails')
->orderBy('created_at', 'DESC')->paginate();
//transform the data in the Paginate instance
$quotes->getCollection()->transform(function($quote){
$quote->quotesum = $quote->quotedetails->sum('total');
unset($quote['quotedetails']);
return $quote;
});
The data in the paginated result is transformed and you have what you requested.
Pro: More possibilities to do more with your response without adding any query.
Con: More data in the result mean more work on the collection especially in case of large data.
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