CODEWITHSUNDEEP
phplaraveleloquent
user4898812
user4898812

Reputation:

Laravel Eloquent; Show the most recent record

I'm building a reporting system, to which I import, on a daily basis, a bunch of files into my database. The issue encountered is I may have duplicate but valid records in my database.

In my database I would have the following, example, data set.

ID        NAME         MATERIAL      PRICE        CREATED_AT
1         Foo          FP1234        3.45         2016-01-01 15:00:00
2         Bar          FP8177        7.21         2016-01-01 15:00:00
3         Foo          FP1234        4.87         2016-01-02 15:00:00
4         Baz          FP1231        3.00         2016-01-02 15:00:00

I would like to know how I can return the above result set, but exclude row 1 because it is the same as row 3, and row 3 is more recent. Adding on to my reasoning, row 1 and 3 are the same, but since row 3 has the more recent created_at timestamp, I would only want to see this.

The tricky thing is I need to be able to paginate on top of this, and using Laravel's Collection method forPage is something I was hoping to avoid.

Edit

I can use the Laravel orderBy and groupBy if running a search with an equal too operand, however I need to account for like as well.

Upvotes: 3

Views: 1974

Answers (2)

Alexey Mezenin
Alexey Mezenin

Reputation: 163768

Use groupBy() and orderBy() methods. You can try something like this:

Model::orderBy('created_at', 'desc')->groupBy('name')->get()

Upvotes: 1

Parvez Rahaman
Parvez Rahaman

Reputation: 4397

You can try any of these..

// For getting result with laravel collection.
YourModel::orderBy('CREATED_AT', 'desc')->groupBy('NAME')->get();

// For getting result without laravel collection.
YourModel::orderBy('CREATED_AT', 'desc')->groupBy('NAME')->get()->toArray();

// For getting result with laravel pagination.
YourModel::orderBy('CREATED_AT', 'desc')->groupBy('NAME')->paginate(5);

Upvotes: 2

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