Return all filename from database by specific field name

Question

I'm trying to write function which will get all filename for particular job_id.
DATABASE: analytics_report
| id | job_id | filename |
|----|--------|----------|
| 1  | abcd   | file1    |
| 2  | abcd   | file2    |
| 3  | bcde   | file3    |

CONTROLLER:
public function downloadZip(Request $request)
{
@$job_id = $request->job_id;
$filename = array();

}

Example:
If user selects job_id = abcd then filename variable in controller should have an array of files from Database.
Thanks!

Clement Sam · Answer

You could use the DB class to do this:
public function downloadZip(Request $request)
{
    $job_id = $request->job_id;
    $filenames = DB::table('analytics_report')->select('filename')
                  ->where('job_id', $job_id)->get()->toArray();

}

Alternatively, if you already have a model for the analytics_report and assuming your model is AnalyticsReport, you could that with:
public function downloadZip(Request $request)
{
    $job_id = $request->job_id;
    $filenames = AnalyticsReport::select('filename')
                  ->where('job_id', $job_id)->toArray();

}

Songoudaha Gabeta Soro · Answer

In namespace
use DB;

in download method:
public function downloadZip(Request $request)
{
  $job_id = $request->job_id;
  $filenames = DB::table('analytics_report')->where('job_id',$job_id)->pluck('filename');
}

xXx_HAWK · Answer

Use This Query
   public function downloadZip(Request $request)
{
  $job_id = $request->job_id;
  $file = DB::table('table_name')->where('job_id','=',$job_id)->select('filename')->get()->toArray();
}

Alok p · Answer

In namespace add
use DB;

then in method add table name,
public function downloadZip(Request $request)
{
  $job_id = $request->job_id;
  $filename = DB::table('table_name')->where('job_id',$job_id)->get()->pluck('filename')->toArray();
}

Return all filename from database by specific field name

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