Hi support team,

My problem is in order to avoid using hard code to define which columns' name and labels to be displayed in view, i am thinking is it possible to use dynamical way to loop out all desired columns and label name from queryB. My queryA from is Select * from tableA and queryB is Select col_name, field_name from tableB where.... QueryB will select all those desired columns and field name. What i have done so far was only display one column in view. Let say my queryB returns 5 records and i want to loop 5 times to display that 5 columns and field name, how could i done it? Thanks.

<?php 
    $col_name = $this->dataStore("queryB")->data()[0]["col_name"];
    $field_name = $this->dataStore("queryB")->data()[0]["field_name "];

    Table::create(array(
        "dataStore"=>$this->dataStore('queryA'),
        "columns"=>array(
            $col_name=>array(
                "label"=>$field_name 
            ),
        ),
        "cssClass"=>array(
            "table"=>"table table-bordered"
        )
    ));
?>

Show error Undefined index: col_name

I have that column named col_name but still show same error.

Below is my Report and Report view. Thanks.

Report:

function setup() 
    {    
        $this->src('a') 
        ->query("SELECT col_name, fieldname FROM queryB where guid = @guid ")
        ->params(array(
            '@guid' => $this->params["guid"],
        ))
        ->pipe($this->dataStore("queryB"));

        $this->src('a')
        ->query("SELECT * FROM queryA")
        ->pipe($this->dataStore("queryA"));
    }

Report view:

        $columns = array();
        $data_field = $this->dataStore('queryB')->data();

        foreach($data_field as $row)
        {
            $columns['col_name'] = array(
                "label" => $row["field_name"],

            );
        }

        Table::create(array(
        "dataStore" => $this->dataStore('queryA'),
        "columns" => $columns,
        "cssClass" => array(
            "table" => "table table-striped table-bordered table-hover table-condensed",
            "td" => "right-align",
            "tf" => "tfcolor",
            "th" => "color",
            "td" => function($row,$columnName){
            switch($columnName)
                {
                case "field_name":
                    return "col-color";
                default:
                    return "";
                }
            },
        ),
    ));

It works, thanks a lot.

My Current view is like this

class MyReport extends \koolreport\KoolReport {

use \koolreport\laravel\Friendship;
use \koolreport\cloudexport\Exportable;   
//We leave this blank to demo only
function setup()
{
 $min = $this->params["min"];
 $agen = $this->params["agen"];
 $sec = $this->params["sec"];
 $sub_sec = $this->params["sub_sec"];
 
 $this->src("mysql")->query(MySQL::type(
        DB::table("demands")->join('approved_project_info', 'approved_project_info.unapprove_project_id', '=', 'demands.project_id')
       ->where('ministry', $min)            

        ))
    ->pipe($this->dataStore('demands'));
   
}

} My view code is here, <!-- MyReport.view.php --> <?php use \koolreport\widgets\koolphp\Table; ?>

     <?php
     
    Table::create([
        "dataSource"=>$this->dataStore("demands")
    ]);
    ?>

But i need to show only two column from the view which is (demand_type and project_code) my database name is (demands) what should i need to do for change in view.