You can pick PK 'idCotizacion_Encabezado' as first column and again select this column as third column with different alias. e.g.

SELECT
idCotizacion_Encabezado,
serie,
idCotizacion_Encabezado as idCotizacion,
idClientes,
fecha,
total,
estatus
FROM table

And make the 1st column hidden:true, 3rd column as editable:false

Thank you, it worked.

Set table.field prefix with that column.

$col = array();
$col["title"] = "No"; // caption of column
$col["name"] = "idCotizacion";
$col["dbname"] = "<TABLE>.idCotizacion";

This will be replaced in WHERE clause.

Hi Abu!

I was using the autofilter in the 3rd column but I get the following error:

Couldn't execute query. Unknown column 'idCotizacion' in 'where clause' – SELECT count(*) as c FROM (SELECT idCotizacion_Encabezado, serie, idCotizacion_Encabezado AS idCotizacion, idClientes, fecha, total, estatus FROM cotizacion_encabezado WHERE 1=1 AND ( `idCotizacion` LIKE '%5%' )) pg_tmp

Source code:

$col = array();
$col["title"] = "Id"; // caption of column
$col["name"] = "idCotizacion_Encabezado"; // grid column name, must be exactly same as returned column-name from sql (tablefield or field-alias)
$col["width"] = 60;
$col["search"]= true;
$col["hidden"] = true;
$col["editable"] = false;
$cols[] = $col;

$col = array();
$col["title"] = "Serie"; // caption of column
$col["name"] = "serie"; // grid column name, must be exactly same as returned column-name from sql (tablefield or field-alias)
$col["editable"] = true;
$col["width"] = 40;
$col["align"] = "center";
$col["edittype"] = "select"; // render as select
$str2 = "A:A;B:B";
$col["editoptions"] = array("value" => $str2);
$col["formatter"] = "select"; // display label, not value
$cols[] = $col;

$col = array();
$col["title"] = "No"; // caption of column
$col["name"] = "idCotizacion"; // grid column name, must be exactly same as returned column-name from sql (tablefield or field-alias)
$col["width"] = 60;
$col["align"] = "center";
$col["editable"] = false;
$col["search"]= true;
$cols[] = $col;

Select command:

$g->select_command = "SELECT idCotizacion_Encabezado, serie, idCotizacion_Encabezado AS idCotizacion, idClientes, fecha, total, estatus FROM cotizacion_encabezado";

Why I'm getting this error?

Thank you Abu!

It worked, I only modified the following line of your code becuase the real name of the ID in the table is idCotizacion_Encabezado.

$col["dbname"] = "<TABLE>.idCotizacion_Encabezado";

I put it as you told me and I got the same error but it was caused by the incorrect column name idCotizacion doesn´t exist.

Thank again!