Rufino asked 6 months ago
Tengo un proyecto MVC y requiro integrar grid, para integrarlo tambien se debe realizar tal y como se muestra en la documentacion:
include_once(“../../config.php”);
// include and create object
include(PHPGRID_LIBPATH.”inc/jqgrid_dist.php”);
$db_conf = array(
“type” => PHPGRID_DBTYPE,
“server” => PHPGRID_DBHOST,
“user” => PHPGRID_DBUSER,
“password” => PHPGRID_DBPASS,
“database” => PHPGRID_DBNAME
);
$g = new jqgrid($db_conf);
o se integra de otra forma porque al realizarlo de esta manera en un controlador me marca el error que no encuentra la clase jqgrid
Uncaught Error: Class “Controllers\Comercial\jqgrid” not found
Abu Ghufran Staff answered 6 months ago
If the path of the jqgrid_dist.php is corrrect, you can try appending root namespace \ before jqgrid():
$g = new \jqgrid($db_conf);
I don’t know which MVC you are using, however you can also check this laravel guide and take help:
Rufino answered 4 weeks ago
Error al llamar on_data_display en MVC
Cual es la forma correcta de llamar la funcion valida_accion
$col1 = array();
$col1[“title”] = “Accion”;
$col1[“name”] = “Accion”;
$col1[“dbname”] = “estatus”;
$col1[“width”] = “30”;
$col1[“align”] = “center”;
$col1[“search”] = false;
$col1[“sortable”] = false;
$col1[“editable”] = false;
$col1[“visible”] = array(“xs”,”sm”,”md”, “lg”, “xl”);
$col1[“on_data_display”] = array( “valida_accion”, “”);
$cols1[] = $col1;
estoy en un proyecto mvc ya intente con
$col1[“on_data_display”] = array( $this->valida_accion, “”);
la funcion es publico
public function valida_accion($data) {
return $buttons;
}
