Chalaka asked 11 years ago
i have below code written inside grid
$g->select_command = "SELECT tbl_doner.Don_Id, tbl_doner.Don_Name,tbl_doner.Don_Contact,tbl_doner.Don_Email,tbl_grant.Gra_Type,tbl_grant.Gra_Amount FROM tbl_doner, tbl_grant WHERE
tbl_doner.Don_Id = tbl_grant.Don_Id AND Prj_Code = '$PrjCode'";
$out = $g_->render("list_APDonarInfo");
but echo $out not showing the grid with required records.
Appreciate your help
Chalaka
Abu Ghufran answered 11 years ago
Q) How to load grid based on $_POST data from other page?
The grid is loaded with 2 server calls.
1) load the columns of grid.
2) do an ajax call, to load data of grid.
Now, if you want to pass data from external form, it is available for step1. But not there in 2nd ajax call, as it is not posted.
Solution is to put the POST variable in session and use it from session for step2.
e.g.
if (!empty($_POST["personid"]))
{
$_SESSION["personid"] = $_POST["personid"];
}
$pid = $_SESSION["personid"];
and use `$pid` in your SQL.
Jhon Leguizamo answered 9 years ago
Hi! And how make if the $_SESSION var change the value? Reload the grid dont work.
Thanks you
Regards
Abu Ghufran answered 9 years ago
If i understand the issue correctly, you can assign some (prefixed) unqiue var name so that it does not get changed other than this page. If there is some misunderstanding please explain further with generting steps.
