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What would you like to do? problem using value from POST input
* PHP Grid Component
* @author Abu Ghufran <> -
* @version 2.0.0
* @license: see license.txt included in package
$db_conf = array("type"=> PHPGRID_DBTYPE, "server"=> PHPGRID_DBHOST,"user"=> PHPGRID_DBUSER,"password"=> PHPGRID_DBPASS,"database" => PHPGRID_DBNAME);
if (!empty($_POST["name"]))
{$_SESSION["name"] = $_POST["name"];
$SelectedName = $_SESSION["name"];
$searchstring = "SELECT client_id,name FROM clients WHERE name LIKE '%{$SelectedName}%'";
$g = new jqgrid($db_conf);
$g->select_command =$searchstring;
$g->table = "clients";
$col = array();
$col["title"] = "Id"; // caption of column
$col["name"] = "client_id";
$cols[] = $col;
$col = array();
$col["title"] = "Client";
$col["name"] = "name";
$cols[] = $col;
$out = $g->render("list1");
<!DOCTYPE html><head> <meta charset="utf-8">
<link rel="stylesheet" type="text/css" media="screen" href="../../lib/js/themes/redmond/jquery-ui.custom.css"></link>
<link rel="stylesheet" type="text/css" media="screen" href="../../lib/js/jqgrid/css/ui.jqgrid.css"></link>
<script src="../../lib/js/jquery.min.js" type="text/javascript"></script>
<script src="../../lib/js/jqgrid/js/i18n/grid.locale-en.js" type="text/javascript"></script>
<script src="../../lib/js/jqgrid/js/jquery.jqGrid.min.js" type="text/javascript"></script>
<script src="../../lib/js/themes/jquery-ui.custom.min.js" type="text/javascript"></script>
echo $out;
echo $searchstring;
<form action="" method="post" name="myForm">
Name:<input id="name" type="name" name="name">
<input type="submit" value="Submit">

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@rob2009 rob2009 commented Jun 19, 2019

This uses the demo database that comes with and uses the $_POST handling suggested in FAQ>Misc
As it stands the search returns just the 2 entries that match.
If you comment this out: line 16 $SelectedName="Ana"; and repeat the search with form input name=Ana, in that case the search returns all items in the database.

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