Display Update button & save values in database - save

Question

We are displaying table grid as below image, in last column you can see commission as 10 & 0.
i need to provide an Update button below that and provide an option to edit 10 & 0 , so if we click on "update" button, it should save in database.
we are using below code to display table grid.
php
function getDesignerCollection()
{
$user_home = new USER();
$stmt = $user_home->runQuery("SELECT * FROM tbl_users");
$stmt->execute(array(":uid" => $_SESSION['userSession']));
$row = $stmt->fetch(PDO::FETCH_ASSOC);
$stmt->execute();
$i=0;
while($data = $stmt->fetch())
{
$responce[$i]=array($data['userID'],
ucfirst($data['name']),
$data['commission1'],
$data['joining_date']);
$i++;
}
echo json_encode($responce);
}
script
var __TEST_DATA__=eval('<?php echo getDesignerCollection();?>');
var grid_demo_id = "myGrid" ;
var dsOption= {
fields :[
{name : 'userID' },
{name : 'name' },
{name : 'commission1'},
{name : 'joining_date'}
],
recordType : 'array',
data : __TEST_DATA__
}
function my_renderId(value ,record,columnObj,grid,colNo,rowNo)
{
var no= record[columnObj.fieldIndex];
return "<input type='checkbox' value='"+record[0]+"' name='userID'/>";
}
var colsOption = [
{id: 'userID' , header: "Customer Id" , width :"90",renderer : my_renderId},
{id: 'name' , header: "Designer Name" , width :"140"},
{id: 'commission1' , header: "commission1" , width :"120"},
{id: 'joining_date' , header: "Customer Since" , width :"120"}
];
now to provide an update button , i am trying below code :
form
<form method="post" name="update" action="updateform.php" />
commission1:
<input type="text" name="commission1" />
<input type="submit" name="Submit" value="update" />
</form>
updateform.php
$sql = "UPDATE tbl_users SET commission1 = '$commission1' WHERE userID= :uid";
$result = $conn->query($sql);
where i need to use form code to display "update" button below values 10 & 0
I am new to php & tried lot before posting here, but did't got any idea related to my code.