Drop down selection after saving - php

Question

I do have a two drop downs called as source and campaign and these two drop down showing the data that are coming from the data base.i do have others input fields as well.my concern is that i want to save this data after filling it in the given input and selecting drop downs data it must be saved but after saving the data the drop downs must show the selection that i had selected while clicking on save button but it is showing default one.
my code is as follows:
This is for Source:
$result= mysql_query("SELECT * FROM infosources where kunnr = '".$_SESSION["kunnr"]."' order by sort_order asc");
$model["source"]=array();
while($row = mysql_fetch_array($result)){
array_push($model["source"],$row);
}
This is for campaign:
$result= mysql_query("SELECT * FROM s_campaigns WHERE kunnr ='".$_SESSION["kunnr"]."' and active = 'true' order by name asc");
$model["campaign"]=array();
while($row = mysql_fetch_array($result)){
array_push($model["campaign"],$row);
}
and my dropdown is as follows:
<select name="srcid"> <?php foreach($model["source"] as &$obj){?>
<option value=<?php echo $obj["srcid"];?>> <?php echo $obj["srcname"];?> </option>
<?php }?></select>
and the other drop down is
<select name="camp_id"> <?php foreach($model["campaign"] as &$obj){?>
<option <?php if($model["selected"]==$obj[""]){?>selected <?php }?> value=<?php echo $obj["id"];?>> <?php echo $obj["name"];?> </option>
<?php }?></select>
please suggest me on this...

Answer 1

for that you have to decide the value in view at time of refreshing your page

Answer 2

this is a very simplivied example and is just working if you send the form to the same file ... if you send the form to any other file or if you pass the $selectedCampId variable from the other file back to this html form
<?php
// fetch the database to get all possible campaings always... before form was saved and also afterwards
$result= mysql_query("SELECT * FROM s_campaigns WHERE kunnr ='".$_SESSION["kunnr"]."'
and active = 'true' order by name asc");
$model["campaign"]=array();
while($row = mysql_fetch_array($result)){
array_push($model["campaign"],$row);
}
// initiate $selectedCampId .... if the form is sent ... this variable will be filled with the campaign_id so that we know which option was selected... otherwise it wil remain empty..
$selectedCampId= '';
// if the save button was pushed, the form method is POST and the camp_id is not empty
// save the value of the camp_id input field to the variable so that in the next step
// we know which one was selected
if(!empty($_POST['camp_id'])){
// validate the input and save the data to database
// check which campaign id is selected and write it to variable
$selectedCampId= $_POST['camp_id'];
}
?>
<?php // the form method is POST otherwise use $_GET to fetch the camp_id after the form was sent ?>
<form method="post">
<select name="camp_id">
<?php foreach($model["campaign"] as &$obj): ?>
<option
<?php // if form is sent and $selectedCampId is not empty ... echo selected = "selected" ---- otherwise echo empty string ?>
<?php echo ($obj['id'] == $selectedCampId) ? 'selected = "selected"' : ""; ?>>
<?php echo $obj["name"];?>
</option>
<?php endforeach;?>
</select>
</form>