Problem updating Select dropdown

Please see my code below

I have a select box populated by table names from a MySQL database. I chose a table from the dropdown list and that table name is displayed. I then click the Get Records button and the table name in the select box reverts to its previous value. If I then click the Next record button the table name is displayed correctly in the drop down select box. Can anyone suggest what I have wrong in my code that means the wrong name is displayed in the select box. I have been at this for a couple of hours and can’t spot it.

TIA

[php]

<?php $query="SELECT tbl_name,opt_name FROM tables"; $result = mysql_query ($query); echo ""; while ($temp = mysql_fetch_assoc($result)) { if ($temp['tbl_name'] == $_SESSION['rec_table']){ echo "".htmlentities($temp['opt_name']).""; } else { echo "".htmlentities($temp['opt_name']).""; } } echo ""; ?>
	<!--Choose starting record-->   
<label class="cs1">Select first record number:</label>
<input type="text" name="rec_from" value="1" size="6" > 
<input id="get_records"  type="submit" name="get_records" value="Get Records"/><br />
<input style="margin:5px 0 5px 10px" name="next_rec" id="next_rec" type="submit" value="Next Record" />
<input style="margin:5px 0 5px 90px" name="new_rec" type="submit" value="Create New Master Person" >

</form>
<?php

// Display records in table

if (isset($_POST[‘get_records’]) || isset($_POST[‘next_rec’]))
{
if (isset($_POST[‘get_records’])){
$table = ($_POST[‘source’]);
$table = mysql_real_escape_string($table);
$new_persona[‘rec_table’]=$table;
$_SESSION[‘rec_table’] = $table;
$record = ($_POST[‘rec_from’]);
$record = mysql_real_escape_string($record);
$record = $record - 1;
$_SESSION[‘last_rec’] = $record;
}

	if (isset($_POST['next_rec'])){
		$last_rec = $_SESSION['last_rec'];
		$record = $last_rec + 1;
		$table = mysql_real_escape_string ($_SESSION['rec_table']);
		$_SESSION['last_rec'] = $record;}

Code continues here with other bits that aren’t relevant to the problem.
[/php]

Colin

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