issue with displaying proper output in PHP - php

Question

I trying to make search option based on a SELECT box and user input (both are mandatory). But in the following code both correct and wrong input are displaying as wrong input. Can someone please explain what is wrong in the code.
Here is HTML
<form action="" method="POST">
<select name="selectOpt">
<option>Select a list</option>
<option value="one">ID</option>
<option value="delName">Dealer Name</option>
<option value="medName">Medical Name</option>
</select>
<input type="text" name="uinput" placeholder="Enter Search Key"/>
<button type="submit" name="submit">search</button>
</form>
PHP
if(isset($_POST['submit'])){
if(!empty($_POST['selectOpt']) && !empty($_POST['uinput'])){
if($_POST['selectOpt']=='one'){
$id = $_POST['selectOpt'];
if (!preg_match("/^[0-9]*$/",$id)){
echo "not valid";
}else{
echo "valid";
}
}
}else{
echo "Enter Value";
}
}

Answer 1

Your pattern /^[0-9]*$/, which is written to require a sequence of zero or more digits, doesn't match any of the possible values for your select box, which are all alphabetic strings.

Answer 2

<form action="save.php" method="POST">
<select name="selectOpt">
<option>Select a list</option>
<option value="one">ID</option>
<option value="delName">Dealer Name</option>
<option value="medName">Medical Name</option>
</select>
<input type="text" name="uinput" placeholder="Enter Search Key"/>
<button type="submit" name="submit">search</button>
</form>
<?php
if(isset($_POST['submit'])){
if(!empty($_POST['selectOpt']) && !empty($_POST['uinput'])){
if($_POST['selectOpt'] == "one"){
$id = $_POST['selectOpt'];
if (preg_match("/^[0-9]*$/",$id)){
echo "not valid";
}else{
echo "valid";
echo $input = $_POST['uinput'];
}
}
else{
echo "PLEASE Enter Proper and Valid Search ";
}
}
else{
echo "PLEASE Enter Value ";
}
}
?>