A simple program that I wrote in PHP that will ask the user to give a number and then the program will check the civil status based on the given number by the user.
I am currently accepting programming work, IT projects, school and application development, programming projects, thesis and capstone projects, IT consulting work, computer tutorials, and web development work kindly contact me in the following email address for further details. If you want to advertise on my website kindly contact me also in my email address also. Thank you.
My email address is the following jakerpomperada@gmail.com, jakerpomperada@aol.com, and jakerpomperada@yahoo.com.
My mobile number here in the Philippines is 09173084360.
My telephone number at home here in Bacolod City, Negros Occidental Philippines is +63 (034) 4335675.
Here in Bacolod I also accepting computer repair, networking and Arduino Project development at a very affordable price.
My personal website is http://www.jakerpomperada.com
Sample Program Output
Program Listing
index.php
<!DOCTYPE html>
<html lang="en">
<head>
<style>
body {
font-family: arial;
}
</style>
<?php
$display="";
$civil=$_POST['civil'];
if(isset($_POST['clear'])){
$civil="";
$display="";
}
?>
<meta charset="UTF-8">
<title>Civil Status Checker Using PHP</title>
</head>
<body>
<h2>Civil Status Checker Using PHP</h2>
<h4>Created By Mr. Jake R. Pomperada</h4>
<form action="" method="post">
<b>Give a Number</b><br/><br>
<input type="text" name="civil" value="<?php echo $civil; ?>" autofocus><br/><br>
<input type="submit" name="submit" value="Submit">
<input type="submit" name="clear" value="Clear">
</form>
<br>
<?php
if(isset($_POST['submit'])){
$civil=$_POST['civil'];
if ($civil == 1) {
$display = "You are Single.";
echo $display;
} else if ($civil == 2) {
$display = "You are Married.";
echo $display;
} else if ($civil == 3) {
$display = "You are Annulled.";
echo $display;
}else if ($civil == 4) {
$display = "You are Widow.";
echo $display;
} else if ($civil == 5) {
$display = "You are Separated.";
echo $display;
}
else {
$display = "Invalid Year Level. Try Again";
echo $display;
}
}
?>
</body>
</html>
