mysql_query how to display html for if condition - php

Question

So I have this rustic and basic php that will check a value in my database and if its corrects it will pull out the rest of that row I guess but my problem is how to echo certaine html if the result is okay and how to echo some other html if else.
btw atm else is not working, in other words if you input a code that is not on my db it will not show anything
<?php
$db_hostname = 'localhost';
$db_database = 'codedb';
$db_username = 'root';
$db_password = '';
$table = 'users';
$field = 'code';
$test = 'first_name';
// Connect to server.
$connection = mysql_connect($db_hostname, $db_username, $db_password) OR DIE ("Unable to
connect to database! Please try again later.");
// Select the database.
mysql_select_db($db_database)
or die("Unable to select database: " . mysql_error());
$query = "SELECT * FROM $table WHERE $field = '{$_GET["qcode"]}'";
$result = mysql_query($query);
if ($result) {
while($row = mysql_fetch_array($result)) {
$name = $row["$field"];
$test = $row["$test"];
echo "Hello: $name $test";
}
}
else {
echo "Im sorry you buddy, you are not a winner this time! $test";
}
mysql_close($connection);
?>

Answer 1

You can use mysql_num_rows(), it's the best way.
if( mysql_num_rows($result) > 0 ){
/* Anything you want here on success */
} else {
/* Anything you want here on failure */
}

Answer 2

if(mysql_num_rows($result) > 0) { echo "output"; } else { echo "error msg"; }

Answer 3

Use mysql_num_rows against the query.
if(mysql_num_rows($result)){ } else { }

Answer 4

Instead of echoing every html value you can simply enclose html in if statements
like:
<?php
$query = "SELECT * FROM $table WHERE $field = '{$_GET["qcode"]}'";
$result = mysql_query($query);
if ($result) {
while($row = mysql_fetch_array($result)) {
$name = $row["$field"];
$test = $row["$test"];
?>
<b>Hello: <?php echo $name $test"; ?></b>
<?php
}
}
else {
?>
<b> Im sorry you buddy, you are not a winner this time! </b> <?php echo $test";
}
mysql_close($connection);
?>

Answer 5

First you have to correct syntax of database selection like this
mysql_select_db($db_database,$connection);
Insted of
mysql_select_db($db_database);
Check Manual