Comparing database stringvalue with new stringvalue - php

Question

Here is what I'm trying to do: When user adds a contact to his list, the number of this contact gets run by with the numbers in the database and it gives feedback if the user is already in the database or not. Right now I always get back "User is in database" even though he isn't. Then again I'm not that well acquainted with php. I changed the code a bit again, now it doesn't work at all, because it doesn't like the part
$number = ($_GET["number"] from $DB_Table);
Full code
<?php
$DB_HostName = "localhost";
$DB_Name = "db";
$DB_User = "user";
$DB_Pass = "pw";
$DB_Table = "contacts";
$number = ($_GET["number"] from $DB_Table);
$fnumber = ($_GET["fnumber"]);
if ($number == $fnumber) {
echo "This user is already in database";
} else {
echo "This user isn't in the database";
}
$con = mysql_connect($DB_HostName,$DB_User,$DB_Pass) or die (mysql_error());
mysql_select_db($DB_Name,$con) or die(mysql_error());
mysql_close($con);
?>

Answer 1

I don't actually see you executing the database query. You could do something like this:
<?php
$DB_HostName = "localhost";
$DB_Name = "db";
$DB_User = "user";
$DB_Pass = "pw";
$DB_Table = "contacts";
$con = mysql_connect($DB_HostName,$DB_User,$DB_Pass) or die (mysql_error());
mysql_select_db($DB_Name,$con) or die(mysql_error());
$fnumber = mysql_real_escape_string($_GET["fnumber"]);
$result = mysql_query("SELECT * FROM $DB_Table WHERE Something = '$fnumber'", $con);
if ($result) {
// Check the number of rows in the result set
if (mysql_num_rows($result) > 0) {
echo "This user is already in database";
}
else echo "This user isn't in the database";
}
mysql_close($con);
?>

Answer 2

This is not valid PHP code: $number = ($_GET["number"] from $DB_Table);
$_GET["number"] represents the value of the "number" parameter that you find in the url of your page.
Example: http://example.com/index.php?number=7 so $_GET["number"] is 7.
In your code, $DB_Table is a just a string ("contact") and "from" does not fit there using php syntax.
mysql_select_db($DB_Name,$con) or die(mysql_error());
is valid PHP but you are not doing anything with what you get from the database. I suggest you at least take a look at this tutorial php mysql select