Update on Simple php login.

gavin101

OK.

I am using a simple username and password form to pass details to a php file to verify a user. In my msql database I also have a field called 'url' that I want to pull out and display as a clickable link for each user (the url field is different for each user). Below is what I have so far. What do I need to add to get the url field for a particular user to print on the page - and where do I put it. If someone could copy and paste the below code to demonstrate I would much appreciate it!

Form results pass to the following php file:

<?php

$connection = mysql_connect("localhost", "myusername", "mypassword")
or die ("Couldn't connect to server.");

$db = mysql_select_db("clients", $connection)
or die ("Couldn't select database.");

$sql = "SELECT id
FROM login
WHERE username='$username' and password='$password'";

$result = mysql_query($sql)
or die("Couldn't execute query.");

$num = mysql_numrows($result);

if ($num == 1) {

echo "<P>You are a valid user!<br>";
echo "Your username is $username<br>";
echo "Your password is $password<br>";

} else if ($num == 0) {

echo "You are not authorized!";

}

//what do I add to the above code to print the url field (located in the same table) when the user is brought to this page?

</body>

Anon

just add the url field to your select statement and use mysql_fetch_array() to get the value of that column.

$sql = "SELECT id, url ...
...
while($row = mysql_fetch_array($result)) {
$url["url"];
echo "Your URL is $url<br>";
}

see http://www.phpbuilder.com/manual/function.mysql-fetch-array.php

p.s. Why are displaying the password in plain text in the browser?

Anon

ummm....typo

inside while loop should be:

$url=$row["url"];

p.p.s. what happens if $num==2? (you prolly should remove the if part of the else statement)