Listing Multiple Values in one variable

[sgtpepper]

I have a field in my DB called 'email'. There can be more than one value for a user.

For example: A user may have more than one email address and the second email address in another row in the DB.

So, my code looks like this:

$query="SELECT * FROM users WHERE username = '$_GET[username]'";

$result=mysql_query($query) or die (mysql_error());
$num=mysql_numrows($result);
while ($row = mysql_fetch_array($result)) // this will grab the results from query
{
$email = $row[email];
$website = $row[website];
}
?>

Email addresses are <? echo $email?>

The problem is that <? echo $email?> is only showing one value. How can I get it to show the other row values???

[leatherback]

Hi,

This really depends on how you want to use it.

Easiest is to place a dot just before the =:

$email .= $row[email];

You might want toplace a space or a comma between them:

$email .= $row[email]. ", ";

J.

[weevil]

Try this:

<?
$query="SELECT * FROM users WHERE username = '$_GET[username]'"; 

$result=mysql_query($query) or die (mysql_error()); 
$num=mysql_numrows($result); 
while ($row = mysql_fetch_array($result)) // this will grab the results from query 
{ 
if ($email) {
  $email .= " and $row[email]";
   $verbiage = "es are";
} else {
  $email = $row[email];
  $verbiage = " is";
}
$website = $row[website]; 
} 
?> 

Email address<? echo "$verbiage $email?> 

[sgtpepper]

Using this code:

<?
$query="SELECT * FROM users WHERE username = '$_GET[username]'";

$result=mysql_query($query) or die (mysql_error());
$num=mysql_numrows($result);
while ($row = mysql_fetch_array($result)) // this will grab the results from query
{
if ($email) {
  $email .= " and $row[email]";
   $verbiage = "es are";
} else {
  $email = $row[email];
  $verbiage = " is";
}
$website = $row[website];
}
?>

Email address<? echo "$verbiage $email" ?>

Displays, and tsting2 and testing and tsting2 and testing and tsting2 and testing and tsting2 and testing and tsting2 and testing and tsting2 and testing and tsting2 and testing and tsting2 and testing and tsting2 and testing and tsting2 and testing and tsting2 and testing and tsting2testing, tsting2, testing, tsting2, testing, tsting2, testing, tsting2, testing, tsting2, testing, tsting2, testing, tsting2, testing, tsting2, testing, tsting2, testing, tsting2, testing, tsting2, testing, tsting2, testing, tsting2, and testing and tsting2 and testing and tsting2 and testing and tsting2 and testing and tsting2 and testing and tsting2 and testing and tsting2 and testing and tsting2 and testing and tsting2 and testing and tsting2 and testing and tsting2....

It adds and testing and tsting2 (the values for email) with every browser refresh.

[sgtpepper]

The code from weevil is working well....but how can I stop it from adding more results when I hit 'refresh' in the browser?

[leatherback]

Is it at all possible that you have a line like :

$email = $_POST[email];

somewhere, and later on you submit email again? Or do you store the data somewhere? Because otherwise.. It shouldn't do this..

Check your code carefully wether you are not appending the email to an existing variable..

(You can always stop this by re-defining the variable just before the while loop:

$email="";

J.

[sgtpepper]

This is the entire script

<?php
require 'db_connect.php';

if ($logged_in == 1) {
        echo 'Logged in as '.$_SESSION['username'].', <a href="logout.php">logout</a>';
} else {
        echo 'Not logged in. <a href="login.php">Login</a>';
}

?>



<?
//$query="SELECT * FROM users WHERE username = '$_GET[username]'";
$query="SELECT * FROM users WHERE username = '$_SESSION[username]'";

$result=mysql_query($query) or die (mysql_error());
$num=mysql_numrows($result);
while ($row = mysql_fetch_array($result)) // this will grab the results from query
{
if ($email) {
  $email .= " and $row[email]";
   $verbiage = "es are";
} else {
  $email = $row[email];
  $verbiage = " is";
}
}
?>

Email address<? echo "$verbiage $email" ?><br>

There shouldnt be anywhere else $email is defined...

hmm.....

[leatherback]

Hm, I am not sure why this is happening..

But this should solve the problem:
It is always wise (Actually.. In most program languages obligatory) to define a variable before using it. So before running the query add a line:

$email = '';

That should take care of the problem (unless the email is stored / retrieved from the session array.. But I think that unlikely).

J.