PHP trouble using $_POST in a loop - php

Question

edit - I solved my "add friend" button issue, now I'm trying to get the userid from the loop below. I want to be able to get the userid of the name that the user looks up (the name that gets submitted to findUsers function, $friend). So basically I want to be able to use result['userid'] and be able to submit that into a database.
I commented in the code where I'm having trouble getting the value for the userid to set.
<input type="hidden" name="userId" value="' . $result['userid'] . '" />
Is there a certain way to use hidden inputs, or is the value just not being set correctly?
<?php
include_once 'config.php';
class Friends{
function addFriend($userId) {
return $userId; //this is supposed to return the value of the user's id selected in the loop below via the if statements towards the bottom.
}
function findUsers($friend){
$search = mysql_query("SELECT * from users where username='$friend'");
if (mysql_num_rows($search) > 0){
// $this->addFriend($friend);
$userLocation = mysql_query("select * from userinfo where username='$friend'");
$locationResult = mysql_fetch_array($userLocation);
$locationResultArray = $locationResult['userlocation'];
$locationExplode = explode("~","$locationResultArray");
if (mysql_num_rows($search)) {
// Table column names
echo '<table><tr><td>Username</td><td>Location</td></tr>';
while($result = mysql_fetch_array($search)) {
echo '<tr>
<td>'. $result['username'] . '</td>
<td>' . $locationExplode[0] . ', ' . $locationExplode[1] . '</td>
<td>
<form method="post" name="friendRequest" action="">
<input type="hidden" name="userId" value="' . $result['userid'] . '" />
<input type="submit" name="addFriend" value="Add Friend" />
</form>
</td></tr>';
}
}
}
}
}
$friends = new Friends();
if (isset($_POST['userId'], $_POST['addFriend'])) {
echo "friend button pressed"; //this message is displayed
if ($friends->addFriend($_POST['userId'])) {
echo "userID set"; //this message is displayed
echo $_POST['userID']; //this is not displayed
} else {
// some error code here
}
}
// Edit this to test here
// $friends->findUsers('<username>');
?>

Answer 1

That way to add friend is incorrect way, because when you click the "Add friend" button, that will send a $_POST['addFriend'] and then in the loop the check are going to add all users as friend.
The correct code is here:
<?php
function addFriend($userId){
// check is 'userId' exist, if not, then return 0;
}
if (isset($_POST['userId'], $_POST['addFriend'])) {
if (addFriend($_POST['userId'])) {
// some display code here
} else {
// some error code here
}
}
while($result = mysql_fetch_array($search)) {
?>
<tr><td>
<form method="post" name="friendRequest" action="">
<input type="hidden" name="userId" value="<?php echo $result['userid']; ?>" />
<input type="submit" name="addFriend" value="Add Friend" />
</form>
</td></tr>
<?php } ?>
EDIT1:
You can't use the code above into a function. I fixed a lot of bug that I can see in your code, but still look strange.
I don't get what you want to do with your code, but I made this:
<?php
function addFriend($userId) {
return 1; //using 1 for testing purposes
}
function findUsers($friend) {
$search = mysql_query('SELECT `userid`, `username`, `userlocation` FROM `users` JOIN `userinfo` ON `users`.`username` = `userinfo`.`username` WHERE `user`.`username` = ' . $friend);
if (mysql_num_rows($search)) {
// Table column names
echo '<table><tr><td>Username</td><td>Location</td></tr>';
while($result = mysql_fetch_array($search)) {
$locationExplode = explode('~', $result['userlocation']);
echo '<tr>
<td>'. $result['username'] . '</td>
<td>' . $locationExplode[0] . ', ' . $locationExplode[1] . '</td>
<td>
<form method="post" name="friendRequest" action="">
<input type="hidden" name="userId" value="' . $result['userid'] . '" />
<input type="submit" name="addFriend" value="Add Friend" />
</form>
</td></tr>';
}
}
}
if (isset($_POST['userId'], $_POST['addFriend'])) {
if (addFriend($_POST['userId'])) {
echo "test"; //I'm simply trying to get the input to work, can't get it to post. Just using this for a test.
} else {
// some error code here
}
}
// Edit this to test here
// findUsers('<username>');
?>
EDIT2:
Well, you just need to put my functions code into the class and then use the other code outside the class, like this:
<?php
include_once 'config.php';
class Friends{
function addFriend($userId) {
return 1; //using 1 for testing purposes
}
function findUsers($friend) {
$search = mysql_query('SELECT `userid`, `username`, `userlocation` FROM `users` JOIN `userinfo` ON `users`.`username` = `userinfo`.`username` WHERE `user`.`username` = ' . $friend);
if (mysql_num_rows($search)) {
// Table column names
echo '<table><tr><td>Username</td><td>Location</td></tr>';
while($result = mysql_fetch_array($search)) {
$locationExplode = explode('~', $result['userlocation']);
echo '<tr>
<td>'. $result['username'] . '</td>
<td>' . $locationExplode[0] . ', ' . $locationExplode[1] . '</td>
<td>
<form method="post" name="friendRequest" action="">
<input type="hidden" name="userId" value="' . $result['userid'] . '" />
<input type="submit" name="addFriend" value="Add Friend" />
</form>
</td></tr>';
}
}
}
}
$friends = new Friends();
if (isset($_POST['userId'], $_POST['addFriend'])) {
if ($friends->addFriend($_POST['userId'])) {
echo "test";
} else {
// some error code here
}
}
// Edit this to test here
// $friends->findUsers('<username>');
?>
EDIT3:
That's because the function addFriend is incorrect... You need to pass the user ID value as argument and then display it like this:
function addFriend($userId) {
return $userId; //this is supposed to return the value of the user's id selected in the loop below via the if statements towards the bottom.
}