Arrays?

[teccus]

I am using mysql with a table with 2 collums, PK_ID and User_ID.

The PK_ID is the ID of the current user and User_ID is a list of IDs of their contacts.

I.e

PK_ID | User_ID
96 | 23
96 | 21
96 | 12

I am using the following code to count how many users they have on their list, then storing it into an array so I can print each user_id to the screen.

The array I am using doesn't seem to store more than 1 result in it when it should be storing 3.

Any ideas why?

$countUsers = mysql_query ("SELECT count(*) from UsersContactlist WHERE PK_ID='$PK_ID'");
$count = mysql_result($countUsers, 0);

//get all users PK_IDs on contact list
//$gatherUsers = mysql_query ("SELECT User_ID from UsersContactlist WHERE PK_ID='$PK_ID'");
$gatherUsers = mysql_query ("SELECT User_ID from UsersContactlist WHERE PK_ID='$PK_ID'");

if (mysql_num_rows($gatherUsers) > 0)
{
$users=mysql_fetch_array($gatherUsers);

//loop to get User_IDs
for ($i=0; $i<$count; $i++) 
{
echo $users[$i];
echo "<br>";
//$users = mysql_result($gatherUsers, $i);
}

}

[nnichols]

When posting code please put it between [ PHP ] and [ /PHP ] tags as it makes it loads easier to read -

$sql = "SELECT User_ID from UsersContactlist WHERE PK_ID='$PK_ID'";
$gatherUsers = mysql_query ($sql)
	or die('MySQL error: '. mysql_error() . '<br>Query: ' . $sql); 

// don't need this if but you can use it with an else to display an error or something
if (mysql_num_rows($gatherUsers) > 0) {

while ($row = mysql_fetch_array($gatherUsers)) {

	echo $row['User_ID'];
	echo '<br>';

} // end while

} // end if
else {
	echo 'Some error message';
}

Hope this helps 😉

[ahundiak]

$gatherUsers = mysql_query ("SELECT User_ID from UsersContactlist WHERE PK_ID='$PK_ID'");

while($user=mysql_fetch_array($gatherUsers))
{
  $users[] = $user;  /* This builds the array */
}
for($i = 0; $i < count($users); $i++)
{
  echo $users[$i];
}

[teccus]

Hi, the second code is more what i'm looking for, but when I execute it, it echo's ArrayArray, rather than the contents of the array.

Any ideas?

[nnichols]

That is because you are trying to echo an array.

while($user=mysql_fetch_array($gatherUsers))
{
  $users[] = $user;  /* This builds the array */  /* Nick: This builds an array of arrays */
}

This code builds a three dimensional array. So to print the list of uids you need to use -

for($i = 0; $i < count($users); $i++)
{
  echo $users[$i][0];
}

Hope this helps 😉

[teccus]

Ok the code is starting to work now. I've tried implimenting the convert User ID to Username but it doesn't work - probably as its using simplar code to my first bit.

I need all the user IDs pulled out from the database to be converted into the Username.

Could anyone help??

What I have is:

$gatherUsers = mysql_query ("SELECT User_ID from UsersContactlist WHERE PK_ID='$PK_ID'");

while($user=mysql_fetch_array($gatherUsers))
{
 $users[] = $user;  /* This builds the array */
}

for($i = 0; $i <= count($users); $i++)
{

$sqll = "SELECT Username from Users WHERE PK_ID='$users[$i][0];'";
			$convertID = mysql_query ($sqll) or die('MySQL error: '. mysql_error() . '<br>Query: ' . $sql); 
			while ($row2 = mysql_fetch_array($convertID)) {

			$userUsername = $row2['Username'];
			echo '<br><font size="1" face="Verdana"><b>' . $userUsername . '<br>';
}

[teccus]

Sorry, the working code I currently have minus the new implimentations of converting User ID to Username is:

Just thought it might help when trying to understand what i've got working so far:

$gatherUsers = mysql_query ("SELECT User_ID from UsersContactlist WHERE PK_ID='$PK_ID'");

while($user=mysql_fetch_array($gatherUsers))
{
 $users[] = $user;  /* This builds the array */
}

for($i = 0; $i <= count($users); $i++)
{
  echo $users[$i][0];
}

[nnichols]

You should be using a join in your query instead of running multiple queries -

$sql = "SELECT u.Username, u.User_ID 
	FROM UserContactlist uc, Users u
	WHERE uc.PK_ID=$PK_ID
	AND uc.User_ID=u.PK_ID";

$gatherUsers = mysql_query ($sql);

while($user=mysql_fetch_array($gatherUsers)) {

echo '<br><font size="1" face="Verdana"><b>' . $row['Username'] . '<br>';

}

Hope this helps 😉

[teccus]

Hey mate. Thanks for all the help your giving me.

The code you've given me doesn't seem to work

$sql = "SELECT u.Username, u.User_ID FROM UserContactlist uc, Users u WHERE uc.PK_ID=$PK_ID AND uc.User_ID=u.PK_ID";

$gatherUsers = mysql_query ($sql);

while($user=mysql_fetch_array($gatherUsers)) {
	//$users[] = $user;
    echo '<br><font size="1" face="Verdana"><b>' . $row['Username'] . '<br>';

}

I get the error on this line

while($user=mysql_fetch_array($gatherUsers)) {

[nnichols]

That's probably me and my naming conventions. Try -

$sql = "SELECT u.Username, uc.User_ID
        FROM UserContactlist uc, Users u
        WHERE uc.PK_ID=$PK_ID
        AND uc.User_ID=u.PK_ID";

$gatherUsers = @mysql_query ($sql)
    or die('MySQL error: ' . mysql_error() . '<br>Query: ' . $sql);

while($row=mysql_fetch_array($gatherUsers)) {
    echo '<br><font size="1" face="Verdana"><b>' . $row['Username'] . '<br>';

}

[teccus]

Thanks.. seems to be working fine now 🙂