Resolved Birthdays in Current Month vb 4.2

Feckie · 10-06-2012, 09:55 PM

I am trying to display the user birthdays this month in vbulletin 4.2

I have done the script below, although there is a birthday set this month
it shows " No Birthdays This Month "

Can anyone help please

Code:

<?php


include("connect.php");


$result = mysql_query("SELECT userid,username,birthday FROM user WHERE MONTH(birthday) = MONTH(NOW())");
    if( mysql_num_rows( $result ) != 0 ) {
        while ( $row = mysql_fetch_array($result))
{
    $userid = $user['userid'];
    $username = $user['username'];
    $birthday = $user['birthday'];
    $birthday = explode("-", $birthday);
    $birthmonth = $birthday[0];
    if ($birthmonth = $month)
    {
        echo "<a href='member.php?u=$userid'>$username</a>,&nbsp;";
    }

}
}
else{
        echo "No Birthdays This Month";
    }
?>

sunfighter · 10-07-2012, 03:01 AM

Shouldn't the section containing

Code:

$userid = $user['userid'];

Be

Code:

$userid = $row['userid'];

Feckie · 10-07-2012, 07:32 AM

Quote:

Originally Posted by sunfighter

Shouldn't the section containing

Code:

$userid = $user['userid'];

Be

Code:

$userid = $row['userid'];

Tried that, it made no difference

sunfighter · 10-07-2012, 02:33 PM

What did you try, please post, because there are at lest three of them. And to save time echo out what you get for $birthday, $birthmonth, and $month.

Fou-Lu · 10-07-2012, 02:43 PM

What is the datatype of the birthday field? Is it a datetime (or any derivative of date and time)?

PHP Code:


			
if ($result = mysql_query("SELECT userid,username,birthday FROM user WHERE MONTH(birthday) = MONTH(NOW())"))
{
    if (mysql_num_rows($result) > 0)
    {
        $i = 0;
        while ($row = mysql_fetch_row($result))
        {
            list($userid, $username, $birthday) = $row;
            if ($i++ > 0)
            {
                print ', ';
            }
            printf('<a href="member.php?u=%d">%s</a>', $userid, $username);
    }
    else
    {
        print 'No birthdays this month';
    }
}
else
{
    print 'Error in query: ' . mysql_error();
}

I would be surprised if a birthday isn't a date or date type datatype, but given that this is using old mysql code and not using mysqli or PDO, its hard to say what datatype has been chosen.

Feckie · 10-07-2012, 07:22 PM

Fou-Lu I got yours to work with a few adjustments, Thanks
ie: it was "birthday_search" not "birthday" My Bad

Next Question:
How would I add the actual date after each name ...

It now shows as follows

name
name
name

Code:

if ($result = mysql_query("SELECT userid,username,birthday_search FROM user WHERE MONTH(birthday_search) = MONTH(NOW())"))
{
    if (mysql_num_rows($result) > 0)
    {
        $i = 0;
        while ($row = mysql_fetch_row($result))
        {
            list($userid, $username, $birthday_search) = $row;
            if ($i++ > 0)
            {
                print '<br /> ';
            }
            printf('<a href="member.php?u=%d">%s</a>', $userid, $username);
    } }
    else
    {
        print 'No birthdays this month';
    }
}
else
{
    print 'Error in query: ' . mysql_error();
}

Fou-Lu · 10-07-2012, 09:05 PM

PHP Code:


			
            printf('<a href="member.php?u=%d">%s</a> %s', $userid, $username, $birthday_search);

Looks like its a date datatype, so that should show without needing formatting.

Feckie · 10-07-2012, 11:47 PM

Cheers Mate that works perfect

Quote:

Originally Posted by Fou-Lu

PHP Code:


			
            printf('<a href="member.php?u=%d">%s</a> %s', $userid, $username, $birthday_search);

Looks like its a date datatype, so that should show without needing formatting.