While loop within a while loop logically not working in PHP

Question

I have this while loop within another while loop in my PHP:

$selecty = mysql_query("SELECT * FROM followers WHERE userid='".$_SESSION['id']."'");

$rowsy = mysql_num_rows($selecty);

echo '<td>'. $table["username"]. '</td>';
echo '<td>';

while ($tables = mysql_fetch_assoc($selecty)) {
    if($tables['followerid']!=$table['id']) {
        echo '<a href="#" data-userid="'.$table['id'].'" class="follow">'.'</a>'; 
    } else {
        echo '<a href="#" data-userid="'.$table['id'].'" class="following">'.'</a>';
    }
}
echo '</td>';
echo "<tr>";

This is more of a logic question and whether or not a nested while loop is the right way to do it. What I'm trying to say is if the 'followerid' from 'user followers table' is not the same as the 'id' from users table (which is from the previous loop) - echo the follow button, else echo the following button.

This is working file while I have data in the followers table but If I don't nothing shows (as there are no rows) - How could I implement this in my PHP? So also if there are no rows in 'followers table' echo follow button?

Before the if($tables['followerid']!=$table['id']) statement, echo the two value and see what you get. — Osiris
– Osiris, Commented Dec 24, 2012 at 13:32
sorry, I think I included them by mistake off the previous loop — Andrew Charlton
– Andrew Charlton, Commented Dec 24, 2012 at 13:37
Please, don't use mysql_* functions in new code. They are no longer maintained and are officially deprecated. See the red box? Learn about prepared statements instead, and use PDO or MySQLi - this article will help you decide which. If you choose PDO, here is a good tutorial. — Naftali
– Naftali, Commented Dec 24, 2012 at 14:06

echo_Me · Accepted Answer · 2012-12-24 14:06:34Z

1

you can try do it like that

$selecty = mysql_query("SELECT * FROM followers WHERE userid='".$_SESSION['id']."'");

$rowsy = mysql_num_rows($selecty);

echo '<td>'. $table["username"]. '</td>';
echo '<td>';

while ($tables = mysql_fetch_assoc($selecty)) {
    if($tables['followerid']!=$table['id'] and $tables['followerid'] != '') {
        echo '<a href="#" data-userid="'.$table['id'].'" class="follow"></a>'; 
    } else if($tables['followerid'] =$table['id'] and $tables['followerid'] !='') {
        echo '<a href="#" data-userid="'.$table['id'].'" class="following"></a>';
    } else {
        echo what you like here when $tables['followerid'] = ''
    }
}

echo '</td>';
echo "<tr>";

edit

      class="follow">'.'</a>'
                      ^------------you dont have to make point and single quotes here

   $selecty = mysql_query("SELECT * FROM followers WHERE    userid='".$_SESSION['id']."'");

 $rowsy = mysql_num_rows($selecty);
 echo '<table><tr>';
  echo '<td>'. $table["username"]. '</td></tr>';

 if ($tables['followerid'] !== ''){
 while ($tables = mysql_fetch_assoc($selecty)) {
 echo '<tr><td>';
if($tables['followerid']!=$table['id'] and $tables['followerid'] != '') {
    echo '<a href="#" data-userid="'.$table['id'].'" class="follow"></a></td></tr>'; 
} else if($tables['followerid'] =$table['id'] and $tables['followerid'] !='') {
    echo '<a href="#" data-userid="'.$table['id'].'" class="following"></a></td></tr>';
} else {
    echo "what you like here  </td></tr>";
}
}
}
else {

 echo "do your code here " ; 
}

echo "</table>";

edited Dec 24, 2012 at 14:06

answered Dec 24, 2012 at 13:38

echo_Me

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8 Comments

echo_Me Over a year ago

buttons ? what buttons do u mean ? u have links <a> u can do them in else

Andrew Charlton Over a year ago

sorry - it still isn't showing the links - it shows the <td>username<td> but not the following <td>

echo_Me Over a year ago

what this variable $table?

Andrew Charlton Over a year ago

$table is from the previous while loop getting the 'id' from mysql of users

echo_Me Over a year ago

replace this and see if the link appears or not <a href="#" data-userid="'.$table['id'].'" class="follow">gggggggg</a> replace it in the first link and in second link

|

ethrbunny · Accepted Answer · 2012-12-24 13:58:25Z

0

Put a boolean (FALSE) at the start of the 'followers' loop such that if it gets crossed make it TRUE. If you get outside the loop and it's still FALSE then add the button anyway.

$trip = FALSE;

while ($tables = mysql_fetch_assoc($selecty)) {
  if($tables['followerid']!=$table['id']) {
    echo '<a href="#" data-userid="'.$table['id'].'" class="follow">'.'</a>'; 
  } else {
    $trip = TRUE;
    echo '<a href="#" data-userid="'.$table['id'].'" class="following">'.'</a>';
  }
}

if( !$trip ) echo '<a href="#" data-userid="'.$_SESSION['id'].'" class="follow">'.'</a

edited Dec 24, 2012 at 13:58

answered Dec 24, 2012 at 13:34

ethrbunny

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7 Comments

Andrew Charlton Over a year ago

this kinda works...but if a user is not following somebody it repeats it twice - so class="follow"

Andrew Charlton Over a year ago

or if a user follows all of the users it shows both the follow class and following class

ethrbunny Over a year ago

Add the $trip to the if{} clause also.

ethrbunny Over a year ago

Add a 2nd type of $trip - or a counter.

Andrew Charlton Over a year ago

is it something to do with nesting the while loop why it isnt working?

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