PHP - echo link ONLY ONCE if string is not in array? link echoed multiple times?

Question

I have a problem here that Ive described in the unanswered question mysql query returning false even when values DO exist in table? Trying to find if not in table?

Basically, I am trying to echo a special image along with a link IF a username string is in an array that means a conversation has been started with them. This all happens when a search bar is entered in.

I have succeeded in storing the users who either SHOULD NOT BE PRINTED or PRINTED WITH A SPECIAL IMAGE here:

$convoArray = array();

                echo '<form class="usersearchpm" method="post"><input class="searchbarpm" name="searchbarpm"></input></form>';
                $con = mysqli_connect("localhost","username","password","sqlserver");

                    //$num = mysqli_query($con, "SELECT * FROM `pm_messages` WHERE user_from=".$account['id']."");
                $numCon = mysqli_query($con, "SELECT * FROM `conversation` WHERE user_one=".$account['id']."");
                    $numrows = mysqli_num_rows($numCon);
                while ($u = mysqli_fetch_assoc($numCon))
                    {
                    //get other users usernames to echo link
    $getUserTwo = mysqli_query($con, "SELECT * FROM `accounts` WHERE id=".$u['user_two']."");
        $s = mysqli_fetch_assoc($getUserTwo); //s[username] is a user that DOES have convo



                    array_push($convoArray, $s['username']);
                    //echo "<a href='message.php?id={$s['id']}'><li><img class = 'dmCircle' src = '../images/chatCircle.png'/>{$s['username']} </li></a>";
                }

This works. Problem comes with the search - this code prints each user the the size of the convo array because the echo is called each time a for loop circles through -

if(isset($_POST['searchbarpm'])){
//  foreach($convoArray as $name)
//  {
//      echo $name;
//  }
//$sess->getUsers();
    $dbh = mysqli_connect("localhost","username","password","sqlserver");
                    $query = $_POST['searchbarpm'];
                    $q = mysqli_query($dbh, "SELECT * FROM sqlserver.accounts WHERE username LIKE '%".$query."%'");
                    //display all the results
                    while($row = mysqli_fetch_assoc($q)){

//                      $checkConvo = mysqli_query($dbh, "SELECT * FROM sqlserver.conversation WHERE user_one=".$user_id." AND user_two=".$row['id']."");

                        if($row['id']!= $user_id) { //only output users they dont have convo going with because theyre already printed!!!

                        for($i=0;$i<sizeof($convoArray);$i++)
                        {
                            if($row['username']==$convoArray[$i])
                            {
                                echo 'match!!';
                                echo $row['username'];
                                echo "<a href='message.php?id={$row['id']}'><li><img class = 'dmCircle' src = '../images/chatCircle.png'/> {$row['username']}</li></a>";
                            }
                            else 
                            {
                                echo "<a href='message.php?id={$row['id']}'><li><img class = 'dmCircle' src = '../images/noChatCircle.png'/> {$row['username']}</li></a>";
                            }
                        }

                        }
                    }
}//

This code does recognize when the user searched for is in the array and DOES print a special image, however again it does this multiple times.

I need help and cannot finish my project without this. Been struggling for a week here.

How can I print the users in the array with a special image? Or have the search not turn them up at all?

What am I doing wrong?

So the problem is with the 2nd code in if else condition, it is echoing match everytime for you? — Dharam
– Dharam, Commented Jul 1, 2016 at 0:24
@Dharam Ok that works for the ones with a match but how do I avoid echoing the ones without a match multiple times? — blue
– blue, Commented Jul 1, 2016 at 0:31
don't run that for loop it is wrong, do an array search, posting an answer for you. But you may need to modify it as per your array values — Dharam
– Dharam, Commented Jul 1, 2016 at 0:33

Dharam · Accepted Answer · 2016-07-01 00:36:27Z

Use this [Assuming that $convoArray contains the usernames are their array values]

if(in_array($row['username'], $convoArray) {
    echo 'match!!';
    echo $row['username'];
    echo "<a href='message.php?id={$row['id']}'><li><img class = 'dmCircle' src = '../images/chatCircle.png'/> {$row['username']}</li></a>";
}
else {
    echo "<a href='message.php?id={$row['id']}'><li><img class = 'dmCircle' src = '../images/noChatCircle.png'/> {$row['username']}</li></a>";
}

Instead of the following code

for($i=0;$i<sizeof($convoArray);$i++)
{
    if($row['username']==$convoArray[$i])
    {
        echo 'match!!';
        echo $row['username'];
        echo "<a href='message.php?id={$row['id']}'><li><img class = 'dmCircle' src = '../images/chatCircle.png'/> {$row['username']}</li></a>";
    }
    else 
    {
        echo "<a href='message.php?id={$row['id']}'><li><img class = 'dmCircle' src = '../images/noChatCircle.png'/> {$row['username']}</li></a>";
    }
}

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