Echo content inside foreach only once

Question

I have written the following PHP, to basically show a LI if the status is set to Offline or Delay:

<?php
    if(count($rows) > 0) {  
        foreach($rows as $row)
            {
                $service = $row->service;
                $details = $row->status_details;
                $status = $row->status;
                // status image check       
                    if($status == "ONLINE")
                    {
                        $symbol = "online.png";
                    }
                    if($status == "OFFLINE")
                    {
                        $symbol = "offline.png";
                    }
                    if($status == "DELAY")
                    {
                        $symbol = "delay.png";
                    }
        ?>

        <li class="<?php echo $status; ?>">
            <img src="<?php echo $moduleLoc, $symbol;?>" alt="Service Status <?php echo $service; ?>" /> <span class="details"><?php echo $details; ?></span></li>
        </li>
        <?php   

                }

        ?>
        <?php } ?>

Using CSS, if the class of "ONLINE" is then shown, then LI is hidden.

However, I want to add a statement to be shown if the status is OFFLINE or DELAY, I've written the following, which is working but due to the foreach statement, it's showing for every LI that has either OFFLINE or DELAY. I want it to be shown only once:

<?php
    if(($status == "OFFLINE") || ($status == "DELAY")){
        echo '<li class="systemsGo"><img src="' . $moduleLoc . 'online.png" alt="Online Update" /><span class="details">All Other Applications Are Working Well</span></li>';   
    }
?>

Answer 1

Right, I've come up with one method.

You first need to define a variable as an array:

$get_status = array();

Then in the foreach loop, you need to combine the status of all the results in this newly defined array, like so:

foreach ($rows as $row)
{
    $get_status[] = $status;
}

After the foreach loop, you'll need to extract the array, like so:

extract($get_status);

then you can check if all the statuses in the array are all the same and display a message, like so:

if (count(array_unique($get_status)) === 1 && end($get_status) === 'ONLINE')
{
    echo 'All systems are online';
}

---

Full code and tested:

if (count($rows) > 0) 
{
    $get_status = array();

    foreach ($rows as $row)
    {
        $service = $row->service;
        $details = $row->status_details;
        $status  = $row->status;
        $symbol  = '';
        
        // status image check       
        if ($status == 'ONLINE')
        {
            $symbol = 'online.png';
        }
        if ($status == 'OFFLINE')
        {
            $symbol = 'offline.png';
        }
        if ($status == 'DELAY')
        {
            $symbol = 'delay.png';
        }
        
        $get_status[] = $status;
    ?>
        <li class="<?php echo $status; ?>">
            <img src="<?php echo $moduleLoc, $symbol;?>" alt="Service Status <?php echo $service; ?>" />
            <span class="details"><?php echo $details; ?></span>
        </li>
        
    <?php   
    }
    
    extract($get_status);
    
    if (count(array_unique($get_status)) === 1 && end($get_status) === 'ONLINE')
    {
        echo 'All systems are online';
    }
    else if (count(array_unique($get_status)) === 1 && end($get_status) === 'OFFLINE')
    {
        echo 'All systems are offline';
    }
    else if (count(array_unique($get_status)) === 1 && end($get_status) === 'DELAY')
    {
        echo 'All systems are delayed';
    }
    else
    {
        echo 'Not the same';
    }
}

Hope this helps

P.S: I've made a few minor tweaks such as replacing double quotes with single quotes and also defined $symbol outside of your if statements, just incase the $status doesn't match (will throw a PHP error otherwise)