Question

there are currently three posts but when onclick alerts only one post.If I echo the posts insite the loop,then all three posts are shown,however if I alert them,then onlyone posts is show.plz help or suggest any alternative approach.

  $sql=mysqli_query($db3,"SELECT * from user where  id='$id'");
            $num_rows=mysqli_num_rows($sql);

            while($row=mysqli_fetch_array($sql)){

                $posts=$row['posts'];
            }


            ?>
        <span onclick=u(<?php echo $posts; ?>)> <?php echo $num_rows  ?> </span>

        <script type="text/javascript">
            function u(posts) {
                alert(posts);
            }
        </script>
        <?php

Update

Here is second query
After using the array approach ,If i use fancy box with $num rows to show all posts in the fancybox .I am again getting the only one result on the fancybox.Plz help

            $posts[] = $row['posts'];

}
foreach ($posts as $af){

                echo "<div id='#modelbox_id'>$af</div>";}

            ?>


            <a href="#modelbox_id" class="modelbox"><?php   echo $num_rows; ?></a>

            <?php

Answer 1

It is because you are assigning a single post row into the $posts variable, and then overwriting that variable with a new value on each iteration of the while loop.

Try something like this instead:

$posts = array();
while($row=mysqli_fetch_array($sql)){
    $posts[] = $row['posts'];
}

Then you will need to print each value of the $posts array.

For example, you could implode the array into a string:

<span onclick=u(<?php echo implode(', ', $posts); ?>)> ...