Why this onclick alerts only one result from the while loop - php

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>";}
?>
<?php echo $num_rows; ?>
<?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); ?>)> ...

Answer 2

Instead try something like this
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>