如何通过单击在新选项卡中打开数据库中的几个URL

Question

I found the answer how to open few pages in new tabs with a single click, but I don't know how to place urls from mysqli database using fetch.

mysqli statement is ...

$pick_site = $mysqli->prepare("SELECT url FROM sites where chosen = ? ORDER BY name ASC");
$pick_site->bind_param('s', $yesterday);   
$pick_site->execute(); 
$pick_site->store_result();
$pick_site->bind_result($list_sites);

while ($pick_site->fetch_array()) {
    $mysites = $list_sites;
    }

here is working javascript code for opening tabs

<a id="test" href="#"> CLick </a>
<script type="text/javascript">

  document.getElementById("test").onclick = function(){
   window.open("http://www.google.com",'_blank');
   window.open("http://www.p3php.in",'_blank');
}
</script>

Thank you very much, Ivan.

Answer 1

Just echo links like that :

<a id="test" href="#"> CLick </a>
<script type="text/javascript">

  document.getElementById("test").onclick = function(){

    <?php while ($pick_site->fetch_array()) { ?>    
     window.open("<?= $link ?>",'_blank');
    <?php } ?>
}
</script>