Drop down list in PHP and MySQL

Question

I need to make drop down list as a link to different pages. How do I do that using PHP, MySQL and HTML.

    <?php
mysql_connect('localhost','root','');
mysql_select_db('test');

$sql="select first_name from users";
$result=mysql_query($sql);

echo "<select First_name=''>";  
echo "<a href='index.html'>";
while($row=mysql_fetch_array($result)){

    echo ":<option value='".$row['first_name']."'>".$row['first_name']."</option>";

    }
    echo"</a>";
    echo"</select>";

?>

If you're writing new code, **_please_ don't use the `mysql_*` functions**. They are old and broken, were deprecated in PHP 5.5 (which is so old it no longer even receives security updates), and completely removed in PHP 7. Use [`PDO`](https://secure.php.net/manual/en/book.pdo.php) or [`mysqli_*`](https://secure.php.net/manual/en/book.mysqli.php) with _prepared statements_ and _parameter binding_ instead. See http://stackoverflow.com/q/12859942/354577 for details. — ChrisGPT was on strike, Mar 02 '17 at 04:02
when click one of the name in the dropdown list i need to get details that belongs to that name.how do i do that. — Dumindu Lakal Nagasinghe, Mar 02 '17 at 04:13
actually those data on data base i need to create item likem button — Dumindu Lakal Nagasinghe, Mar 02 '17 at 04:27
when click that item in drop down i need to get those data from database — Dumindu Lakal Nagasinghe, Mar 02 '17 at 04:27
you are echoing `a` inside `select`, only options are valid. — Suchit kumar, Mar 02 '17 at 04:29
Get rid of `echo "";` and `echo"";`. Make the `value` of the `option` the link and have JS send the user to the page when it is selected, or you could send it via a `form` to a PHP and have it redirect with `header`. — chris85, Mar 02 '17 at 04:47

score 0 · Answer 1 · answered Mar 02 '17 at 04:32

0

You can't use links on the option tag, in order to do that, you need to use javascript. You can try to do something like this:

echo "<select name=\"First_name\" onchange=\"document.location='?'+this.value\">";

answered Mar 02 '17 at 04:32

Rafael Garcia

41
3

score 0 · Answer 2 · answered Mar 02 '17 at 04:42

PHP is a server-side script and does not manipulate a page after a user has adjusted it. Like real time. Only javascript and others do that. PHP creates a page with what you want to see but if you need to change something bases on a dropdown use java. Here is a function that can do that. It unhides a div tag that can have your info you need.

<script type="text/javascript">
window.onload = function() {
    var eSelect = document.getElementById('dropdown');
    var divtag1 = document.getElementById('divtag1');
    var divtag2 = document.getElementById('divtag2');
    eSelect.onchange = function() {
        if(eSelect.selectedIndex === 1) {
            divtag1.style.display = 'block';
        }
        if(eSelect.selectedIndex === 2) {
            divtag2.style.display = 'block';
        }//or if you want it to open a url
        if(eSelect.selectedIndex === 3) {
            window.open("https://yourwebsite.com", "_NEW");
        }
    }
}
</script>
 echo "<div id=\"divtag1\" style=\"display:none;\">/*your code*/
</div>";
echo "<div id=\"divtag2\" style=\"display:none;\">/*your code*/
</div>";

Drop down list in PHP and MySQL

2 Answers2