Skip to main content

drop-down navigation menu with HTML PHP MySQL



EDIT FIXED by user @jeroen You have to move the li and a tags inside the while loop.





Thank you for this, F4LLCON





So the final code (not so clean, but works now):







<?php

$query = mysql_query("SELECT * FROM `apps` ");

while ($query_row = mysql_fetch_assoc($query))

{

?>

<li><a href="">

<?php

echo $query_row['TITLE'];

?>

</a>

</li>

<br />

<?php

}

?>







I want a drop-down menu that will show the TITLES out of my MySQL database. I know how to do everything concerning the ID etc.





The only problem is that the PHP version of the drop-down menu will SELECT all the TITLES in my database and parse it as ONE link.





So instead of







Home

>About

>Contact

>Another







it will look like







Home

>About

Contact

Another







With other words







Home

>About Contact Another







It will not make multiple drop-down links but only one drop-down link





If you for example do:







<div>

<?php

echo $query_row['TITLE'];

?>

</div>







It will make a individual <div></div> for every TITLE in my database, so I thought this method would also work for the drop-down links..





Does anybody know how to fix it so it will make individual <li></li> for every TITLE?





Here is a normal drop-down menu:







<li id="media"><a href=""></a>

<ul>

<li id="1a"><a href="">About</a></li>

<li id="1b"><a href="">Contact</a></li>

<li id="1b"><a href="">Another</a></li>

</ul>

</li>







And here is the PHP drop-down menu:







<li id="media"><a href=""></a>

<ul>

<li>

<a href="">

<?php

$query = mysql_query("SELECT * FROM `apps` ");

while ($query_row = mysql_fetch_assoc($query))

{

echo $query_row['TITLE'];

?>

<br />

<?php

}

?>

</a>

</li>

</ul>

</li>




Comments

  1. You have to move the li and a tags inside the while loop.

    echo '<li><a href="">' . $query_row['TITLE'] . '</a></li>';

    ReplyDelete
  2. <li id="media"><a href=""></a>
    <ul>
    <?php
    $query = mysql_query("SELECT * FROM `apps` ");
    while ($query_row = mysql_fetch_assoc($query))
    {
    echo "<li><a href=''>" . $query_row['TITLE'] ."</a></li>";
    }
    </ul>
    </li>

    ReplyDelete

Post a Comment

Popular posts from this blog

[韓日関係] 首相含む大幅な内閣改造の可能性…早ければ来月10日ごろ=韓国

div not scrolling properly with slimScroll plugin

I am using the slimScroll plugin for jQuery by Piotr Rochala Which is a great plugin for nice scrollbars on most browsers but I am stuck because I am using it for a chat box and whenever the user appends new text to the boxit does scroll using the .scrollTop() method however the plugin's scrollbar doesnt scroll with it and when the user wants to look though the chat history it will start scrolling from near the top. I have made a quick demo of my situation http://jsfiddle.net/DY9CT/2/ Does anyone know how to solve this problem?

Why does this javascript based printing cause Safari to refresh the page?

The page I am working on has a javascript function executed to print parts of the page. For some reason, printing in Safari, causes the window to somehow update. I say somehow, because it does not really refresh as in reload the page, but rather it starts the "rendering" of the page from start, i.e. scroll to top, flash animations start from 0, and so forth. The effect is reproduced by this fiddle: http://jsfiddle.net/fYmnB/ Clicking the print button and finishing or cancelling a print in Safari causes the screen to "go white" for a sec, which in my real website manifests itself as something "like" a reload. While running print button with, let's say, Firefox, just opens and closes the print dialogue without affecting the fiddle page in any way. Is there something with my way of calling the browsers print method that causes this, or how can it be explained - and preferably, avoided? P.S.: On my real site the same occurs with Chrome. In the ex