Listing MySQL table

[LukeBateson]

Right, it's hard to explain, but I'll try my best.

Litterally, I have a 'tab' box on my homepage.

 

And a table called 'news'. Now, the table has the rows "shorttitle" "shortstory" and "image" in it.

I want the title the three latest titles to be on the three tabs, and the story and image in the respective content areas. (The table auto increments the ID, and so order by ID descending limit of 3 yes?)

 

Now, the code for the tabs is:

<?php
$query = "SELECT * FROM news where published = '1' ORDER by id DESC LIMIT 3";
$row_news = mysql_fetch_array($query);
?>
<div id="myTabs">
         <ul>
            <li><a href="#firsttab"></a><? echo $row_news['shorttitle']; ?></li>
            <li><a href="#secondtab">newsID2</a></li>            
            <li id="last"><a href="#thirdtab">newsID3 </a></li>            
        </ul>
<div id="firsttab" class="tab_content">
<div class="rafat">
<img src="<? echo $row_news['image']; ?>" />
<p> <? echo $row_news['shortstory']; ?> </p>
</div>
</div>
<div id="secondtab" class="tab_content">
<div class="rafat">
<img src="newsIMAGE2" />
<p> newsSTORY2 </p>
</div>
</div>
<div id="thirdtab" class="tab_content">
<div class="rafat">
<img src="newsIMAGE3" />
<p> newsSTORY3 </p>
</div>
</div>
</div>

 

Now, i've filled in the first tab with the MYSQL queries, but i'm unsure how to fill in the other two, because of the way it's coded.. I can't do a 'repeating' code, because of the IDs of the divs, and the three tabs are bunched together..

 

Anyone any ideas?

Much apreciated.

 

Thanks

Luke

[Psycho]

Dump the results for all three records into an array, then use the array for the output

 

<?php
$query = "SELECT *
          FROM news
          WHERE published = '1'
          ORDER by id DESC
          LIMIT 3";
$row_news = mysql_fetch_array($query);
$news = array();
while($row = mysql_fetch_assoc($row_news))
{
    $news[] = $row;
}
?>
<div id="myTabs">
    <ul>
        <li><a href="#firsttab"><?php echo $news[0]['shorttitle']; ?></a></li>
        <li><a href="#secondtab"><?php echo $news[1]['shorttitle']; ?></a></li>            
        <li id="last"><a href="#thirdtab"><?php echo $news[2]['shorttitle']; ?></a></li>            
    </ul>
    <div id="firsttab" class="tab_content">
        <div class="rafat">
            <img src="<?php echo $news[0]['image']; ?>" />
            <p><?php echo $news[0]['shortstory']; ?></p>
        </div>
    </div>
    <div id="secondtab" class="tab_content">
        <div class="rafat">
            <img src="<?php echo $news[1]['image']; ?>" />
            <p><?php echo $news[1]['shortstory']; ?></p>
        </div>
    </div>
    <div id="thirdtab" class="tab_content">
        <div class="rafat">
            <img src="<?php echo $news[2]['image']; ?>" />
            <p><?php echo $news[2]['shortstory']; ?></p>
    </div>
</div>

[LukeBateson]

I got this error

 

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/lukebate/public_html/**/includes/functions.php on line 68

Warning: mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource in /home/lukebate/public_html/**/includes/functions.php on line 70

[Psycho]

Change this

$row_news = mysql_fetch_array($query);
$news = array();
while($row = mysql_fetch_assoc($row_news))

 

To this

$result = mysql_query($query) or die(mysql_error());
$news = array();
while($row = mysql_fetch_assoc($result))

[LukeBateson]

Brilliant!

Thanks!