[RESOLVED] Selectively displaying multiple records

Maharg105

Hi Guys,

I'm currently having a problem for which I have no doubt there is a ridiculously easy solution. But I'm afraid I appear to be lacking the terminology to find it with any searches.

I currently have seven divs acting as links on my site, they are labelled a, b, c, d, e, f, g. I have a buttons table in my database that contains an id column (corresponding to their character designation) and a picture_link column (containing the url of the image that should be used as the div background.

I can retrieve the records from the database using this code

mysql_query("SELECT id, pic_link FROM buttons")
or die(mysql_error());

but I am struggling to implement them into the site. This is how I would like it to work.

<div class="image_holder" style="background-image: url($picture_link Where ID=a);">
Image 1
</div>

<div class="image_holder" style="background-image: url($picture_link Where ID=b);">
Image 2
</div>

etc.

Any help you could provide would, as always, be massively appreciated.

Thanks,

paulnaj

Hi,

This should get you going...

<?php
$image_directory = './images'; // Change this as required
$mysql_result = mysql_query("SELECT id, pic_link FROM buttons") or die(mysql_error()); 
while($assoc = mysql_fetch_assoc($mysql_result)){
?>
<div class="image_holder" style="background-image: url(<?php echo $image_directory.'/'.$assoc['pic_link']; ?>);">Image 1</div> 
<?php
}
?>

... hope it helps. P

Pikachu2000

Maharg105;10945205 wrote:
Hi Guys,

I'm currently having a problem for which I have no doubt there is a ridiculously easy solution. But I'm afraid I appear to be lacking the terminology to find it with any searches.

I currently have seven divs acting as links on my site, they are labelled a, b, c, d, e, f, g. I have a buttons table in my database that contains an id column (corresponding to their character designation) and a picture_link column (containing the url of the image that should be used as the div background.

I can retrieve the records from the database using this code
mysql_query("SELECT id, pic_link FROM buttons")
or die(mysql_error());
but I am struggling to implement them into the site. This is how I would like it to work.
<div class="image_holder" style="background-image: url($picture_link Where ID=a);">
Image 1
</div>

<div class="image_holder" style="background-image: url($picture_link Where ID=b);">
Image 2
</div>

etc.
Any help you could provide would, as always, be massively appreciated.

Thanks,

Depending on the layout, it might work to use an array to build the divs and fill the pic url value at the same time. If you need text in each div as well, it could be stored in the DB and sent to the div also. I didn't test this, but you get the idea.


$query = "SELECT id, pic_link FROM buttons";
$result = mysql_query($result) or die(trigger_error(mysql_error())); // don't echo mysql_error to browser, log it instead.
while( $array = mysql_fetch_assoc($result) ) {
     echo( "<div id= \"" . $array['id'] . "\" class=\"image_holder\" style=\"background_image:url(" . $array['pic_link'] . ");\"></div>\n" ); // <-- my syntax may be off here
}