Ie error images

[Xagrik]

Hi everyone!!

 

I'm developing a website in php/mysql.

The database have 4 image fields, but in some cases i just use 2 or 3, the other image fields are empty.

The page where i show the rows os the database was developed with the 4 fields in it.

In firefox everything is allright, but in IE the empty image fields are shown as the image below.

 

Can anyone help me to solve this problem?

 

[attachment deleted by admin]

[trq]

We need to see the relevent code. That image indicates your browser cannot find the image.

[Xagrik]

The code:

 

<a href="<?php  echo $row_motos['img1']; ?>" rel="lightbox"><img src="<?php echo $row_motos['img1']; ?>" width="250" height="187" border="0" /></a> <a href="<?php  echo $row_motos['img2']; ?>" rel="lightbox"><img  src="<?php echo $row_motos['img2']; ?>" width="250" height="187" border="0" /></a>

 

The problem is when the img2 field is blank should not appear anything, in firefox works fine, but in IE shows that image :S

[ChadNomad]

What does this output?

 

print_r($row_motos);

[Xagrik]

Nothing, it stays equal :s

[ChadNomad]

The reason your getting the "X" image with Internet Explorer is because the image doesn't exist (or your linking to it wrong). That's just the way the browser behaves.

 

Are you sure you aren't trying to display an image depending on whether it's set in your array?

 

You could do something like:

if (isset($row_motos['img1']))
{
// print the HTML for img1
}
else
{
// print the HTML for img2
}

 

??

[Xagrik]

I'm not sure, but i think that i could not use the code you gave me, because the structure of my code is already set to show all the image fields, and if i put a default image, it will link to a blank image (i'm using lightbox to enlarge the images).

I just don't finde a way to hide that image if it doesn't exist in the database :s

[ChadNomad]

You really do need to show more code...

 

If you want to hide the image use CSS. For example:

 

<img  src="<?php echo $row_motos['img2']; ?>" width="250" height="187" border="0" style="visibility:hidden;" />

[wrathican]

if i were you i would use a while statement.

 

<?php

$query = "";
$result = mysql_query($query);

while $row = mysql_fetch_array($result)) {
//gets vars
$image = $row['col'];
//display images here like so
?>
<img src="/images/<?php echo $image; ?>" alt="image" />
<?php
}
<?

 

and please, if your going to make the site standards compliant DO NOT forget to add the alt="" tag.

[Lamez]

view your source code, and see what image it is trying to show.