How can i display image from a link instead of link?

Hello,

I have got links field in the database. The entries are like this:

http://webpicshot.com/images/0vhoxwbxxoaicir13la.jpg;http://webpicshot.com/images/i2qf04wgmwstdukqpdt.jpg;http://rapidshare.com/files/106224452/TenaciousDFhergently.avi ;

PHP:

I have got image link urls and normal urls. Every url is seperated with ";"

What can I do to show the image for the image url instead of the url?

I want to display image instead of the url.

I just use

$row['links']

PHP:

to show the links.


Thanks

 

You need to split up the url's and then display the images.

Something like:

$imageArray = array('jpg','gif','png');

$urls  = explode(';',$row['links']);

foreach($urls as $url)
{

if(in_array(substr($url,3),$imageArray))
{

echo '<img src="'.$url.'" />';

}

}

PHP:

 

Thank you for your reply. I tried your code but it did not work. It is just saying Array instead of image or links.


Thanks

 

if(in_array(substr($url,3),$imageArray))

for the line above replace the number 3 with minus 3 (-3)
tell me what happens

 

$imageArray = array('jpg','gif','png');
$urls  = explode(';',$row['links']);
foreach($urls as $url)
{
if(in_array(substr($url,(-3)),$imageArray)){
echo '<img src="'.$url.'" />';
}
}

PHP:

This worked kind of. The problem is if there is normal link other than image in the field , it does not display.

Any idea how can i show normal links as well as image links?


Thanks

 

ok add the following lines to the code

if(in_array(substr($url,(-3)),$imageArray)){
echo '<img src="'.$url.'" />';
}
else
{

echo '<a href='.$url.'>'.$url.'</a>';

}

tell me what happens

 

Thank you, it worked.

 

Last one more question. What about if i want to show the first image and rest of the urls as links. Say there are 4 urls. First 2 are images and last 2 are links.

I want to show first image url as image and other 3 urls as url.

this is what i have got now.

$imageArray = array('jpg', 'gif', 'png'); 

$urls  = explode(';', $row['links']); 

foreach ($urls AS $url) 
{ 
    if (in_array(substr($url, -3), $imageArray)) 
    { 
        echo "<img src=\"$url\" /><br /><br />"; 
    } 
else
{

echo '<a href='.$url.'>'.$url.'</a><br />';

}
}

PHP:

 

ok try this code

$counter = 0;

foreach($urls as $url)
{

if(in_array(substr($url,-3),$imageArray))
{
if($counter==0)
{
echo "<img src=\"$url\" /><br /><br />";
$counter++;

}
else
{

echo '<a href='.$url.'>'.$url.'</a><br/><br/>';

}
}

 

Thank you very much. it worked, this code does not make my site run slow, does it?

Thanks

$imageArray = array('jpg', 'gif', 'png'); 
$urls  = explode(';', $row['links']); 
$counter = 0;
foreach ($urls AS $url) 
 {     
 if (in_array(substr($url, -3), $imageArray))     
 {
 if($counter==0)
 {
 echo "<img src=\"$url\" /><br /><br />";     
 }
 $counter++;
 }
 else
 {
 echo '<a href='.$url.'>'.$url.'</a><br />';
 }
 }

PHP: