[Resolved] Opening a Image in a new Layer?

[Sayian]

How is this done?

If you want to open an image in a layer , with a X to close the layer , instead of having the image open in a browser window? an example of this can be found at www.cpixel.com , click on any profile , then try to enlarge a thumbnail .. ?

[thorpe]

this sort of thing is client-side, nothing to do with php.

[gkelly]

when the user clicks on the image it is activating some DHTML script.

I would look into a simple popup DHTML Window script which can be found at:

http://www.dynamicdrive.com/

and then modify simply use php to pass the address of the image into the dhtml script.

~Eugene Kelly

[Sayian]

Ok , ive looked into the dhtml script and found one .. But here is where my question with php comes in.

My php code used a php file to grab the image and then its displayed.

Such as show_image.php?id=2&picture=4

will return user's 2's 4th image , but this is returned in a php file.

Any idea how to use this with somthing like that dhtml script?

[gkelly]

You are probably already thinking this to but I would guess that in your loop where you have the line:

show_image.php?id=2&picture=4

you encapsulate it with some kind of code like the following:

<A HREF="javascript:popUp('show_image.php?id=2&picture=4')">show_image.php?id=2&picture=4
</A>

The DHTML script will have its own way of calling the popup. Theoretically you should be able to include your image link inside some kind of call function provided by the DHTML script.

Hope this helped.

~Eugene Kelly

[Sayian]

I tried that and it was giving me many error's .. here is the string it's calling from.

$return .= "<td width=\"33%\" align=\"center\"><a href=\"viewimage.php?$imgsrc\" target=\"_blank\"><img src=\"$img_src\" width=100 height=100 border=0 alt=\".: $array[username] :.\" title=\".: $array[username] :.\"></a>";

But why am i trying to turn this into a javascript popup? I dont think this will be able to work with the DHTML script , do you?

[gkelly]

I really need more code to look at to be able to help.

Why am i trying to turn this into a javascript popup?

What we want to do is pass the image information into the dhtml popup window.

Suppose you found a dhtml popup window that reads from say file x.php.

In file x.php we have a string that pulls up an image based off information that was passed into it.

File x.php will also use a resize function like the one I attached. This will make sure all images fit into the window nicely.

~Eugene Kelly

[Sayian]

Ok i am starting to understand what your saying now. ive been reading the php manual on passing info to other scripts. Im still a little confused .. let me show you what i want to use.

Here is the php code , making showimage.php display the image first.

$img_src  = $base_url."/image.php?id=" . $a_row["user_id"] . "&p=" . $a_row['total_file_count'];
            $imgsrc = "id=" . $a_row["user_id"] . "&p=" . $a_row['total_file_count'];
            $img_src1 = $base_url."/show_image.php?id=" . $a_row["user_id"] . "&p=" . $a_row['total_file_count'];
            $delete_url = $base_url . "/delete_image.php?id=" . $a_row['total_file_count'] . "&user=" . $a_row['user_id'];

        $return .= "<td width=\"33%\" align=\"center\"><a href=\"viewimage.php?$imgsrc\" target=\"_blank\"><img src=\"$img_src\" width=100 height=100 border=0 alt=\".: $array[username] :.\" title=\".: $array[username] :.\"></a>";

So when you browse a profile on my site , and click on sub-image ( I run a personals site ) it opens the sub image in viewimage.php.

Here is the code the DHTML code want's to use.
http://www.dynamicdrive.com/dynamicindex4/thumbnail.htm

Thats what i want to use. Im starting kind of to understand what your getting at .. but still am very confused.

Thanks and Regards.

[gkelly]

Something like this.

Why return? Just make the function void (no return value) and have it output the string.

Cleaned code up a little. Much easier to read. The javascript call is just an example of where you would want the DHTML function call to be. Then you call the link inside that with the parameters you want to pass to that link. Let PHP do all the work of getting the image and resizing the image. All DHTML is going to do for you is build a nice little pretty window for displaying the image.

<?

$img_src = ''.$base_url.'/image.php?id='.$a_row[user_id].'&p='.$a_row[total_file_count].'';
$imgsrc = 'id='.$a_row[user_id].'&p='.$a_row[total_file_count].'';
$img_src1 = ''.$base_url.'/show_image.php?id='.$a_row[user_id].'&p='.$a_row[total_file_count].'';
$delete_url = ''.$base_url.'/delete_image.php?id='.$a_row[total_file_count].'&user='.$a_row[user_id].'';

echo '<td width="33%" align="center"><a href="javascriptopUp(viewimage.php?'.$imgsrc.')" target="_blank"><img src="'.$img_src.'" width=100 height=100 border=0 title=".: '.$array[username].' :.\"></a>";

?>

~Eugene Kelly

[Sayian]

i think i must use the return element. It gave me a t varible error and another t error as well. Let me show you the whole function. It's kind of long. Im sorry but im getting more and more confused. Im kinda new to php and trying to learn and keep up with what your telling me.

function _get_subpics($id, $album_id=0)
{
  global $tb_users, $tb_albums, $tb_upload_image_details, $base_url, $HTTP_GET_VARS;

  $img_src      = $base_url."/images/notfound_image.gif";
  if ($album_id == 0)
  {
    $album_result = mysql_query("SELECT * FROM $tb_upload_image_details WHERE main='n' AND user_id='$id'");
  }
  elseif ($album_id == -2)
  {
    $album_result = mysql_query("SELECT * FROM $tb_upload_image_details WHERE main='y' AND user_id='$id'");
  }
  else
  {
    $album_result = mysql_query("SELECT * FROM $tb_upload_image_details WHERE album_id='$album_id' AND user_id='$id'");
  }
  $sql          = "select id,username,image,image_url,concat(id,'.',image_ext) as theimage,image_status,total_files from $tb_users where id = '$id'";
  $query        = mysql_query($sql);
  $array        = mysql_fetch_array($query);

  if($array["image_status"] == "approved" || $array["image_status"] == "queued")
  {
    if ($array["total_files"] > 0)
    {
      if ($array["image"] == "here")
      {
        if (mysql_num_rows($album_result) != 0)
        {
          $return = '<table width="100%" border="0" align="center" cellpadding="0" cellspacing="0"><tr>';
          $i = 1;
          while ($a_row = mysql_fetch_array($album_result))
          {
            $img_src  = $base_url."/image.php?id=" . $a_row["user_id"] . "&p=" . $a_row['total_file_count'];
            $imgsrc = "id=" . $a_row["user_id"] . "&p=" . $a_row['total_file_count'];
            $img_src1 = $base_url."/show_image.php?id=" . $a_row["user_id"] . "&p=" . $a_row['total_file_count'];
            $delete_url = $base_url . "/delete_image.php?id=" . $a_row['total_file_count'] . "&user=" . $a_row['user_id'];

        $return .= "<td width=\"33%\" align=\"center\"><a href=\"viewimage.php?$imgsrc\" target=\"_blank\"><img src=\"$img_src\" width=100 height=100 border=0 alt=\".: $array[username] :.\" title=\".: $array[username] :.\"></a>";
        if ( (!isset($_GET['v'])) AND (!isset($_GET['i'])) AND (!isset($_GET['s'])) AND (!isset($_GET['z'])))
        { 
                        if (!isset($_GET['pid']) && $album_id != -2)
                        $return .= "<br><a class=link01 href='" . $delete_url . "'>x</a>";
        }
        $return .= "<td>";
        ((($i%4)==0) ? $return .= "</tr><tr><td colspan=3></td></tr><tr>" : '');
        $i++;
      }
      $return .= '</tr></table>';
    }
  }
}
  }

  return $return;
}

That is the function that displays the user's subpic's as smaller images ( thumbnails) and onclick , loads them into viewimage.php

Sorry for my confusion , this is the most complicated issue ive ever delt with.

Thanks and Regards.

[gkelly]

This just fried my brain 🙂

Your gonna have to give me some time on this (1-2 weeks kind of busy) I will try to find a dhtml script like the one you want and implement an example.

~Eugene Kelly

[Sayian]

everyone i talk to says the php code i use is very confusing ... =( Is thier no way to make the $return use the java script pop up?

[gkelly]

You could probably leave out the return and echo out the link in the javascript popup function inside the php function.

instead of doing...

return "blahblahblah";

do

echo "blahblahblah";

If you have a variable equal to that php function erase it and just put the function where the link would go on you website.

you probably have.

$var=function();
echo "the link is: $var";

instead get rid of $var.
and say:

echo "the link is:";
function();

~Eugene Kelly

[Sayian]

Im lost , thanks for your help and your time tho , i just cant seem to figure this out.

[gkelly]

I figured it out and actually did something similiar to what you are looking for.

This will allow someone to hover over an image and have dhtml create a popup of a larger image for that person.

The image will load when the person hovers over the smaller image.

For this to work I used the attached scripts.

I modified the dhtml script which came off 'SimplytheBest Scripts'. What we are doing in this is dynamically building references to other larger photos from the database.

Then when we hover over a picture we tell the script to build a window for the larger picture that matches the reference tag which I made as the id so we have no duplicates.

Its not hard so if you have not found a solution yet spend like 5 minutes looking at this.

~Eugene Kelly