how do I add a link to an image inside of an array ($) variable?

[zoldos]

Below is some code I'm working with. I am still a newbie to PHP.

I want to insert a bit of code in the red highlighted section, directly before $user that will display an image file (.gif) that I specify and pass it along with the rest of the variables. I have not the slightest clue how to do it. Any help or suggestions would be most appreciated! Thanks!

         $icon_sql="select icon_name,icon_alt from se_fileicons where file_type like '%$type%'";
	
		$tmp=$database->database_query($icon_sql);
		$num=$database->database_num_rows($tmp);
		
		$file_id=mysql_insert_id();
		[COLOR=red]$actions->actions_add($user, "newfileupload", Array($user->user_info[user_id],$user->user_info[user_username], $user->user_displayname, $file_id, $title,$icon,$icon_alt), Array(), 0, false, "user", $user->user_info['user_id'], $user->user_info['user_privacy']);[/color]

 

[zoldos]

Nevermind, I found another way to do it.

 

[jpadie]

just so that you know we would never have been able to answer this question as you did not provide the code for the class instantiated as an object under the $actions variable. without understanding the actions_add method you were asking us to guess what might have been inside the proverbial never-opened black box.

 

[zoldos]

The more I looked at the code and from the lack of replies, I started thinking this was the case. Thanks for the info!