brem13 Posted March 28, 2009 Share Posted March 28, 2009 hey, im calling an album name from a database and putting it in a link for a image display but if the album has more than one word in it, it doesnt get put in the link, does anyone know why? $album = $qry[profalbum]; $pic = $qry[profpic]; if($pic == ""){ echo "<img src=../../../../../../../../../ques.jpg>"; } else{ echo "<img src=".$folder."/".$album."/thumbs/".$pic.">"; } $album is 'My Album' in the database but 'My' only shows in the link and doesnt include the rest, like it should be .../My%20Album/thumbs/pic.jpg Link to comment https://forums.phpfreaks.com/topic/151552-solved-only-taking-one-word-from-database/ Share on other sites More sharing options...
brem13 Posted March 28, 2009 Author Share Posted March 28, 2009 anyone? please????? Link to comment https://forums.phpfreaks.com/topic/151552-solved-only-taking-one-word-from-database/#findComment-795979 Share on other sites More sharing options...
corbin Posted March 28, 2009 Share Posted March 28, 2009 You need quotes around the name. echo "<img src=\"".$folder."/".$album."/thumbs/".$pic."\">"; Link to comment https://forums.phpfreaks.com/topic/151552-solved-only-taking-one-word-from-database/#findComment-795982 Share on other sites More sharing options...
brem13 Posted March 28, 2009 Author Share Posted March 28, 2009 thank you Link to comment https://forums.phpfreaks.com/topic/151552-solved-only-taking-one-word-from-database/#findComment-795988 Share on other sites More sharing options...
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