ThePHPNewbie Posted August 18, 2013 Share Posted August 18, 2013 Hi There, I have got this code: function displayPosts() { $query = mysql_query("SELECT * FROM social ORDER BY id DESC") or die(mysql_error()); while( $result = mysql_fetch_assoc($query) ) { echo ' <div id="socialnewsfeeditem"> <p id="newsfeeditemperson"> '.$result["fullname"].' </p> <p id="newsfeeditemcontent"> '.$result["content"].' </p> <p> <center> <a href="social.php?id='.$result["id"].'&do=like">Like!</a> </center> </p> <div id="likes"> <p>'.self::displayLikes($result["id"]).'</p> </div> </div> '; } } function displayLikes($id) { $query = mysql_query("SELECT * FROM sociallike WHERE postid = '$id'") or die(mysql_error()); } function like($id, $username) { $query2 = "INSERT INTO sociallike VALUES('$id', '$username')"; mysql_query($query2); } But when I try to displayLikes the all come up all over the place on the page and I have no clue why it is doing this. I will upload the whole (functions.php) functions and the (social.php) php file on here. Many Thanks for Your Help social.php functions.php Quote Link to comment Share on other sites More sharing options...
trq Posted August 18, 2013 Share Posted August 18, 2013 But when I try to displayLikes the all come up all over the place on the page and I have no clue why it is doing this. displayLikes does not display anything. You might need to describe what you are actually doing. Quote Link to comment Share on other sites More sharing options...
ThePHPNewbie Posted August 18, 2013 Author Share Posted August 18, 2013 displayLikes does not display anything. You might need to describe what you are actually doing. Aaaa sorry! I was trying a couple of other things! I'll just undo a couple of things and upload. Sorry for wasting your time! Quote Link to comment Share on other sites More sharing options...
ThePHPNewbie Posted August 18, 2013 Author Share Posted August 18, 2013 displayLikes does not display anything. You might need to describe what you are actually doing. function displayLikes($id) { $query = mysql_query("SELECT full_name FROM sociallike WHERE postid = '$id'") or die(mysql_error()); while( $result = mysql_fetch_assoc($query) ) { echo $result['full_name']."likes"; } } Theres the corrected code. Apologies again. Quote Link to comment Share on other sites More sharing options...
Solution ThePHPNewbie Posted August 18, 2013 Author Solution Share Posted August 18, 2013 Hello? Quote Link to comment Share on other sites More sharing options...
Recommended Posts
Join the conversation
You can post now and register later. If you have an account, sign in now to post with your account.