ebz451
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Everything posted by ebz451
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How can I fix this undefined variable in my code?
ebz451 replied to ebz451's topic in PHP Coding Help
Got it, thanks. I ended up defining it as a null -
How can I fix this undefined variable in my code?
ebz451 replied to ebz451's topic in PHP Coding Help
Thanks for your replies guys, here's what I did with your advice: I'm going to try this right now... this is the same as writing $display_block = ''; right? Just tried this and it makes the notice go away but the script only shows one data entry from the table (the old script shows the notice and all data entries in the table)... Thats what I was thinking but I dont know how to edit it out without using an echo command... -
Hey Everyone, I writing on here because I am taking a course on PHP and learning it by using the textbook PHP 6 Fast & Easy Web Development by Julie Meloni and came across a problem on an assignment that I dont know how to fix. Any who, in the assignment I was working on I wrote a script that shows data I have in a table in my broswer (which btw works) BUT it always gives me the following message: Notice: Undefined variable: display_block in 127.0.0.1\sel_byid.php on line 29. I have checked my code over and and am sure I copied it right but still get the notice. Here's the code: <?php $db_name = "testDB"; $table_name = "my_music"; $connection = @mysql_connect("localhost", "brenda", "mouse") or die(mysql_error()); $db = @mysql_select_db($db_name, $connection) or die(mysql_error()); $sql = "SELECT * FROM $table_name ORDER BY title"; $result = @mysql_query($sql, $connection) or die(mysql_error()); while($row = mysql_fetch_array($result)) { $id = $row['id']; $format = $row['format']; $title = stripslashes($row['title']); $artist_fn = stripslashes($row['artist_fn']); $artist_ln = stripslashes($row['artist_ln']); $rec_label = stripslashes($row['rec_label']); $my_notes = stripslashes($row['my_notes']); $date_acq = $row['date_acq']; if($artist_fn != "") { $artist_fullname = trim("$artist_fn $artist_ln"); }else{ $artist_fullname = trim("$artist_ln"); } if($date_acq == "0000-00-00") { $date_acq = "[unknown]"; } $display_block .= "<P><strong>$id. $title</strong>on $rec_label, by $artist_fullname<br>$my_notes<em>(acquired: $date_acq, format: $format)</em></P>"; } ?> <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <title>My Music (Ordered by Title)</title> </head> <body> <H1>My Music: Ordered by Title</H1> <?php echo "$display_block"; ?> <P><a href="my_menu.html">Return to Menu</a></P> </body> </html> I've been trying to fix it for like a week now and the only thing I understand from the notice is that this line: $display_block .= "<P><strong>$id. $title</strong>on $rec_label, by $artist_fullname<br>$my_notes<em>(acquired: $date_acq, format: $format)</em></P>"; is trying to concatentate to a variable which does not exist ($display_block) and I dont know how to fix it... Can anyone help me?