joshgarrod Posted June 13, 2009 Share Posted June 13, 2009 Hi all, I am up against my biggest challenge yet! I have been asked to code a blog and the only thing that i can't get working poperly is the comment system. The following code is an include on the article page: <?php error_reporting(E_ALL); $con = mysql_connect("host","user","pass"); if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db("quest", $con); $db = mysql_select_db("quest", $con); if ($_POST['submit']) { $comment = mysql_real_escape_string($_POST['comment']); $name = mysql_real_escape_string($_POST['name']); $submitted = mysql_real_escape_string($_POST['timedate']); $live = "No"; $articleref = mysql_real_escape_string($_POST['articleref']); $page = mysql_real_escape_string($_POST['page']); $SQL = " INSERT INTO comments "; $SQL .= " (comment, name, submitted, live, articleref) VALUES "; $SQL .= " ('$comment', '$name', '$submitted', '$live', '$articleref') "; $result = mysql_db_query($db,$SQL,$cid); if (!$result) { echo("ERROR: " . mysql_error() . "\n$SQL\n"); } //show comments (if any) code $query = mysql_query("SELECT * FROM comments WHERE `articleref` = '$page'"); if(mysql_num_rows($query)==0){ echo "<p>There are currently no comments for this article - please use the form below if you wish to make a comment</p>"; }else{ while($info = mysql_fetch_array($query)){ $comment = $info ['comment']; $name = $info ['name']; $submitted = $info ['submitted']; $live = $info ['live']; if ($live == "Yes") { echo "<p>Comment by - $name on $submitted</p> <p>$comment</p><hr class=\"hr\" width=90%>"; } else { } } } //header("location:thanks_comment.php"); //exit(); } ?> From this I am getting the following errors: Notice: Undefined variable: cid in C:\Users\website\comments.php on line 26 Warning: mysql_db_query(): supplied argument is not a valid MySQL-Link resource in C:\Users\website\comments.php on line 26 ERROR: INSERT INTO comments (comment, name, submitted, live, articleref) VALUES ('comment text, User Name', 'Tuesday 12th June 2009', 'No', '1') Link to comment https://forums.phpfreaks.com/topic/162035-blog-comment-system-not-working/ Share on other sites More sharing options...
keeps21 Posted June 13, 2009 Share Posted June 13, 2009 You're using $cid in your query on line 26 but $cid hasn't been definied anywhere, in other words $cid doesn't exist. Link to comment https://forums.phpfreaks.com/topic/162035-blog-comment-system-not-working/#findComment-855004 Share on other sites More sharing options...
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