wee493 Posted November 1, 2009 Share Posted November 1, 2009 I'm getting a Resource id #4 with the following code and can't find the source of the issue. The following PHP receives a list of posts from a database and echo's them. Everything echos fine and the Resource id #4 error only shows up once, so I don't think it has anything to do with the while loops echoing the posts. $error = ''; $ppp = '5'; $reg_num = ($page - 1) * $ppp; if ($user != 'guest' && isset($_SESSION['user'])){ $has_feeds = "SELECT feed FROM subscriptions WHERE username = '$user' AND type = 'blog'"; if (mysql_num_rows($has_feeds) == '0'){ // If user has no feeds // $error = '<p align="center"><font color="red" size="4"><ul><strong>You are not subscribed to any feeds</strong></ul></font><font color="green" size="4"><strong>You can edit your feeds <a href="manage_feeds.php">here</a>.</strong></font>'; } else { $user_feeds = array(); // define an empty array while ($r = mysql_fetch_array($has_feeds)){ $user_feeds[] = $r['feed']; // add an element to the array } // end while // // Get posts that user has hidden // if ($user != 'guest'){ $sql = mysql_query("SELECT * FROM user_hide_posts WHERE user = '$user'"); if (mysql_num_rows($sql) != '0'){ $exempt_p = array(); // define an empty array while ($r = mysql_fetch_array($sql)){ $exempt_p[] = $r['pid']; } } else { $exempt_p = array(1, 2, 3); } $qry = mysql_query("SELECT post_per_page FROM userinfo WHERE username = '$user'"); // Get users Post Per Page Setting // $ppp = mysql_result($qry, 0); } // end hidden posts // $user_feed = array(); // define an empty array foreach ($user_feeds as $f){ $user_feed[] = $f; // add an element to the array } // End foreach // $sql = mysql_query("SELECT * FROM posts WHERE feed_id IN(" . implode(',',$user_feed) . ") AND status = 'publish' AND id NOT IN (" . implode(',',$exempt_p) . ") ORDER BY `posts`.`date` DESC LIMIT $reg_num, $ppp"); // Set getting post SQL statement // $sqlf = mysql_query("SELECT id FROM posts WHERE feed_id IN(" . implode(',',$user_feed) . ") AND status = 'publish' AND id NOT IN (" . implode(',',$exempt_p) . ") ORDER BY `posts`.`date` DESC LIMIT $reg_num, 1"); // SQL to see if it is first post // } } else { $sql = mysql_query("SELECT * FROM posts WHERE feed_id > '0' AND status = 'publish' ORDER BY `posts`.`date` DESC LIMIT $reg_num, $ppp"); // Set getting post SQL statement // $sqlf = mysql_query("SELECT id FROM posts WHERE feed_id > '0' AND status = 'publish' ORDER BY `posts`.`date` DESC LIMIT $reg_num, 1"); // SQL to see if it is first post // } echo $sql; while ($r = mysql_fetch_array($sql)){ $title = $r['title']; $id = $r['id']; $content = $r['data']; $link = $r['permalink']; $feed_id = $r['feed_id']; // begin shares // $share_digg = $r['share_digg']; $share_delicious = $r['share_delicious']; $share_reddit = $r['share_reddit']; $share_facebook = $r['share_facebook']; $share_twitter = $r['share_twitter']; // end shares // $date = Timesince($r['date']); $first1 = @mysql_result($sqlf, 0); if ($first1 == $id){ $first1 = 'first '; } else { $first1 = ''; } // Get link to website homepage & favicon // $query = mysql_query("SELECT feed_domain FROM sc_feeds WHERE feed_id = '$feed_id'"); $post_domain = @mysql_result($query, 0); $query2 = mysql_query("SELECT feed_icon FROM sc_feeds WHERE feed_id = '$feed_id'"); $favicon = @mysql_result($query2, 0); include('../functions/remove_unwanted_post_content.php'); ?> <div class="<?PHP echo $first1; ?> article" id="<?PHP echo $id; ?>" style="display: block"> <h3><a href="<?PHP echo $link; ?>" onmousedown="javascript: MyAjaxRequest('main', './functions/count_visit.php?state=visit&feed=<?PHP echo $feed_id; ?>&post=<?PHP echo $id; ?>&type=blog&user=<?PHP echo $user; ?>');return true"><?PHP echo $title; ?></a></h3> <p class="meta"><a href="http://<?PHP echo $post_domain; ?>" onmousedown="javascript: MyAjaxRequest('main', './functions/count_visit.php?state=visit&feed=<?PHP echo $feed_id; ?>&post=<?PHP echo $id; ?>&type=external&user=<?PHP echo $user; ?>&dest=http://<?PHP echo $post_domain; ?>');return true"> <?PHP echo " <img src=\"$favicon\" style=\"border: 0px; vertical-align: text-bottom; padding:0px; background: none; margin: 0px; width:16px; height: 16px;\" /> "; echo ucfirst($post_domain); ?></a> | <?PHP echo $date; ?> ago <?PHP if ($user != 'guest'){ ?> <span class="hide_post"><a href="javascript: MyAjaxRequest('mn', './functions/hide_post.php?state=hide&post=<?PHP echo $id; ?>'); toggle(<?PHP echo $id; ?>);" id="hide_post"><img src="img/delete.png" alt="Hide Post" /></a></span> <?PHP } ?> </p> <?PHP echo $content; ?> <!-- Share Links --> <div class="share_post"> <!-- <span class="right"><a href="#" onclick="javascript: MyAjaxRequest('email', './functions/share.php?state=email&post=<?PHP //echo $id; ?>'); return false;"><img src="img/email-48x48.png" style="border: 0px; padding:0px; margin: 0px; width:35px; height:35px;" /></a></span> --> <span class="right"><a href="<?PHP echo $share_delicious; ?>"><img src="img/delicious-48x48.png" style="border: 0px; padding:0px; margin: 0px; width:35px; height:35px;" /></a></span> <span class="right"><a href="<?PHP echo $share_digg; ?>"><img src="img/digg-48x48.png" style="border: 0px; padding:0px; margin: 0px; width:35px; height:35px;" /></a></span> <span class="right"><a href="<?PHP echo $share_facebook; ?>"><img src="img/facebook-48x48.png" style="border: 0px; padding:0px; margin: 0px; width:35px; height:35px;" /></a></span> <span class="right"><a href="<?PHP echo $share_reddit; ?>"><img src="img/reddit-48x48.png" style="border: 0px; padding:0px; margin: 0px; width:35px; height:35px;" /></a></span> <span class="right"><a href="<?PHP echo $share_twitter; ?>"><img src="img/twitter-48x48.png" style="border: 0px; padding:0px; margin: 0px; width:35px; height:35px;" /></a></span> <div id="email"></div> </div> <!-- End Share Links --> </div> <?PHP } // End While // ?> Link to comment https://forums.phpfreaks.com/topic/179779-cant-find-the-source-of-resource-id-4/ Share on other sites More sharing options...
Alex Posted November 1, 2009 Share Posted November 1, 2009 You're trying to echo a mysql resource. echo $sql; Link to comment https://forums.phpfreaks.com/topic/179779-cant-find-the-source-of-resource-id-4/#findComment-948505 Share on other sites More sharing options...
wee493 Posted November 1, 2009 Author Share Posted November 1, 2009 You're trying to echo a mysql resource. echo $sql; Wow, I feel stupid. What a noob move, haha. Thanks! Link to comment https://forums.phpfreaks.com/topic/179779-cant-find-the-source-of-resource-id-4/#findComment-948510 Share on other sites More sharing options...
Alex Posted November 1, 2009 Share Posted November 1, 2009 Np. Just make sure to mark the topic as solved, there's a button on the bottom left. Link to comment https://forums.phpfreaks.com/topic/179779-cant-find-the-source-of-resource-id-4/#findComment-948515 Share on other sites More sharing options...
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