a1amattyj Posted February 15, 2008 Share Posted February 15, 2008 Hi, Im currently in the process of making an administration panel. I want admins to be able to add comments to each "report". So, after submitting the text to a form, it directs to the following code.. <? if($_GET['id'] && $_GET['comment'] && !$_GET['email'] && !$_GET['mail']){ require("config.php"); $id = $_GET['id']; $query = "SELECT * FROM reports WHERE id='$id'"; $comments = $_POST['comments']; $result = mysql_query("UPDATE reports SET admin_comments='$comments' WHERE comments='NULL'") or die(mysql_error()); $row = mysql_fetch_array( $result ); } ?> I get this error : Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/easyfor1/public_html/report/admin/report_list.php on line 454 Finally, is it possible that the update ADD's to what is already in the field? Thanks Link to comment https://forums.phpfreaks.com/topic/91323-mysql-update-function/ Share on other sites More sharing options...
craygo Posted February 15, 2008 Share Posted February 15, 2008 You get an error because you can't fetch an array from an update query. if you want to add to whats there $id = $_GET['id']; $query = "SELECT * FROM reports WHERE id='$id'"; $res = mysql_query($query) or die(mysql_error()); $r = mysql_fetch_assoc($res); $comments = stripslashes($r['admin_comments']; $comments .= "\n ".addslashes($_POST['comments']); $result = mysql_query("UPDATE reports SET admin_comments='$comments' WHERE comments='NULL'") or die(mysql_error()); Ray Link to comment https://forums.phpfreaks.com/topic/91323-mysql-update-function/#findComment-468016 Share on other sites More sharing options...
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