john_wenhold Posted February 3, 2010 Share Posted February 3, 2010 I'm getting an 'error querying database' saying when I hit submit with the entire form filled out. Anyone know what the problem is? Note: the 'xxx' in $dbc are my personal info that I just took out for security sake. <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd"> <html xmlns="http://www.w3.org/1999/xhtml" lang="en" xml:lang="en"> <html> <head> <meta http-equiv="Content-Type" content="text/html; charset=ISO-8859-1" /> <title> TheTextNook Sign up - Results </title> <link type="text/css" rel="stylesheet" href="signup.css" /> </head> <body> <div id="allcontent"> <h1> <span id="title"> TextNook </span> <span id="home"> <a href="home.html"> Home </a> </span> </h1> <h2> TextNook Sign Up </h2> <?php $firstname = ($_POST['firstname']); $lastname = ($_POST['lastname']); $username = ($_POST['username']); $password1 = ($_POST['password1']); $password2 = ($_POST['password2']); $phonenumber = ($_POST['phonenumber']); $sex = ($_POST['sex']); $month = ($_POST['month']); $day = ($_POST['day']); $year = ($_POST['year']); $categories = ($_POST['categories']); $days = ($_POST['days']); $text = ($_POST['text']); $comments = ($_POST['comments']); $submit = ($_POST['signup']); if (isset($submit)) { $output_form = false; if (isset($firstname) && !empty($firstname) && isset($lastname) && !empty($lastname) && isset($username) && !empty($username) && isset($password1) && !empty($password1) && isset($password2) && !empty($password2) && isset($phonenumber) && !empty($phonenumber) && isset($month) && !empty($month) && isset($day) && !empty($day) && isset($year) && !empty($year) && isset($categories) && !empty($categories) && isset($days) && !empty($days) && isset($text) && !empty($text) && isset($sex) && !empty($sex) && ($password1 == $password2)) { $dbc = mysql_connect ('server', 'xxxx', 'xxxx', 'xxxxxx') or die ('Error connecting to MySQL server.'); mysql_select_db ('thetextnook', $dbc); $query = "SELECT * FROM textnook WHERE user_name = '{$_POST['username']}'"; $data = mysql_query ($query); if (mysql_num_rows($data) == 0) { $query = "INSERT INTO textnook ( first_name, last_name, user_name, password, phonenumber, sex, month, day, year, category, days, text, comments) VALUES ( '$firstname', '$lastname', '$username', '$password1', '$phonenumber', '$sex', '$month','$day', '$year', '$categories', '$days', '$text','$comments')"; $result = mysql_query ($query, $dbc) or die ('Error querying database.'); echo 'Sign Up Confirmed' . '<br /> <br />'; echo 'Thanks for joining TextNook!' . '<br /> <br />'; echo 'You have signed up to receive text messages for the following categories:' .implode(", ",$_POST['categories']) . '<br /> <br />'; echo 'You have signed up to receive text messages on the following days:' .implode(", ",$_POST['days']) . '<br /> <br />'; echo 'And you have signed up to receive the following amount of text message(s) per category per day:' . $_POST['text'] . '<br /> <br />'; echo 'You can change these selections at anytime by logging into your account.' . '<br /> <br />'; echo 'Coming Soon: You will be able to choose individual stores and restraunts instead of entire categories.</span>' . '<br /> <br /> <br /> <br /> <br /> <br /> <br /> <br />'; mysql_close ($dbc); Link to comment https://forums.phpfreaks.com/topic/190848-i-need-help-with-this-code-cant-figure-out-whats-wrong/ Share on other sites More sharing options...
jl5501 Posted February 3, 2010 Share Posted February 3, 2010 For debug purposes, it would be useful to print out mysql_error() on the die() call. Obviously not the way to go for a finished product, but useful whilst developing to say or die(mysql_error()); Link to comment https://forums.phpfreaks.com/topic/190848-i-need-help-with-this-code-cant-figure-out-whats-wrong/#findComment-1006428 Share on other sites More sharing options...
schilly Posted February 3, 2010 Share Posted February 3, 2010 well you likely have an error in your query. change $result = mysql_query ($query, $dbc) or die ('Error querying database.'); to $result = mysql_query ($query, $dbc) or die ("MYSQL ERROR: $query - " . mysql_error()); and see what it says. Link to comment https://forums.phpfreaks.com/topic/190848-i-need-help-with-this-code-cant-figure-out-whats-wrong/#findComment-1006432 Share on other sites More sharing options...
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