kjtocool

How do you decide which functions to use, mysql_ or mysqli_ ? I see most people on these forums using mysql_ , though I personally tend to use mysqli_ . In general, I was wondering your thoughts on which to use, and when.

I was going on what he said here, where he makes it sound like he wants only the last 25 rows. You are right though, my code only is faulty.

If the table is auto_incremented, does it not make more sense to simply get the last incremented value, and use it to get the exact values to not delete?

Try simplifying: <?php if (isset($_GET['username'])) { $username = $_GET['username']; $query = "SELECT * FROM users WHERE Username = '$username' LIMIT 1"; $result = mysql_query($query) if ($result) { if (mysql_num_rows($result)) { $array = mysql_fetch_assoc($result); $pID = $array['ID']; $puser = $array['Username']; echo $puser . " " . $pID; } else { echo "No users found with id $id<br />"; } } else { echo "Query failed<br />$sql<br />" . mysql_error(); } } else { echo "No ID passed"; } ?> You had at least 1 syntax error in the last block of code you posted.

Or <?php $databaseConnect = mysqli_connect("localhost", "username", "password", "database"); $result = mysqli_query($databaseConnect, "SELECT LAST_INSERT_ID()"); $last_row_to_delete = $result - 26; $query = "DELETE FROM table_name WHERE column_id >= $last_row_to_delte"; if (mysqli_query($databaseConnect, $query)) echo "Success"; else echo "Failure"; ?>

It all depends on the ID. Assuming you used an auto_incrementing ID, starting at 1, which also serves as a primary key, and you know the ID you want to not delete, you could do something like this: DELETE FROM table_name WHERE column_id > 74 If you don't know the ID, you should first do a statement to select the last inserted value, then subtract 26.

First off, your using mysql_ functions, which I am not familiar with. I tend to use the updated mysqli_ functions. If I were using those, I would try the following: <?php $databaseConnection = mysqli_connect("localhost", "username", "password", "database") Or die("Unable to connect to the database."); $query = "SELECT phpbb_user_group.user_id, phpbb_users.username FROM phpbb_user_group, phpbb_users WHERE phpbb_user_group.group_id = 6 AND phpbb_users.user_id = phpbb_user_group.user_id"; $result = mysqli_query($databaseConnect, $query); $row = mysqli_fetch_assoc($result); while ($row) { $username = $row['username']; $user_id = $row['user_id']; echo "Username = " . $username . " - User_ID = " . $user_id; $row = mysqli_fetch_assoc($result); } ?>

The $userID and $user_ID issue was what was giving me the error. Thanks! I think you're right about the join, I'll try that out and see if I can't get it to work. EDIT: The join worked great. Thanks again for your help.

Couldn't you do something like: SELECT phpbb_user_group.user_id, phpbb_users.username FROM phpbb_user_group, phpbb_users WHERE phpbb_user_group.group_id = 6 AND phpbb_users.user_id = {$i[user_id]}

Hey guys, any help is greatly appreciated. In short, I make a $databaseConnect variable earlier in the code. I then run a query on the database, and get a results set. I enter into a while loop to go through the results, and each time through the loop, I want to query the database again to find something else. I use the same $databaseConnect variable. I am getting an error when trying to query the database the second time through: <?php $query = "SELECT * FROM Comments WHERE article_ID = $article_ID"; $result = mysqli_query($databaseConnect, $query); $row = mysqli_fetch_assoc($result); while ($row) { $userID = $row['user_ID']; $commentID = $row['comment_ID']; $comment = $row['comment']; $query2 = "SELECT username FROM Users WHERE user_ID = $user_ID"; $result2 = mysqli_query($databaseConnect, $query2); $row2 = mysqli_fetch_assoc($result2); $userName = $row2['username']; mysqli_free_result($result2); echo '<table width="70%" border="0" align="center" cellpadding="0" cellspacing="1">'; echo '<tr>'; echo '<td bgcolor="#F4FAFF" class="style6">Author: ' . $userName . ' ( ' . $userID . ' )</td>'; echo '</tr>'; echo '<tr>'; echo '<td bgcolor="#F4FAFF" class="style6">' . $comment . '</td>'; echo '</tr>'; echo '</table><br />'; $row = mysqli_fetch_assoc($result); } mysqli_free_result($result); mysqli_close($databaseConnect); ?> Any explanation as to why I can't query while still using the results of the same query would be great, as well as how I can get around this. KJ

I never define $userID. I defene $_SESSION['userID'] in loginVal.php here: if (mysqli_num_rows($result) == 0) echo "Username or password is incorrect."; else { $row = mysqli_fetch_assoc($result); $_SESSION['userID'] = $row['user_ID']; echo "Successful Login"; echo '<META HTTP-EQUIV="refresh" CONTENT="0.0; URL=http://www.thelink.com/options.php?' . SID . '">'; } In any case, apparently my problem was that this: if (!isset($userID)) should have been: if (!isset($_SESSION['userID'])) So it works now. Thanks a bunch!

Hi there, I am having a problem creating a session and holding it across pages. I start with a simple form that requests username and password. When the user clicks submit, it sends them to loginVal.php: <?php session_start(); ?> <!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=iso-8859-1" /> <title>Xtinct CMS</title> <link href="main.css" rel="stylesheet" type="text/css"> <style type="text/css" > <!-- .style1 { font-family: Arial; font-size: 16px; color: #0F45FF; font-weight: bold; } .style6 { font-family: Arial; font-size: 12px; } --> </style></head> <body> <table width="670" border="0" align="center" cellpadding="0" cellspacing="0"> <form id="form1" name="form1" method="post" action=""> <tr> <td background="images/top.jpg" height="50"> </td> </tr> <tr> <td background="images/middle.jpg" height="200" valign="top"> <table width="670" border="0" cellspacing="0" cellpadding="0"> <tr> <td width="75"> </td> <td><p align="center"><span class="style1">Login</span></p> <p align="left" class="style6"> <?php $userName = $_POST["userName"]; $_SESSION['userName'] = $userName; $passwordHash = md5($_POST["password"]); $_SESSION['passwordHash'] = $passwordHash; $databaseConnect = mysqli_connect("localhost", "username", "password", "database_name") Or die("Unable to connect to the database."); $query = "SELECT user_ID FROM Users WHERE username = '$userName' AND password = '$passwordHash' LIMIT 1"; $result = mysqli_query($databaseConnect, $query); if (mysqli_num_rows($result) == 0) echo "Username or password is incorrect."; else { $row = mysqli_fetch_assoc($result); $_SESSION['userID'] = $row['user_ID']; echo "Successful Login"; echo '<META HTTP-EQUIV="refresh" CONTENT="0.0; URL=http://www.thelink.com/options.php?' . SID . '">'; } mysqli_free_result($result); mysqli_close($databaseConnect); ?> </p> <br /> </td> <td width="75"> </td> </tr> </table> </td> </tr> <tr> <td background="images/bottom.jpg" height="35"> </td> </tr> </form> </table> </body> </html> As you can see, I use session_start() to start the page. I then get values for username and password, hash the password, and make sure the login is valid. I pass values to a few $_SESSION variables, and if the login was valid, I redirect them to another page, options.php. On options.php I run the following check: <?php session_start(); if (!isset($userID)) { header("Location: http://www.thelink.com/index.php"); } ?> Even though I set the $_SESSION['userID'] variable, it always re-directs. While troubleshooting, I tried echoing the SID in the loginVal.php file, and it came up as an empty string. So for whatever reason, I am not generating a session ID. I am new to sessions, can someone tell me what I am doing wrong? KJ