rallokkcaz

You'd have too make it so in the database there's a field that indicates if the user is online or not. It would be like: online = true when they sign in, and online = false when they sign out or the session expires. That's the only way todo it because sessions are local. Well there are other ways, but this is the most practical. -Zack

Okay guys, mysql_num_rows() is acting really strange, when ever i call it in my loop, it doesn't allow me to display other records that i need to come up. For instance I'm checking to see if a category has a subcategory but it isn't showing the categories that don't have subcategories. Here's the code: $sub_cats = mysql_query("SELECT * FROM `categories` WHERE sub = '". $result['name'] ."'"); while($row = mysql_fetch_array($sub_cats)) { $query = mysql_query("SELECT * FROM `categories` WHERE sub = '". $row['name'] ."'"); $rows_count = mysql_num_rows($query) or die(mysql_error()); if($rows_count > 0) { echo "<div class=\"category\" onclick=\"getCategories('". $row['name'] ."'); return false;\">". $rows_count ." ". UCWords($row['name']) ."</div>"; } else { //Just to check if it would even show the rows without sub-categories. echo "DSAKFKASDSA"; } } Anyone have any idea's why this is happening, or is this just some dumb mistake i'm making because it's been awhile since i've used PHP. -Zack

That didn't work. ???

Is it possible to do something like "mysite.com/user:1". I've tried stuff like: RewriteEngine on RewriteRule ^user:([0-9]+)$ profile.php?id=$1 [L] But it doesn't work, any ideas on what i'm doing wrong, or solutions other than the one i'm doing.

foreign key anyone!?!?!?!?!? ????????

you would probly have to set some variable that makes it so the user can only post 2 posts every minute or somthing like that im not sure how it would look in code but i might work

anyone?

so how exactly would the foreign key be coded into mysql??

no thats for my news table so how exactly would i use a foreign key?

k its got id user content topic time

yup, i had the LIMIT part in the wrong place and got it too work. and can someone answer my question?

i had to go man it was 2 am and i had school sorry

i currently have a news system, and want to allow people to add comments on what they think about the article. would i have to add another table, or another row into my database. any help? Thanks

so how would i make it show it shows links on the bottom and top of the page for pages?

it means that your missing { } 's

anyone

now i get this error You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '-1,0' at line 1 here is the code <?php include 'header.php'; //connect to and select db $conn = mysql_connect('localhost','username','password') or die(mysql_error()); $db = mysql_select_db('pokebash_users',$conn) or die(mysql_error()); //get a list of the info from the table to make the linkies $page = $_GET['page']; // page number $X = ($page - 1); $Y = (5 * $page); $sql = "select * from news ORDER BY id desc Limit $X,$Y "; $result = mysql_query($sql, $conn) or die(mysql_error()); // for each row fetched from the results... while ($list = mysql_fetch_array($result)) { //make the custom linkie echo " <fieldset style='padding:5px; width:500px;'> <legend>"; echo $list['topic'] ; echo " "; echo "Posted at: "; echo $list['time'] ; echo " </legend> <table cellpadding='10px'> <tr> <td> "; echo $list['content'] ; echo " </td> </tr> </table> </fieldset><br> <a href='http://www.getmetola.com/post.php?id={$list['id']}'>View Topic</a> <br> <br> "; } // end while // if the user clicked on a linkie and therefore an id exists (the is_numeric is a basic security precaution) ... if (is_numeric($_GET['id'])) { // for easier var syntax handling $id = $_GET['id']; $sql = "select topic from news where id = '$id'"; $result = mysql_query($sql, $conn) or die(mysql_error()); $userfound = mysql_num_rows($result); if ($userfound) { $user = mysql_fetch_array($result); } // end if found } // end if there was a linkie clicked ?>

how would i make it so when i grab info from the data base, instead of it making one long page of info it spreads it into different pages <?php include 'header.php'; //connect to and select db $conn = mysql_connect('localhost','username','password') or die(mysql_error()); $db = mysql_select_db('pokebash_users',$conn) or die(mysql_error()); //get a list of the info from the table to make the linkies $sql = "select * from news ORDER BY id desc"; $result = mysql_query($sql, $conn) or die(mysql_error()); // for each row fetched from the results... while ($list = mysql_fetch_array($result)) { //make the custom linkie echo " <fieldset style='padding:5px; width:500px;'> <legend>"; echo $list['topic'] ; echo " "; echo "Posted at: "; echo $list['time'] ; echo " </legend> <table cellpadding='10px'> <tr> <td> "; echo $list['content'] ; echo " </td> </tr> </table> </fieldset><br> <a href='http://www.getmetola.com/post.php?id={$list['id']}'>View Topic</a> <br> <br> "; } // end while // if the user clicked on a linkie and therefore an id exists (the is_numeric is a basic security precaution) ... if (is_numeric($_GET['id'])) { // for easier var syntax handling $id = $_GET['id']; $sql = "select topic from news where id = '$id'"; $result = mysql_query($sql, $conn) or die(mysql_error()); $userfound = mysql_num_rows($result); if ($userfound) { $user = mysql_fetch_array($result); } // end if found } // end if there was a linkie clicked ?> does anyone know how to do that???

hey i would but i g2g sorry guys night!

the field name is id

i get this error You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'ORDER BY id desc' at line 1 what does that mean

so how would that look like?

i wanna make it so when i list the topics in my news page it shows them the oppisite way instead of the the articles going 1, 2, 3. i want them to appear 3, 2, 1. here if the code for the view page. <?php include 'header.php'; //connect to and select db $conn = mysql_connect(INFOICANTSHOWGOESHERE!) or die(mysql_error()); $db = mysql_select_db('pokebash_users',$conn) or die(mysql_error()); //get a list of the info from the table to make the linkies $sql = "select * from news LIMIT 3"; $result = mysql_query($sql, $conn) or die(mysql_error()); // for each row fetched from the results... while ($list = mysql_fetch_array($result)) { //make the custom linkie echo " <fieldset style='padding:5px; width:500px;'> <legend>"; echo $list['topic'] ; echo " "; echo "Posted at: "; echo $list['time'] ; echo " </legend> <table cellpadding='10px'> <tr> <td> "; echo $list['content'] ; echo " </td> </tr> </table> </fieldset><br> <a href='http://www.getmetola.com/post.php?id={$list['id']}'>View Topic</a> <br> <br> "; } // end while // if the user clicked on a linkie and therefore an id exists (the is_numeric is a basic security precaution) ... if (is_numeric($_GET['id'])) { // for easier var syntax handling $id = $_GET['id']; $sql = "select topic from news where id = '$id'"; $result = mysql_query($sql, $conn) or die(mysql_error()); $userfound = mysql_num_rows($result); if ($userfound) { $user = mysql_fetch_array($result); } // end if found } // end if there was a linkie clicked ?> anyone know or is my description not good enough?

i do have a login page with sessions and all that stuff its just that all i want is to know how i would make the code i posted above work :-\

i wann make it so only the user admin can view the page, and if your not admin then it sends you somewhere else.