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Okay, for this page, I need to check if the user is logged into my site or not. If the person is, $_SESSION['username'] will return a true value. If the value is true, I want to print the content of the page. If it returns false, I want to print an error message like "You are not currently...

Thanks, it's now working. :)

I get an error now. 9 $time = date("U"); 10 $display = $time-900; 11 $res = mysql_query("SELECT COUNT (*) as cnt FROM messages WHERE date >= $display"); 12 $info=mysql_fetch_assoc($res); 13 $entries = $info['cnt']; Warning: mysql_fetch_assoc(): supplied argument is not a valid MySQL result...

Okay, the following code is for a type of chat room in which, either messages that are less than 15 minutes old, or the most recent 25 messages will be printed. However, all that is being printed is the HTML tag and the Backround tag, which are not in the PHP coding. And, yes, I do have a...

I have recurring issues with incorporating SQL syntax with PHP. This is what I have, followed by the error I'm getting. elseif ($_GET[action] == 'edit' & $_POST[username] != '') { $query="UPDATE users SET username='$_POST[username]', PASSWORD='$_POST[PASSWORD]', handle='$_POST[handle]'...

Oh, thank you very much, it worked :D

I did that and got this error:

Yes, that helps, but now I'm ending up with an SQL error in syntax.

I'm not sure why this code isn't working. $data="('$_POST[username]', '$_POST[PASSWORD]', '$_POST[handle]', '$_POST[rank]', '$_POST[group]', '$_POST[gender]', '$_POST[av]')" $result3 = mysql_query("INSERT INTO users $data") or die(mysql_error()); echo "Account successfully updated...

I'm sort of confused. I'm a beginning programmer, and I first learned PERL. I'm wondering if PHP has branching for if, elsif, elsif, elsif, else type of thing where there are mulitple situations, and if so what is the syntax for it?

Oh thanks, I guess my table setup failed :)

I can't get this function to work. A sample code: <? $link = mysql_connect("localhost", "users", "users"); mysql_select_db("users", $link); $result = mysql_query("SELECT * FROM user", $link); $num_rows = mysql_num_rows($result); echo "$num_rows Rows\n"; ?> And I'm getting the following...

Not what I meant, but that works just as well, I think. Like, if I wanted to print the number nine, I'd just use $rec[9] correct?

I guess I'm not describing it very well. I want ($username,$rank,$handle,$password,$group,$word) to be in a separate file, so that if I need to add a new variable, ex: ($username,$rank,$handle,$password,$group,$word,$NEW), then I wouldn't want to go through every program and add ,$NEW to every...

Any help?

In this case, $group will be used to determine whether or not a user has access to a given page.

Oh, and it's an account file.

pollock77|Admin|John Smith|abc123|admin|Programmer

I see why you would be confused, but that's not what I meant. I want one file to be ($username,$rank,$handle,$password,$group,$word) Then, I want to call it up in this program, something like open(ACCOUNT, $data_file) || die("Could not open file!"); @raw_data=<ACCOUNT>; close(ACCOUNT)...

open(ACCOUNT, $data_file) || die("Could not open file!"); @raw_data=<ACCOUNT>; close(ACCOUNT); foreach $account (@raw_data){ chop($account); ($username,$rank,$handle,$password,$group,$word)=split(/\|/,$account); I want the ($username,$rank,$handle,$password,$group,$word) in a separate...