Senior Coders
A better question would be what are you trying to do? From the looks of your code, you're either trying to print out a list of users and flag which ones have level = 9, or you're selecting a specific user, determining if they have level = 9, and you're doing it wrong.Originally Posted by keyboard1333
- Josh
Administrator
After a user logs in to the members area, if they have are an admin ( levels ==9) I would like it to display welcome admin
global $Moderator;
A better approach would be to determine that during the login, and then your scripts could refer to it whenever needed.
Administrator
Thanks for the suggesting traq, I will do that after i've gotten the actual script to work.
Something in there is causing this errorPHP Code:$username1 = '$username';
$SQL = "SELECT `id` FROM `users` WHERE `username`='$username1' AND `level`='9'";
if( mysql_num_rows( mysql_query( $SQL ) ) === 1 ){
echo "Welcome Admin";
}else{
echo "Welcome";
}
( ! ) Parse error: syntax error, unexpected T_VARIABLE, expecting ',' or ';' in C:\Documents and Settings\Owner\Desktop\canberra amatuer productions\www\Canberra Amatuer Productions\echo time.php on line 104
global $Moderator;
where is line 104? post the dozen or so lines before that.
also, is the user's username literally "$username" ? or is$usernamea variable that holds the actual username?
seems like -everyone- is quoting their variables lately...???
Administrator
Thats the entire code. And $username is the name of the variable that holds the usernamePHP Code:<?php
mysql_connect("localhost", "root", "") or die(mysql_error());
mysql_select_db("keyboard_test") or die(mysql_error());
if(isset($_COOKIE['ID_my_site']))
{
$username = $_COOKIE['ID_my_site'];
$pass = $_COOKIE['Key_my_site'];
$check = mysql_query("SELECT * FROM users WHERE username = '$username'")or die(mysql_error());
while($info = mysql_fetch_array( $check ))
{
if ($pass != $info['password'])
{ header("Location: login.php");
}
else
{
$links = "link2.php";
}
}
}
else
{
header("Location: login.php");
}
?>
<?php
if (isset ($_POST['submit']))
{
$comment = mysql_escape_string (trim (nl2br($_POST['comment'])));
// makes sure they filled it in
if(!$_POST['comment']) {
die('You didn\'t enter a comment.');
}
$date = date("d/m/y");
$cheese = "$comment <br /><hr /> ";
$guoc = "<b>$username</b>";
$sql = mysql_query ("INSERT INTO comments (id,comments,name,date,ip) VALUES ('0','".$cheese."','".$guoc."','".$date."','" . $_SERVER['REMOTE_ADDR']."')");
echo '<br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><center>Your comment has been entered successfully!</center>';
echo '<center><form method="post" action="echo time.php">';
echo '<input type="submit" value="Click me" name="name"> <br />';
echo '</form>';
echo '</center>';
}
else
{
?>
<html>
<head>
</head>
<body>
<?php include("$links");
echo "Hey <b> $username</b>"
$username1 = '$username';
$SQL = "SELECT `id` FROM `users` WHERE `username`='$username1' AND `level`='9'";
if( mysql_num_rows( mysql_query( $SQL ) ) === 1 ){
echo "Welcome Admin";
}else{
echo "Welcome";
}
?>
<br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br />
<p><b><u>COMMENTS</u></b></p>
<?php
// POST data wasnt entered, so display the comments and comment form
// view comments from database
$sql = mysql_query ("SELECT * FROM comments ORDER BY date DESC") or die(mysql_error());;
while ($row = mysql_fetch_array ($sql)) {
?>
<table border="0" cellspacing="0" cellpadding="4" width="500">
<tr>
<td>
<?php
echo $row['name'].'<br />';
?>
</td>
<td>
<?php
echo $row['date'].'<br />';
?>
</td> </tr>
</table><br />
<?php
echo $row['comments'].'<br />';
}
echo '<br /><br />
<form action="echo time.php" method="post">
Comments:<br />
<textarea name="comment" cols="40" rows="7"></textarea>
<input type="submit" value="Submit" name="submit">
</form>';
}
?>
<body style="background-color:lightgreen">
</body>
</html>
global $Moderator;
this line (101):has no ending semicolon.PHP Code:echo "Hey <b> $username</b>"
And if $username is a variable, don't quote itPHP Code:$username // works
"$username" // works (but no reason to do it like this)
'$username' // don't works : /
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