memo script =/

xlordt

im tring to create this memo script on my own so i can get my php skills up .. but i have a problem.. the form does not show...
nothing shows.. can anyone please tell me what am i doing wrong...

//Check for users is logedin 

if (!$_USER['uid'] >= 2){ 
   echo "<center><br><br><br> 
         Access denied!<br /> Please login to use this feature"; 
   exit; 
   } 
  //When user clicks on submit it checks for inputs and inserts in to db 

  if ((($submit) || ($_POST['username']) || ($_POST['memo']))){ 
  echo "<br><br><br>Hhmm you forgot something =) go back and redo your memo"; 

$date = date("D dS M,Y h:i a"); 
$sql = "INSERT INTO gl_memo(username, memo, date) VALUES ('$_POST[username]', 
         '$_POST[memo]', '$_POST[date]')"; 
$result = mysql_query($sql);    

echo "<br><br><br> Memo Sent to " . $_POST['username']; 

  }else{ 
  // Show Form for users inputs 
  $_FORM .= '<BR><BR><TABLE WIDTH=\"55%\"><TR><TD><FORM METHOD=\"POST\" ACTION=\"$PHP_SELF\"> 
             Select user: <br> 
             <SELECT NAME=\"username\"> 
          <OPTION> ' . $_USER['username'] . ' 
          </SELECT> 
          Messege: <br> 
          <TEXTAREA ROWS=\"6\" COLS=\"45\" NAME=\"memo\"></TEXTAREA> 
          <INPUT TYPE=\"submit\" NAME=\"submit\" ALT=\"submit\"> 
          </FORM></TD></TR></TABLE'; 

}    

echo COM_sitefooter(true);    

?>

diego25

You never printed the variable out! i.e. you don't have a

print $_FORM;

anywhere!

Diego

xlordt

ok i fixed that part.. now i have another error.. ggrr...
the error is in the from part of the script...
this is the error that i get..

You have an error in your SQL syntax near 'from, username, memo, date) VALUES ('xlordt', 'digitalx', 'sup supppp' at line 1

and here is part of my fixed code...


 if ((($submit) || ($_POST['username']) || ($_POST['memo']))){

 $_FROM = $_USER['username'];
 $date = date("D dS M,Y h:i a");
 $sql = "INSERT INTO gl_memo(from, username, memo, date) VALUES ('$_FROM', '$username', 
         '$_POST[memo]', '$date')";
 $result = mysql_query($sql) or die(mysql_error());	
 echo "<br><br><br> Memo Sent to " . $username;

  }else{
  // Select users from db
  $result = "SELECT * FROM users";
  $sql = mysql_query($result) or die(mysql_error());
  WHILE ($row = mysql_fetch_array($sql)){
  $username = $row['username'];
  $option_block .= "<option name=\"username\"> ".  $row['username'] . "</option>";

  }
  // Show Form
  echo '<BR><BR><FORM METHOD="POST" ACTION=' . $PHP_SELF . '>
             <TABLE WIDTH=55% align=center><TR><TD><br>
			 From: <b>' . $_FROM . '</b><br>
             Select user: <br> 
            <SELECT NAME="username"><br> ' .
			   $option_block . '<br> 
		    </SELECT><BR><br>
			 Messege: <br>
			 <TEXTAREA ROWS=8 COLS=65 NAME="memo"></TEXTAREA><br>
			 <INPUT TYPE="submit" NAME="submit" ALT="submit" value="submit memo">
			 </FORM></TD></TR></TABLE>
';

?>

now .. if i was to take out.. from then it will work.. but how will a user know where the memo came from.. how can i fix. this.. thanx =( btw.. thanx for the fast replay

xblue

rename the column "from" - this is a reserved word:
http://www.mysql.com/doc/en/Reserved_words.html