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Few Questions

shadowayex

By shadowayex
in PHP

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shadowayex Invested Member

shadowayex

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Is there any way to have things from the fields in my database display on my page automatically. Like lets say a field had names in it. Is there a way to make that name display automatically if that person signs in? And I have it figured out how to sign people in, but how do you make it so the page knows WHO specifically is signed in?

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jhecht Invested Member

jhecht

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Could you explain it a bit more? From what i understand is that you want to display things from a db when a user signs in. If thats the case, then yes, it is possible. you just use SELECT * from tablename SQL statements and the mysql_fetch_assoc() function to create an array with column_name=value pairs. you then just echo out the values(echo $array['column_name'])

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Synook w3schools Guru

Synook

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The way I always do it is just to set a session variable to the user's id, and the script checks that when it needs a user-specific action.

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shadowayex Invested Member

shadowayex

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Ok, I have a log in page made that I learned in an old post. It works. The deal is I want each page to recognize WHO is logged in, and right now, my testing just confirms that someone IS logged in but not WHO. Here's the code from login.php:

<?php$link = mysql_connect('host', 'username', 'password')    or die('Could not connect: ' . mysql_error());mysql_select_db('database') or die('Could not select database');$username = mysql_real_escape_string($_POST['username']);$password = sha1($_POST['password']);$result = mysql_query("SELECT * FROM users WHERE username='$username' AND password='$password'");if ($username == "" || $password == ""){   echo "";}else{  if (mysql_num_rows($result) == 1)   {     $_SESSION['logged_in'] = true;  }  else   {     echo "Error: Incorrect username or password!";  }}if ($username == "" || $password == ""){   echo "";}else{  if (isset($_SESSION['logged_in']))  {     header('location: home.php');  }  else  {     echo "";  }}mysql_close($link);?><!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd"><html xmlns="http://www.w3.org/1999/xhtml" xml:lang="en" lang="en"><head><meta http-equiv="Content-Type" content="text/html;charset=utf-8" /><title>Log In</title></head><body><div><form action="login.php" method="post"><div>Username: <input type="text" name="username" value="" size="20" /><br />Password: <input type="password" name="password" value="" size="20" /><br /><input type="submit" name="submit" value="Login" /></div></form></div></body></html>

Now, I'm sure there are errors and it's more than likely a nightmare to all of you advanced people :) but it works. I had an idea to use the $username variable to recognize the user each page, but if PHP is anything like Javascript, the variables are cleared as soon as you leave the page. So I have no idea how to do it. I thought about a cookie, but then that makes a nightmare for people who have cokies disabled, and honestly I have no idea how to do anything with cookies. What should I do?

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scott100 Dedicated Member

scott100

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Create a session when the login equals true.if (mysql_num_rows($result) == 1) { $_SESSION['logged_in'] = true; $_SESSION['user'] = $username; }//////////////////////////////////Then welcome them with something like this, i hope it works cause i haven't tested it, but should give you an idea.<?php session_start();$user=$_SESSION['user'];if(isset($user)) echo "Welcome $user";?>

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