PHP - Sign In With Facebook/twitter

Okay, here's what I'm doing. It's sort of like a market script.

When someone wants to buy something, they choose from a drop down the item they wish to buy. However, there is a setup cost for it to be started. Then, based upon the length they choose (anywhere from 1 - 12 months), the cost will add up.

My questions a

Is it possible to assign a setup cost? How can I make it so that, while some things will be, say $2 a month, others will be $3 or other amounts?
Here's what I have so far, to try and figure this out. Feel free to laugh... Or if you could point me to an example of something like this, I'd be very appreciative! Thanks!!


function ChangeInitialCost(){
$cic = ($_POST['newpremium']);
if (($cic == "clan") || ($cic == "guild") ||  ($cic == "house") || ($cic == "militia")){

}


And the form I have: (I haven't entered a submit button yet...)
Code: [Select]
<table width="200" border="1">
  <form action="premium.php" method="post" name="purchase">
  <tr>
    <td>Premium:</td>
    <td><select id="newpremium" name="newpremium">
<option value="casino">Casino</option>
<option value="clan">Clan</option>
<option value="guild">Guild</option>
<option value="house">House</option>
<option value="militia">Militia</option>
</select></td></form>
  </tr>
  <tr><form action="premium.php" method="post" name="length">
    <td>Length:</td>
    <td><select id="length" name="length">
<option value="0">-----</option>
<option value="1">One Month</option>
<option value="2">Two Months</option>
<option value="3">Three Months</option>
<option value="4">Four Months</option>
<option value="5">Five Months</option>
<option value="6">Six Months</option>
<option value="7">Seven Months</option>
<option value="8">Eight Months</option>
<option value="9">Nine Months</option>
<option value="10">Ten Months</option>
<option value="11">Eleven Months</option>
<option value="12">Twelve Months</option>
</select></td></form>
  </tr>
  <tr>
  <td>Cost:</td>
  <td>SOME TYPE OF FUNCTION</td>
</table>

I am coding using php and I am adding a pound sign in my mysql table but it outputs as a question mark, how do I make it output the pound sign? thanks.

Hi there, im trying to create a sign up page for my site! I have used confirm password and email fields on the form but the code still shows the error message even if the passwords and emails match! Can anyone advice me please? 

I am having an issue with £ that i have in my database showing on my webpage as a triangle ?.

I have my database encoded as utf8_general_ci and i declare <meta charset="UTF-8"> in my head.

Is there something that i am missing here?

Help is appreciated.

Kind Regards

Adam

Hi there. I have a javascript code that packages an array as a string and sends it to a php mailer, which then separates the string into an array for mailing using a preg_split regular expression that searches for a comma followed by an html tag.

$myWrongArray = preg_split("/,(?=<)/",$myVar); // lookahead splits the string when the comma is before a tag (the opening bracket < )
//The look ahead is (?=<), which interrogates the next character and if it is the angle bracket, <, the preg_split function operates.

Everything works great EXCEPT when the string contains a degree sign or an &, in which case the array is returned up to the degree or & and then stops.

I can’t for the life of me understand it. Any ideas?

Hi. I am brand new here...

I have a script where I'm trying to parse an email and everything is working fine but it looks like the percent sign (%) is getting stripped out of my message.

The percent sign is in the subject message and it gets removed and I also can't use it in any other strings in the script.

I've tried just adding a slash in front like \% but that didn't fix it.

Is there another way to escape a percent sign when using it in a script?

Any help is appreciated.
Thanks.
-Jeff

The Script:

<?php

    echo "<h1>The Hashtag: </h1>";
    echo $_GET['hashtag'];

?>

An example for the URL on localhost:

/search.php?hashtag=#xbox

If I am not using the number sign, then the word "xbox" does get printed on screen, if I do use the number sign then the hashtag "#xbox" does not get printed on screen.   Any suggestions on how to do this?

Hi guys, I've got a snippet of code that's supposed to make a new directory with the user's unique ID (member_id) in the database. I have the code here but I get a myriad of errors including the infamous "at line 1 error, and the directory does not create. Please help me!!

//Create Directory
$queryString2 = $_SERVER['QUERY_STRING'];
$query2 = "SELECT member_id FROM members WHERE $queryString2 = $row[activation]"; 
$result2 = mysql_query($query2) or die(mysql_error());
mkdir("/home/gafferzo/www/members/clubs/" . mysql_result($result2,0), 0777);  

$username = $_POST['username'];
$password = $_POST['password'];
$month = $_POST['month'];
$day = $_POST['day'];
$year = $_POST['year'];

$query = mysql_query("INSERT INTO users VALUES ('','$username','$password','$month','$day','$year')
mysql_query($query);


The code above is a sample of what I have but what I want is to store an entire birthdate in ONE SQL cell. More like this...


$username = $_POST['username'];
$password = $_POST['password'];
$month = $_POST['month'];
$day = $_POST['day'];
$year = $_POST['year'];

$query = mysql_query("INSERT INTO users VALUES ('','$username','$password','$birthdate')
mysql_query($query);


How is this possible? Can I do this and actually use it efficiently in the future?

I am trying to build my own custom login script.

What I am trying to achieve is once a user has logged in depending on wether they have checked the keep me logged in checkbox they have two options. If they haven't checked it then it creates session variables only, and if they have checked it it also creates cookie variable as well as the session variables.

If they then close their browser / tab without logging out and then revisit the site they will get redirected to login page because the active session variable is no longer there. As soon as they land on the loggin page, it automatically checks for the cookie variable and if it exists, it uses it to login and redirect them automatically.

However the problem that I am facing is that the session variable is still being trashed after a default amount of idle time and forcing a login.

My goal is that the user shouldn't have to re-login unless they have either clicked the logout button.

Can someone please have a look through my solution and advise me as to wether this is the correct method that I am implementing, if there is an easier way to achieve what I want, and is this a secure way to handle user logins.

Thanks in advance.

Andrew



Here is the check code I have placed at the top of each admin page.

Code: [Select]
<?php 
session_start();
$url = (!empty($_SERVER['HTTPS'])) ? "https://".$_SERVER['SERVER_NAME'].$_SERVER['REQUEST_URI'] : "http://".$_SERVER['SERVER_NAME'].$_SERVER['REQUEST_URI'];
$uid = $_SESSION['uid'];
if (!isset($uid)) {
header('location:login.php?redirect='.$url);
exit();
}
?>
Next we have the code for the login.php file.

Code: [Select]
<?php include ('functions.php'); ?>
<?php get_header('login'); ?>
    <div id="login-result">
    <?php
connect();
$redirect = htmlspecialchars(mysql_real_escape_string(addslashes($_GET['redirect'])));

if(isset($_COOKIE['remembered'])){

$username = htmlspecialchars(mysql_real_escape_string(addslashes($_COOKIE['remembered']['username'])));
$password = htmlspecialchars(mysql_real_escape_string(addslashes($_COOKIE['remembered']['password'])));

$sql      = "SELECT * FROM usersT WHERE username='$username' AND password='$password'";
$result   = mysql_query($sql);
$count    = mysql_num_rows($result);
$row      = mysql_fetch_array($result);

$uid      = $row['uid'];
$fname    = $row['firstname'];
$lname    = $row['lastname'];
$role     = $row['role'];

   
if($count==1){
$sql2    = "UPDATE usersT SET status = '1' WHERE uid = '$uid'";
$result2 = mysql_query($sql2);

if($result2){

session_register("uid");
session_register("uname");
session_register("ulevel");
$_SESSION["uid"]    = $uid;
$_SESSION["uname"]  = $fname;
$_SESSION["ufullname"]  = $fname . " " .$lname;
$_SESSION["urole"] = $role;
                                        $home = get_option('home');

if(!empty($redirect)) {
header( 'Location: '. $redirect ) ;
exit(); 
}
else {
header( $home ) ;
exit();
}

}

}
else {
echo "<div class=\"error rounded5 shadow\">Invalid username or password!</div>";
}
}
else if (isset($_POST['admin_login'])){

if(isset($_POST["username"]) && isset($_POST["password"])){

$username_p = htmlspecialchars(mysql_real_escape_string(addslashes($_POST["username"])));
$password_p = htmlspecialchars(mysql_real_escape_string(addslashes($_POST["password"])));
$psw        = md5($password_p);

$sql3      = "SELECT * FROM usersT WHERE username='$username_p' AND password='$psw'";
$result3   = mysql_query($sql3);
$count3    = mysql_num_rows($result3);
$row3      = mysql_fetch_array($result3);

$uid      = $row3['uid'];
$fname    = $row3['firstname'];
$lname    = $row3['lastname'];
$role     = $row3['role'];

   
if($count3==1){
$sql4    = "UPDATE usersT SET status = '1' WHERE uid = '$uid'";
$result4 = mysql_query($sql4);

if($result4){

session_register("uid");
session_register("uname");
session_register("ulevel");
$_SESSION["uid"]    = $uid;
$_SESSION["uname"]  = $fname;
$_SESSION["ufullname"]  = $fname . " " .$lname;
$_SESSION["urole"] = $role;
$home = get_option('home');

if(isset($_POST['remember'])) { 
setcookie("remembered[username]", $username, time() + 86400 * 365 * 2);
setcookie("remembered[password]", $psw, time() + 86400 * 365 * 2); 
}

if(!empty($redirect)) {
header( 'Location: '. $redirect ) ;
exit(); 
}
else {
header( $home ) ;
exit();
}

}

}
else {
echo "<div class=\"error rounded5 shadow\">Invalid username or password!</div>";
}
}
}
?>
    </div><!-- / login-results -->
    <div id="login" class="rounded5 shadow">
<form name="loginform" id="loginform" action="<?php $_SERVER['PHP_SELF']; ?>" method="post">
            <p>
                <label for="username">Username<br>
                <input type="text" name="username" id="username" class="rounded5" value="<?php echo $username_p; ?>" size="20" tabindex="10" /></label>
            </p>
            <p>
                <label for="password">Password<br>
                <input type="password" name="password" id="password" class="rounded5" value="<?php echo $password_p; ?>" size="20" tabindex="20" /></label>
            </p>
            <p class="submit">
            Keep me logged in<input type="checkbox" name="remember" id="remember" /><br /><br /><a href="" class="left">Lost your password?</a>
                <input type="submit" name="admin_login" id="admin_login" class="btn rounded10 right" value="Log In" tabindex="100" />
            </p>
            <div class="cleaner"></div><!-- / cleaner -->
        </form>
    </div><!-- / login-->
<?php get_footer('login'); ?>

Finally here is the code I am using for the logout.php page.

Code: [Select]
<?php
session_start();
include ('functions.php');
connect();
$uid    = mysql_real_escape_string($_SESSION['uid']);
$sql    = "UPDATE usersT SET status = '0' WHERE uid = '$uid'";
$result = mysql_query($sql);

if($result) {
session_unset(); 
session_destroy();

if(isset($_COOKIE['remembered'])){ 
setcookie("remembered[username]", $username, time() - 3600);
setcookie("remembered[password]", $psw, time() - 3600); 
header("location: login.php");
}
exit();
}
else {
echo "You couldn't be logged out at this time.";
}
?>

this code is to create a new user based on posted variables. it works fine for creating the necessary tables and granting privileges. but then when i log out as root and try to log back in as the user it's failing. why is it failing? Thanks, G.

Code: [Select]
<?php
extract( $_POST );
include( '../php/userobject.php' );

@ $db = mysqli_connect('localhost', 'root', 'xxxxxx', 'mealchamp');
if ( mysqli_connect_errno() ) {
echo 'Error: could not connect to database as root.<br />';
exit;
}
else
echo 'Connected to database as root.<br />';

$query = "SELECT username FROM userlist WHERE username = $uname";
$result = mysqli_query($db, $query);

if ($result) 
echo 'That user name is already taken.<br/>Please hit back on your browser, and try a different user name.<br/>';
else {
echo 'That username is available.<br/>';
echo "$fname<br/>";
echo "$lname<br/>";
echo "$add1<br/>";
echo "$add2<br/>";
echo "$city<br/>";
echo "$prov<br/>";
echo "$ctry<br/>";
echo "$post<br/>";
echo "$email<br/>";
echo "$uname<br/>";
echo "$pass1<br/>";
echo "$sex<br/>";
echo "$birth<br/>";
echo "$height<br/>";
echo "$weight<br/>";

$query = "INSERT INTO userlist 
(firstname, lastname, address1, address2, city,  province, country, postal, email,  username, password, sex,  birth,  height,  weight) VALUES
('$fname', '$lname', '$add1', '$add2', '$city','$prov', '$ctry', '$post','$email','$uname', '$pass1', '$sex', '$birth', '$height',  '$weight')";
$result = mysqli_query($db, $query);
if ($result)
echo "Userlist was updated successfully<br />";
else
echo "Could not update userlist";

$query = "CREATE TABLE ".$uname."_persfood  
(persfoodid  INT UNSIGNED NOT NULL AUTO_INCREMENT PRIMARY KEY,
 food   VARCHAR(30) NOT NULL)";
$result = mysqli_query($db, $query);
if ($result)
echo "Personal food table was created successfully<br />";
else
echo "Could not create personal food table<br />";

$query = "GRANT select ON mealchamp.userlist TO ".$uname." IDENTIFIED BY '$pass1'";
$result = mysqli_query($db, $query);
if ($result)
echo "User priveleges were granted successfully<br />";
else
echo "Could not grant user priveleges<br />";
$query = "GRANT select, insert, update, delete, index ON mealchamp.".$uname."_persfood TO ".$uname." IDENTIFIED BY '$pass1'";
$result = mysqli_query($db, $query);
if ($result)
echo "User priveleges were granted successfully<br />";
else
echo "Could not grant user priveleges<br />";

mysqli_close($db);

@ $db = mysqli_connect('localhost', $uname, $pass1, 'mealchamp');
if ( mysqli_connect_errno() ) {
echo 'Error: could not connect to database as user.<br />';
exit;
}
else {
echo 'Connected to database as user.<br />';
?>

So this is a simple code that finds out the difference between two dates and displays it in number of days.

$date1=date_create("2013-03-15");
$date2=date_create("2013-12-12");
$diff=date_diff($date1,$date2);
echo $diff->format("%R%a days");

// RESULT
+272 days

 

My first question. Is it possible to remove the + sign in the result above?

Second question. Is it possible to show "months" if it's greater than 30 days? And years if the days are greater than 365?

How would I do this?