PHP - Return Home Link Not Showing

I've been learning programming with the hope of being able to get work at home with decent pay and semi-reliable job offers. However, I'm disabled and it would be difficult to work on-site.     Is it a waste of time, am I going to have great difficulty finding work?
Edited by icor1031, 18 September 2014 - 10:35 PM.

This may sound funny, but I'm not sure what I was trying to get this code to do...
Code: [Select]
        <!-- Find Current Page Name -->
        <?php
          $page = basename($_SERVER['REQUEST_URI']);
echo '<p>$page = ' . $page . '</p>';
//          if ($page == '') {
          if (($page == '/')||($page == '')||($page == 'index.php')){
            $page = "index.php";
echo '<p>You are on the Home Page.</p>';
          }else{
echo '<p>You are on Some Other Page.</p>';
}

What I mean by that is I'm not understanding why I chose...
Code: [Select]
          if ($page == '') {

and then...
Code: [Select]
          if (($page == '/')||($page == '')){

and then...
Code: [Select]
          if (($page == '/')||($page == '')||($page == 'index.php')){


Why would I get those different results from basename($_SERVER['REQUEST_URI'])??

And what should I expect locally on my Laptop with a Virtual Host set up (e.g. "http://local.debbie/index.php") versus using a Test Site (e.g. "http://www.doubledee.byethost2.com/") versus a real site (e.g. "www.debbie.com")?!

Hope I'm making some sense here...


Debbie

Hi. I have problem and I would like to know if anyone here has any ideas on how to fix it.

My home page is where my log in form resides. If I leave the home page to browse other pages of my site I want to stay logged in. However, when I return to the home page I see the log in form. The log in form should not be there if I already logged in. Instead I should see the welcome message. Which does appear after loggin in but does not stay on the home page when I leave the page and come back.
Code: [Select]
<?php
//Starts the PHP Sesssion
session_start();
//Includes the MYSLQ Connection Info from another PHP Page
include ('spartacus_mysql.php');
//Submits Log In Info to Database and encrypts Password
if (isset ($_POST['loginsubmit'])) {
$username = mysql_escape_string($_POST['username']);
$userpassword = mysql_escape_string(md5($_POST['userpassword']));

if (!empty ($username) && !empty ($userpassword)) {
//Runs Query to Selects the Username and Password from the Database
$sql = mysql_query ("SELECT * FROM users
WHERE username='".$username."' AND userpassword='".$userpassword."' LIMIT 1");
//Finds and Matches the Log In Info from the Database to the one submitted
if(mysql_num_rows ($sql) > 0) {
$_SESSION['loggedin'] = true;
$_SESSION['username'] = $username;
//Runs This Message if the right Username/Password is entered
echo
'<div id="welcome_member">
Welcome,&nbsp; '.$_SESSION['username'].'.<br />
You are now logged in!<br/>
The Combat Tips and Downloads is now accessible!<br />
Enjoy!<br />
</div>';
} else {
//Runs This Error if the Username/Password do not match the Database 
echo '<div id="welcome_member">
Your username and/or password is incorrect! <a href="spartacus_home.php">Try Again</a>
</div>';
}
} else {
//Runs This Code if the Log In Form is Left Blank
echo '<div id="welcome_member">
You must enter a username and a password! <a href="spartacus_home.php">Try Again</a> 
or <a href="spartacus_joinsite.php">Register</a>
</div>';
}
} else {
//This is the code for the Log in Form
echo'<div id="join">
<a href="spartacus_joinsite.php">{ JOIN! }</a><br />
</div>
<div class="signin">
<div id="titles">
{ SIGN IN }
</div><br />
<form method="post" action="spartacus_home.php">
<table>
<tr><td>
Username: <input type="text" name="username" />
</td></tr>
<tr><td>
&nbsp;Password: <input type="password" name="userpassword"/>
</td></tr>
<tr><td>
<input class="submit" name="loginsubmit" type="submit" />
</tr></td>
</table>
</form>
</div>';
}
?>
<!--end sign in PHP-->

Here is my code

I have an issue I am not sure how to solve.  What I want to do is to make sure authentication has already occurred .  Or that my $currentMember is set and it if is set for an IF statement to do nothing..  However it it is not set for a redirect to the home page with no echo statements or anything to that nature just a redirect. Am I even close?

Code: [Select]
<?php


include( ABSOLUTE_PATH . 'class/database.class.php' );
include( ABSOLUTE_PATH . 'class/person.class.php' );

//Start Session
session_start();

$currentMember = unserialize($_SESSION['currentMember']);

$db = new Database;
$person = new Person($currentMember->memberid);


if ( $auth_satus == 0 ) {
header('Location:' . APP_ROOT . 'index.php');
}

exit();

?>

             

I am trying to force home landing page text onto this wordpress themes (weblizar Enigma)  standard page (which doesnt seem to allow for this even with settings/reading/static/home page option selected wordpress settings) The landing page just seems to be wigitised.    In a nutshell, I want to have a "page" of text on home page below banner slider and instead of "BLOG" and blog text. Hope this makes sense. Not getting help from theme author and I have been pointed towards .phpfreaks This is the website below.... http://www.cloudchasing.co.uk Thankyou in advance guys

Hi, I have a lovely wordpress theme on my website. Unfortunately I can not get text or the home page itself into the landing page, is there any way to do this at all?  I want to the home page text to sit under the    I have contacted the theme author with no help. I want the text to sit underneath the four image buttons (below banner slider)   This is the website  http://www.cloudchasing.co.uk/   Hope this makes sense, many thanks for help in advance.   Karl

My school is running PHP 4.3.2 on its network, and I'm running PHP 5.3.1 on my localhost machine using Xampp/Apache.

My login system works perfectly at school, however $_SESSION["LoggedIn"] does not register, recognise nor acknowledge on my home system.

It's not the script itself as I've copied the files across, so in theory it should work exactly the same?

I've compared phpinfo()'s and within the session category all values are the same.

The database/MySQL works fine on both networks so it's not a problem to do with this.

Any ideas? Thanks.

This topic has been moved to Miscellaneous.

http://www.phpfreaks.com/forums/index.php?topic=314251.0

Hi.
I had some trouble getting the mail function to work, I guess because I'm developing on LAMP (localhost), but I got it working at the office yesterday by configuring things as outlined here

http://mattsk.blogspot.com/2010/09/configure-lamp-on-ubuntu-1004-to-send.html

This got it working fine.  But today I'm trying it from home and it's not working (mail is not getting sent).  I don't know how the network is set up at work, but at home I'm on wireless lan through a d-link router.  I have a feeling it has something to do with the smtp_port = 465 (see link above), and that port maybe being blocked by the firewall or something.  I'm not really sure how all this stuff works though.  I tried contacting d-link, but they want some money before they'll help me out.  I followed some instructions I found online to forward port 465, which seemed to work (at lease the router told me so).  But the mail is still not sending.  A also tried the "fix-almost-any-computer-problem-ever" solution (i.e. I restarted my computer).  But still nothing.

I hope I've come to the right place for help.  Any help would be great.
Thanks.

Hi all,

fIXED

Warning: mysqli_fetch_array() expects parameter 1 to be mysqli_result, boolean given in /home/myritebook/public_html/footer.php on line 13

I am trying to fix the above error.

Please Help.

code is given below:

<?php
include 'db.php';

if ($conn) {
   echo "Database Found ";
}
else {
 die("Database connection failed");
}
$find_count=mysqli_query($conn,"SELECT * FROM count_number");

while($row=mysqli_fetch_array($find_count));
{
$current_count=$row['count'];
$new_count=$current_count+1;

try{
$sql= "UPDATE count_number SET count = '$new_count' where id=1";  
$update_count=mysqli_query ($conn,$sql);

}

catch(Exception $e){
echo 'error;'.$e->getMessage();
}
}
?>

I have been getting that error and I cannot figure out why it is happening

Here is the error: Parse error: syntax error, unexpected T_ENDWHILE, expecting ',' or ';' in /home/scswc188/public_html/index.php on line 23

Here is my Code (Database Credentials removed for obvious reasons)
<?PHP

// Conect to the Mysql Server
$connect = mysql_connect("IP","USER","PASS");

//connect to the database
mysql_select_db("TABLE");

//query the database
$query = mysql_query("SELECT * FROM users_online WHERE online = 1");

// fetch the results / convert into an array

        WHILE($rows = mysql_fetch_array($query)):

 

 

        $users = $rows['name'];

echo "'<font color='black'>Online:<font color='green'>$users, </font></font>;"

endwhile;

?>

or here http://pastebin.com/ZYh4t2pD

Thanks

Edit: Found the php tag

hello, if any one can help me solve this error.   

						case "unsuspend":
							$setupd = array("status" => "Active");
							$db->update("members", $setupd, "id=" . $mid);
							break;

							case "activate":
                                $setupd = array("status" => "Active");
							    $db->update("members", $setupd, "id=" . $mid);
							{
								if ($settings['referral_contest'] == 'yes') {
									$usr = $db->fetchRow( "SELECT ref1, status FROM members WHERE id=" . $mid );

									if (( $usr['status'] == 'Un-verified' && $usr['ref1'] != 0 )) {
										$referrer = $usr['ref1'];

										if ($referrer != 0) {
											$today_var = $settings['referral_reset'];
											$chk = $db->fetchOne( "SELECT COUNT(*) AS NUM FROM  referral_stats WHERE user_id=" . $referrer );

											if ($chk == 0) {
												$data = array( 'user_id' => $referrer, 'day' . $today_var => 1 );
												$db->insert( 'referral_stats', $data );
											} 
else {
												$db->query( "UPDATE referral_stats SET day" . $today_var . '=day' . $today_var . '+1 WHERE user_id=' . $referrer );
											}
										}
									}
								}
								
							
							break;

						case "delete":
							$userdata = $db->fetchRow("SELECT username, country, ref1 FROM members WHERE id=" . $mid);
							deletemember($mid);

							if ($userdata['username'] != "BOT") {
								$db->query("UPDATE country SET users=users-1 WHERE name=" . $userdata['country'] . "'");
							}


							if ($userdata['ref1'] != 0) {
								$db->query("UPDATE members SET referrals=referrals-1, myrefs1=myrefs1-1 WHERE id=" . $userdata['ref1']);
							}
							break;