freemancomputer

Lines 1-3 are as follows 1 <?php ini_set ("display_errors", "1"); 2 error_reporting(E_ALL); 3 ?> removed line 13 so it is now as follows 10 header("location:login.php"); 11 } 13 ?> 14 <?php 15 $rank=$_SESSION['rank']; The errors are the same the last one being at line 14 now Thank you

I am moving to a different hosting site that is using php 5.2 from one that has 5.3. I am running into an error with my session_start() that worked just fine before. Was wondering if I could get some pointers. These are the errors that I get This is what I have in my header that is included in every page. 4 <? 5 session_start(); 6 if(!$_SESSION['login']){ 7 $_SESSION['rank']; 8 $_SESSION['loggedinusername'] = $loggedinusername; 9 $_SESSION['loggedinuseremail'] = $loggedinuseremail; 10 header("location:login.php"); 11 } 12 ?> 13 14 <?php 15 $rank=$_SESSION['rank']; 16 $loggedinusername=$_SESSION['loggedinusername']; 17 $loggedinuseremail=$_SESSION['loggedinuseremail']; 18 ?>

wow cant believe i missed that one thanks.

Sorry that is in there, just cut it out when I cut out the log in info. $id = mysql_real_escape_string(urldecode($_GET["id"])); $posteduserkey = mysql_real_escape_string(urldecode($_GET["userkey"]));

I am working on user validation through email and I am having some problems with it. I can get it to work but it gives me probloms some times. This is the link that is sent to the users after they sign up mydomain/confirm.php?id=".$id."&userkey=".$userkey and this is what i have for the confirm page <?php //connection info $userquery = mysql_query("SELECT * FROM user"); $active=mysql_result($userquery, "active"); $userid=mysql_result($userquery, "id"); $userkey=mysql_result($userquery, "userkey"); if ($active == 1) { printf("This account has already been activated"); echo "<br />"; } elseif($userkey != $posteduserkey){ printf("We are sorry but the data offered does not match data listed, plese contact system admin."); } else{ $query="UPDATE user SET active='1' WHERE id='$id'"; mysql_query($query) or die (mysql_error()); echo "You have activated your account"; ?> <div align="center"><br /><a href="login.php" class="nav">Back to the log in page</a></div> <?php } ?> This is the fun part. I checked that the id and userkey are correct in the link, and they are. I then replaced the & in the link with & it works in gmail but not yahoo email address. I'm not sure where else to look. Thanks in advance

I am looking to add a comment area to pages that are created with data that is pulled off of a database. I am looking for something like the forums on this site, but not as complex. I was thinking about attempting to use a text file to hold the comments for each ID number that is in my database and then displaying those comments when the ID number is called. I'm not sure how I would go about making a function that would search if there was a text file for that ID number and if not to create one. I'm not even sure if this is the best way to go about it but I know if I put all the comments into a database it will get rather large rather quickly. Any pointers and good tutorials would be welcome. Sorry if it sounds as if this post is rambling, I think I need to step away from the computer for a bit. Thanks

Thank you that worked just right.

I am attempting to have some links show up on my page only if a user is an admin. The way I have the user data base is user name password and rank. rank is a 1 or 2, 2 for admin. This is what I have right now as a test bed, I think the rank variable is not being passed properly but I'm not sure. Thanks in advance. This is what i have for the log in page <?php //connection stuff $sql="SELECT * FROM user WHERE username='$myusername' and password='$mypassword'"; $result=mysql_query($sql); $count=mysql_num_rows($result); $rank=mysql_num_rows('rank'); if($count==1){ session_start(); $_SESSION['login'] = "1"; $_SESSION['rank'] = $rank; header ("Location:index.php"); } else { $errorMessage = "Invalid Login"; session_start(); $_SESSION['login'] = ''; } ?> This is the header that will be on every page to keep the session. <?php ini_set ("display_errors", "1"); error_reporting(E_ALL); ?> <?php session_start(); if(!$_SESSION['login']){ $_SESSION['rank']; header("location:login.php"); } ?> <?php $rank=$_SESSION['rank']; ?> And this is what I have to show the link. <?php if($rank<=2){ ?> <tr> <td><span class="nav"><a href="link">link</a></span><br /></td> </tr> <?php } ?>

Thanks I'll take a look.

I just added a user log on for a site I am building using PHP. I used a simple session based system for this. I can log on just fine. What I am looking for is a way to pass on in ht session what rank a user is so some links/pages will only show for the admin or the super user but not the normal user. I rank the users as Admin 1, superuser 2, and user 3. This is what I have for the session in my header that allows you to be logged in. <? session_start(); if(!session_is_registered(myusername)){ header("location:login.php"); } ?> Here is what I have that starts the session //Connection stuff $query="SELECT * FROM user WHERE username='$myusername' and password='$mypassword'"; $result=mysql_query($query); $count=mysql_num_rows($result); if($count==1){ session_register("myusername"); session_register("mypassword"); header("location:index.php"); } else { echo "Wrong Username or Password"; } ob_end_flush(); ?>

I have been working on a way to split my data into 2 columns as a way to make the page shorter and easier to read. I have been able get it to split like this 1 2 3 4 5 6 But I would rather it go like this, because it would be easier on the eyes. 1 4 2 5 3 6 This is what I have now, I have no idea if I'm even on the right track. <?php include"scripts/connect.php" ; mysql_connect('localhost',$username,$password); @mysql_select_db($database) or die( "Unable to select database"); $query="SELECT * FROM movies WHERE type LIKE 'tv' ORDER BY title"; $result=mysql_query($query); $num=mysql_numrows($result); mysql_close(); ?> <table width="70%" align="left"> <?php $i=0; while ($i < $num/2) { $title=mysql_result($result,$i,"title"); ?> <tr> <td><a class="nav" href=/show.php?title=<?php echo urlencode($title); ?>><?php echo $title; ?></td> <?php while ($i >= $num/2) { ?> <td><a class="nav" href=/show.php?title=<?php echo urlencode($title); ?>><?php echo $title; ?></td> </tr> <?php } $i++; } ?> </table> This will show only the first half of the titles. Thanks in advance.

sorry, I already added that in after it didn't work on the first go around.

I added the ini_set ("display_errors", "1"); error_reporting(E_ALL); to what jcbones gave me and received no errors

jcbones thanks for the reply. When I added that and go to the page I get the page, header, footer and the such just none of the data that should be there. Going to look at it to see if I can figure it out. Thanks again

Right now you can only reach the site from inside of my network. What its showing now is 1 2 3 4 5 6 I would like it to show 1 4 2 5 3 6

I am trying to get the output for a mysql query to be displayed in colums. I have it where it does this but it goes from left to right. I would rather it go from top to bottem, it would look better this way. I will also need it to brake into sevral pages when there are a certain amout of entrys shown. If you know of a good how to on that let me know please. The other problom that I am having is to make the output a link. I was able to do this before I attemped to devide everything but if I add in what I had before it comes back with no data. Here's what I have now that put shows the data from left to right in 2 colums. <?php include"scripts/connect.php" ; mysql_connect(localhost,$username,$password); @mysql_select_db($database) or die( "Unable to select database"); $query="SELECT * FROM movies WHERE type LIKE 'movie' ORDER BY title"; if(mysql_query($query)) { $result=mysql_query($query); $num=mysql_numrows($result); } for($i=0; $i<$num; $i++) { if($i%2==0){ echo "<tr><td>"; } else{ echo "<td>"; } print mysql_result($result,$i,"title"); if ($i%2==0){ echo "</td>"; } else{ echo "</td>"; } } echo "</tr>"; ?> Thanks in advance