Notice: Undefined variable: logged in C:\xampp\htdocs\Site\Site\view_category.php

Question

Can anyone explain where my problem is here? im trying to create a link so that the user can go onto create a forum post

 include_once("DBconnect.php");
 $cid = $_GET['cid'];

 if (isset($_SESSION['uid'])){
     $logged .= "  <a href='create_topic.php?cid=".$cid."'>Click here to create a topic</a>";
 }else{
     $logged .= "  Please log in to create topics in this forum";
 }

You are giving the absolute path but should give relative path of file. — Suman Bogati, Apr 05 '14 at 15:22
What is unclear about the error messages being presented to you? — PeeHaa, Apr 05 '14 at 15:23
@SumanBogati How would a path have anthing to do with an undefined variable? As far as PHP is concerned, it's simply a string. — Fluffeh, Apr 05 '14 at 15:24

score 2 · Answer 1 · edited Jun 20 '20 at 09:12

2

Initialize the variable before you use it

 $logged = ''; //<--- Add here
 if (isset($_SESSION['uid'])){

Since you are using .= , it will check whether $logged has previously been assigned a value to be appended with your new value.. since there is no such value associated with $logged variable , you get that notice.

edited Jun 20 '20 at 09:12

Community

1
1

answered Apr 05 '14 at 15:22

Shankar Narayana Damodaran

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score 1 · Answer 2 · answered Apr 05 '14 at 15:23

1

You are appending to logged but it is not defined try puttting

$logged="";

Before the if statement

answered Apr 05 '14 at 15:23

smistry

1,127
1
9
9

score 1 · Answer 3 · answered Apr 05 '14 at 15:26

1

$logged .= "…" is shortcut for $logged = $logged . "…", which includes reading of $logged. This causes the mentioned notice if $logged is undefined.

Initialize $logged with a value before reading it like with an empty string:

$logged = "";

answered Apr 05 '14 at 15:26

Gumbo

643,351
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Azeem Hassni · Answer 4 · 2014-04-05T16:13:52.380

0

Change Your Code to this

<?php
include_once("DBconnect.php");
 $cid = $_GET['cid'];
$logged = '';
 if (isset($_SESSION['uid'])){
     $logged .= "  <a href='create_topic.php?cid=".$cid."'>Click here to create a topic</a>";
 }else{
     $logged .= "  Please log in to create topics in this forum";
 }
?>

OR Initialize the variable $logged instead of appending value in it . if $logged dose't contain any value before .

<?php
include_once("DBconnect.php");
 $cid = $_GET['cid'];
 if (isset($_SESSION['uid'])){
     $logged = "  <a href='create_topic.php?cid=".$cid."'>Click here to create a topic</a>";
 }else{
     $logged = "  Please log in to create topics in this forum";
 }
// this will print the variable on your page 
 echo $logged;
?>

edited Apr 05 '14 at 16:13

answered Apr 05 '14 at 15:33

Azeem Hassni

885
13
28

But why isnt thg link appearing on my page? – user3501382 Apr 05 '14 at 16:09
you have to print the `$logged` variable in order to render it . i have added a new line in the answer – Azeem Hassni Apr 05 '14 at 16:12
http://stackoverflow.com/questions/22882209/link-isnt-appearing-on-my-page my full code is there, i dont understand why its not appearing everything seems correct to me – user3501382 Apr 05 '14 at 16:14
here is code which will work hopefully http://stackoverflow.com/questions/22882209/link-isnt-appearing-on-my-page/22883488#22883488 – Azeem Hassni Apr 05 '14 at 16:23

Notice: Undefined variable: logged in C:\xampp\htdocs\Site\Site\view_category.php

4 Answers4

Initialize the variable before you use it