Undefined variable

Question

Reaching out for a second pair of eyes.

I just don't get what the issue is here, I've been using the same format throughout out my whole site until now I get this error:

<?php
session_start();

include"lib/config.php";

$logged_user = $_SESSION['username'];

$check_admin = "SELECT * FROM `members` WHERE `is_admin` = ".sql_val($logged_user);
$result = $conn->query($check_admin);

if ($result->num_rows > 0) {

while($row = $result->fetch_assoc()) {
$members_id = $row['members_id'];
$is_admin = $row['is_admin'];
$is_mod = $row['is_mod'];
$username = $row['username'];
$password = $row['password'];
$email = $row['email'];
$login_date = $row['login_date'];
$login_time = $row['login_time'];
$session = $row['session'];
}}

?>


<?php

if ($is_admin == 1){

echo 'Is Admin';
}
else
{

echo 'Is not Admin';
}

?>

this is my output:

Notice: Undefined variable: is_admin in C:\wwwroot\htdocs\snippet\is_admin.php on line 30 Is not Admin

which don't make sence.

place your `if $is_admin` inside the while loop. and if `$logged_user` is a string, you need to quote that in your query and make sure there's a value for it. *blah blah blah*. — Funk Forty Niner, Feb 17 '16 at 18:49
You're defining `$is_admin` inside your while loop, which means it's only available inside there. See [Link](http://stackoverflow.com/questions/7337743/using-a-variable-outside-of-the-while-loop-scope) aswell. — ccKep, Feb 17 '16 at 18:49
*"which don't make sence."* - Yes it does, it makes *perfect* sense. — Funk Forty Niner, Feb 17 '16 at 18:55
@ccKep [That's not accurate, as proven further down on your link.](http://stackoverflow.com/questions/7337743/using-a-variable-outside-of-the-while-loop-scope#comment8849566_7337865) — Litty, Feb 17 '16 at 18:56
@Litty Umm.... did you even look at their track record before pinging me? — Funk Forty Niner, Feb 17 '16 at 19:02
Thank you, everyone, I learnt something new today. This is very new to me, I'm used to the old format with mysql, so still learning. — TrevTech, Feb 17 '16 at 19:06
@Fred-ii- I'm working on the question at hand in the present, not digging up the user's past. No argument though, he should be accepting answers. — Litty, Feb 17 '16 at 19:07

score 1 · Answer 1 · answered Feb 17 '16 at 18:52

You need to define $is_admin on top of file as a null. Because if your if condition will not satisfy then you will not get $is_admin at the end of the file.

Like this :

<?php
session_start();

include"lib/config.php";
$is_admin = "";
$logged_user = $_SESSION['username'];

$check_admin = "SELECT * FROM `members` WHERE `is_admin` = ".sql_val($logged_user);
$result = $conn->query($check_admin);

if ($result->num_rows > 0) {

while($row = $result->fetch_assoc()) {
$members_id = $row['members_id'];
$is_admin = $row['is_admin'];
$is_mod = $row['is_mod'];
$username = $row['username'];
$password = $row['password'];
$email = $row['email'];
$login_date = $row['login_date'];
$login_time = $row['login_time'];
$session = $row['session'];
}}

?>


<?php

if ($is_admin == 1){

echo 'Is Admin';
}
else
{

echo 'Is not Admin';
}

?>

Although this makes the notice go away, keep in mind that this might not "fix the real problem." Shouldn't `$is_admin` be a strict true or false? Allowing it to be a null value means a member wasn't found -- but in this case, a member should *probably* always be found. — Litty, Feb 17 '16 at 19:03

score 0 · Answer 2 · answered Feb 17 '16 at 18:55

When checking if $is_admin == 1, the variable $is_admin is not defined. In your code, this means...

Your SQL query returned no rows, so that while loop is never iterated. Is your table filled? Are you querying the correct table?
$row does not contain a value for $row['is_admin']. Is that column in your table? For this row, is the value NULL?

Undefined variable

2 Answers2