Notice: Undefined variable: ids in

Question

i have a script that works fine for years and suddenly i face this problem : please check the code and mend any solutionplease. any kind of help willbe appreciated.

Notice: Undefined variable: ids in /home/content/a/t/a/ataasaid1/html/category.php on line 4

Notice: Undefined variable: id in /home/content/a/t/a/ataasaid1/html/category.php on line 5

Notice: Use of undefined constant image - assumed 'image' in /home/content/a/t/a/ataasaid1/html/category.php on line 11

Notice: Use of undefined constant video - assumed 'video' in /home/content/a/t/a/ataasaid1/html/category.php on line 15

Notice: Use of undefined constant sound - assumed 'sound' in /home/content/a/t/a/ataasaid1/html/category.php on line 19

the code in file category.php is :

<?
  error_reporting(E_ALL | E_STRICT);
  ini_set("display_errors", 1);    
$id1=$ids;
$id2=$id;
                      include('config.php');

                    $result= mysql_query("SELECT * FROM products where id like '$id1' ;");
                    $row=mysql_fetch_array ($result);
                    $type=$row['type'];
                    if($type==image)
                    {
                    include('categories.php');
}
else if($type==video)
{
include('categories2.php');
}
else if($type==sound)
{
include('categories3.php');
}
?>

i use the url like:

/category.php?id=64&ids=305

thank you in advanced

I don't see where $ids and $id are defined...And, you missed quotes around video and sound : 'video', 'sound'. — Brice, Nov 05 '13 at 13:26
1) Your line 4 and 5 are different from what you posted here. 2) You're trying to access undefined constants in your code. Change `if($type==image)` to `if($type == "video")` (replace all similar occurrences) — Amal Murali, Nov 05 '13 at 13:27

score 0 · Answer 1 · answered Nov 05 '13 at 13:26

0

write like this:

if($type=="video")

else if($type=="sound")

answered Nov 05 '13 at 13:26

Suresh Kamrushi

15,627
13
75
90

Why we shouldn't use as if($type==video) else if($type==sound) – Pavan Kumar Nov 05 '13 at 13:30
PHP take it as constant. – Suresh Kamrushi Nov 05 '13 at 13:31
Ok then is it effects the comparision operation... – Pavan Kumar Nov 05 '13 at 13:32

score 0 · Answer 2 · answered Nov 05 '13 at 13:28

0

Are you looking for

$id1=$_GET["ids"];
$id2=$_GET["id"];

or

$id1=$_POST["ids"];
$id2=$_POST["id"];

???

answered Nov 05 '13 at 13:28

Daniele Vrut

2,835
2
22
32

i use the url like: /category.php?id=64&ids=305 – user2956413 Nov 05 '13 at 13:41
@user2956413: so you have to use $_GET as I suggested. – Daniele Vrut Nov 05 '13 at 13:43

score 0 · Answer 3 · answered Nov 05 '13 at 13:29

This error basically means that you're trying to use the variables $ids and $id, even though they do not exist. You need to define a variable before you can use it. Otherwise, you will get notices like this and your application will end up being a buggy mess. Because those two variables are not set and you are assigning them to other variables:

$id1 = $ids;
$id2 = $id;

Those variables are now useless, which means that your SQL query is also useless:

$result= mysql_query("SELECT * FROM products where id like '$id1' ;");

because $id1 isn't anything.

i use the url like: /category.php?id=64&ids=305 – user2956413 Nov 05 '13 at 13:43 — user2956413, Nov 05 '13 at 13:43

Notice: Undefined variable: ids in

3 Answers3