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error_reporting(E_ALL)......

Posted on 2007-03-31
6
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Last Modified: 2013-12-12
Hello.
I v added error_reporting(E_ALL);  at the top of my page and i m geting some errors like...

Notice: Undefined variable: photo_category_list in C:\Program Files\xampp\htdocs\homebody\homebody.php on line 867

Notice: Undefined index: s_cat in C:\Program Files\xampp\htdocs\homebody\homebody.php on line 867

when..
$result = mysql_query( "SELECT category_id,category_name FROM gallery_category ORDER BY category_name" );
     while( $row = mysql_fetch_array( $result ) )
     {
                    $photo_category_list .= '<option value="'.$row[1].'" '.($row[1]== $_SESSION["s_cat"]? "selected":"").'>'.$row[1].'</option>';    <<<<<<<<<<<<<<867 line
     }

Can you PLS explain me what i m doing wrong??
0
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Question by:NTGrE
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6 Comments
 
LVL 143

Expert Comment

by:Guy Hengel [angelIII / a3]
ID: 18829022
that error means that the array item "s_cat" in the session variable array does not exist.
to get rid of the warning, you could do this:

$result = mysql_query( "SELECT category_id,category_name FROM gallery_category ORDER BY category_name" );
     while( $row = mysql_fetch_array( $result ) )
     {
                    $photo_category_list .= '<option value="'.$row[1].'" '.($row[1]==@ $_SESSION["s_cat"]? "selected":"").'>'.$row[1].'</option>';    <<<<<<<<<<<<<<867 line
     }

however, that depends on the fact if you might want to react differently on the existance or non-existance of that item in the array
0
 
LVL 12

Expert Comment

by:str_kani
ID: 18829108
also you can set something like this to avoid notices...

error_reporting(E_ALL ^ E_NOTICE);

http://au3.php.net/error_reporting
0
 

Author Comment

by:NTGrE
ID: 18829455
angelIII
As far as understand the @ hide the error..Right??
I just wonder if these errors(notices) indicates that something is wrong for security.
example...
Notice: Undefined variable: photo_category_list in ..........etc......
What means UNDEFINED ??? There is a way to define it ???
iv try
$photo_category_list="";
and the error is gone.

str_kani
thnx ...but i dont want to hide the error msg...i want to understand why this error displayed
and if means something for my security..
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LVL 143

Accepted Solution

by:
Guy Hengel [angelIII / a3] earned 500 total points
ID: 18829491
>As far as understand the @ hide the error..Right??
yes

>Notice: Undefined variable: photo_category_list in ..........etc......
>What means UNDEFINED ??? There is a way to define it ???
>iv try
>$photo_category_list="";
>and the error is gone.
exactly. that error will appear when you try to use a variable (ie it's value) before you have declared it like you have found out.

>but i dont want to hide the error msg...i want to understand why this error displayed
good!

actually, your initial code should be written more like this, in regards to the warning:
$s_cat = "";
if (is_set($_SESSION["s_cat"])) { $s_cat = $_SESSION["s_cat"]; }

$result = mysql_query( "SELECT category_id,category_name FROM gallery_category ORDER BY category_name" );
     while( $row = mysql_fetch_array( $result ) )
     {
                    $photo_category_list .= '<option value="'.$row[1].'" '.($row[1]== $s_cat ? "selected":"").'>'.$row[1].'</option>';    <<<<<<<<<<<<<<867 line
     }

0
 

Author Comment

by:NTGrE
ID: 18831467
Thnx angelIII.....
One last Question..
Leaving an UNDEFINED variable can cause me security problems ???
0
 
LVL 143

Expert Comment

by:Guy Hengel [angelIII / a3]
ID: 18832737
>Leaving an UNDEFINED variable can cause me security problems ???
not that I know of, but I am not the (php/net) security guy/specialist
0

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