**php notice:undefined Index: ... How to solve the problem

This problem may be caused by undefined variables:

if (!empty ($current _user_id)) {
    $digg = $this->m_feed_digg->show_entity (Array (' digg_user_id ' = = $current _user_id, ' digg_feed_id ' = = $ result[$feed _key][' feed_id '));
    if (!empty ($digg)) {
        $result [$feed _key][' Is_digg '] = true;
        $result [$feed _key][' digg_id '] = $digg [' digg_id '];
    }else{
        $result [$feed _key][' Is_digg '] = false;
        $result [$feed _key][' digg_id '] =  $digg [' digg_id '];

;
    }

}

The problem above is that $digg is empty, so

$digg [' digg_id '] must be undefined, this should be changed to NULL

----------------------------------------------------------------------

First of all, this is not a mistake, it is warning. Therefore, if the server cannot be changed, each variable should be defined before use.

Method 1: Server configuration Modifications
Modify the php.ini configuration file, error_reporting = E_all & ~e_notice

Method 2: Initialize the variables, standardize the writing (more cumbersome, because there are a lot of variables). But have not found a good definition method, I hope you advise

Method 3: Each file header plus: error_reporting (0); If not, only open php.ini, find display_errors, set to Display_errors = Off. Any future errors will not be prompted.

Method 4: Make a Judgment: isset ($_get["page"]) If-else judgment

or add "@" to indicate this line if there is an error or warning do not output
such as: @ $page =$_get["page"]

Method 5:file1.php file to pay a value $xx variable, with post to file2.php,
If the file2.php does not have the definition of $xx, the direct use of $yy= $xx; The system will be error: "Undifined variaable $xx", if file2.php's files start with $xx= ""; definition, then file1.php $xx value will not pass!

File2.php can do that.
if (!isset ($xx)) $xx = "";

**php notice:undefined Index: ... How to solve the problem