what does it mean "(include_path='.:/usr/share/pear:/usr/share/php')"?

Question

I have file structure on EC2 like : but facing some file referencing problem.

index.php
-db
  -config.php
-cron
  -cron1.php

I have tried file referencing as:

`require_once (dirname(__FILE__).'/db/config.php');`
`require_once (($_SERVER['DOCUMENT_ROOT']).'/db/config.php');`

but cron doesn't run.it gives error in mail as

`PHP Warning:  require_once(/db/config.php): failed to open stream: No such file or directory in /var/www/html/cron/cron1.php on line 3

Warning: require_once(/db/config.php): failed to open stream: No such file or directory in /var/www/html/cron/cron1.php on line 3

PHP Fatal error:  require_once(): Failed opening required '/db/config.php' (include_path='.:/usr/share/pear:/usr/share/php') in /var/www/html/cron/cron1.php on line 3

Fatal error: require_once(): Failed opening required '/db/config.php' (include_path='.:/usr/share/pear:/usr/share/php') in /var/www/html/cron/cron1.php on line 3`

Answer 1

If you look at the PHP constant PATH_SEPARATOR, you will see it being ":" for you.

If you break apart your string ".:/usr/share/pear:/usr/share/php" using that character, you will get 3 parts.

• . (this means the current directory your code is in)
• /usr/share/pear
• /usr/share/php

Any attempts to include()/require() things, will look in these directories, in this order.

It is showing you that in the error message to let you know where it could NOT find the file you were trying to require()

For your first require, if that is being included from your index.php, then you dont need the dir stuff, just do...

require_once ( 'db/config.php');