PHP Cron Job: Including file not working? - php

i run a cron job every night, but for some reason, it is saying that the file i try to include is inexistant:
Warning: require(../includes/common.php): failed to open stream: No such file or directory in /home/fini7463/public_html/cron/journeyNotifications.php on line 2
Fatal error: require(): Failed opening required '../includes/common.php' (include_path='.:/usr/lib/php:/usr/local/lib/php') in /home/fini7463/public_html/cron/journeyNotifications.php on line 2
here's the code:
set_include_path('/home/fini7463/public_html/includes/');
require 'common.php';
the file 'common.php' is located as follows
public_html => cron => journeyNotifications.php
=> includes => common.php
i even set the include path (as shown in the code), but i am still getting this error. what could the problem be?
thanks!

If you do require('../includes/common.php'), the path is traversed relative to the current working directory.
If you do require('common.php'), the file is searched in the include path, and in the directory of the script which calls the require().
To solve this, first change directory in your crontab:
cd /home/fini7463/public_html; php -f cronjob.php

Calling set_include_path() as you do trashes the previous path. The call replaces the previous path with whatever you're passing as an argument, so if any of your code loads other libraries (ie: PEAR/PECL modules), they'll no longer be accessible as you've trashed the include path. You should use:
set_include_path(get_include_path() . PATH_SEPARATOR . '/home/fini7463/public_html/includes/');
That will append your new path to the include path.
As well, you can never quite tell what the working directory will be when cron fires up your script. It could be the home directory of the user you're running the script as, could be /tmp, or some other directory entirely. If you want to use relative paths in the script for anything, you'll have to make sure the working directory is set to a known value. Either by using a 'cd' in the crontab, or using a 'chdir' inside the script before doing anything involving the relative paths.

Related

Warning: require_once(/HTML/Template/ITX.php): failed to open stream: No such file or directory in C:\wamp64\www\mysite\main\login.php on line 13 [duplicate]

I was writing an web app in PHP, when I encountered a strange situation. To illustrate my problem, consider a web app of this structure:
/
index.php
f1/
f1.php
f2/
f2.php
Contents of these files:
index.php:
<?php require_once("f1/f1.php"); ?>
f1.php:
<?php require_once("../f2/f2.php"); ?>
f2.php: blank
now when I try to open index.php in my browser I get this error:
Warning: require_once(../f2/f2.php) [function.require-once]:
failed to open stream: No such file or directory in /var/www/reqtest/f1/f1.php on line 2
Fatal error: require_once() [function.require]:
Failed opening required '../f2/f2.php' (include_path='.:/usr/share/php:/usr/share/pear') in /var/www/reqtest/f1/f1.php on line 2
Is there something obvious I'm missing? how do include paths work in PHP?
Before I asked this question, I attempted to experiment and find out. I set up another test, like so:
/
index.php
f1/
f1.php
f2.php
index.php:
<?php require_once("f1/f1.php"); ?>
f1.php:
<?php require_once("f2.php"); ?>
f2.php: blank
To my surprise (and utter confusion), this worked out fine!
So, what is the secret behind the path resolution?
PS I saw this question, but it still does not answer the second case that I've stated here.
If you include another file, the working directory remains where the including file is.
Your examples are working as intended.
Edit: The second example works because . (actual directory) is in your include path (see your error message).
Edit2:
In your second example, the key point of your interest is this line:
<?php require_once("f2.php"); ?>
At first it will look in the current working dir (/var/www/req_path_test), but does not find f2.php.
As fallback, it will try to find f2.php in your include_path ('.:/usr/share/php:/usr/share/pear'), starting with '.' (which is relative to the actual file, not the including one).
So './f2.php' works and the require does not fail.
When you open index.php, working dir is set to the folder this file resides in. And inside insluded f1.php this working dir does not change.
You can include files by using their absolute paths, relative to the current included file like this:
require_once(dirname(__FILE__).'/../../test/file.php')
But better consider using an autoloader if these files contain classes.
Normaly in you old structure
<?php require_once("f2/f2.php"); ?>
instead of
<?php require_once("../f2/f2.php"); ?>
should work. As far as i know php takes the paths from the initial script
It sounds like your server has the open_basedir setting enabled in the PHP configuration. This makes it impossible to include (and open) files in folders above your in the directory structur (i.e., you can't use ../ to go up in the folder structure).
From the PHP Docs PHP include
Files are included based on the file path given or, if none is given, the include_path specified. If the file isn't found in the include_path, include will finally check in the calling script's own directory and the current working directory before failing.
If the file path is not given then i.e require_once("f2.php");
1st. The include_path is checked
2nd. The calling scripts own directory is checked
3rd. Finally the current working directory is checked
If file not found then PHP throws warning on file include & fatal error on require
If a path is defined — whether absolute (starting with a drive letter or \ on Windows, or / on Unix/Linux systems) or relative to the current directory (starting with . or ..) — the include_path will be ignored altogether. For example, if a filename begins with ../, the parser will look in the parent directory to find the requested file.
If you include/require your file beginning with . or .. or ./ then PHP's parser will look in the parent directory which is the current working directory i.e require_once("../f2/f2.php"), php will check at the root directory as the calling script index.php is in that directory.
Now You have not defined any include path in your PHP script thus it always falls back to the calling script and then into the current working directory.
// Check your default include path, most likely to be C:\xampp\php\PEAR
echo get_include_path();
// To set include path
set_include_path ( string $new_include_path ) : string
The Current Working Directory is derived from your main calling script index.php.
// The Current Working Directory can be checked
echo getcwd();
In the first Example where the required file "../f2/f2.php" is from f1.php
You code does not work because -
The specified path is ignored by PHP as your filename begins with ../
f1/ the calling script's own directory is ignored as well.
The parser directory looks into the parent directory to find the requested file. The current working directory is root directory, this is from where you have initiated the working script index.php. The file is not located at this directory, wrong path given.
Thus you get the Fatal Error
In the Second example you have changed the directory & from f1.php you require_once("f2.php").
Your code works because -
This time you require("f2.php") no leading ../ or ./ This time PHP checks the include_path but does find it there, as you haven't defined it and the file does not reside in the default preset include_path.
This time the calling script f1.php's directory is f1/. and you require file ("f2.php") is located at this directory. PHP This time checks the file in this directory and finds it.
PHP does not have to check the working directory as the file was found.
Thus Your Code Works Fine!

Executing a PHP script containing a path to another PHP script in another location

I am trying to run a PHP code in a PHP file and it has another php script that needs to be executed but PHP cannot find it. The other PHP script is in another path location than the php file that is called from.
I get an error from PHP about the script that is being called from /Users/peter/Documents/firm/worlddiki/na/ejs
The script from which the following is called is located at location
/Users/peter/Documents/firm/worlddiki/na/ejs
ERROR:
Fatal error: require_once(): Failed opening required '/scripts/checkcmtimg.php' (include_path='.:/usr/lib/php6') in /homepages/27/d468941446/htdocs/na/ejs/efranky.php on line 102
Any help would be appreciated in understanding how the php works in terms of paths.
The include path only applies if you give a relative pathname. You gave an absolute pathname (starts with "/"), so PHP was looking for the file in /scripts (that is, at the root directory of your computer).
You have two options:
Use a relative pathname, like "./scripts/checkcmtimg.php" or simply "scripts/checkcmtimg.php".
This will involve the include_path, searching each directory in the include path for that relative path.
Use an absolute pathname, like "/Users/peter/Documents/firm/worlddiki/na/ejs/scripts/checkcmtimg.php".
This will not use the include_path, it will search for the file in just the one location.

PHP: Require path does not work for cron job?

I have a cron job that needs to include this file:
require '../includes/common.php';
however, when it is run via the cron job (and not my local testing), the relative path does not work.
the cron job runs the following file (on the live server):
/home/username123/public_html/cron/mycronjob.php
and here's the error:
Fatal error: require(): Failed opening required '../includes/common.php'
(include_path='.:/usr/lib/php:/usr/local/lib/php') in
/home/username123/public_html/cron/mycronjob.php on line 2
using the same absolute format as the cron job, common.php would be located at
/home/username123/public_html/includes/common.php
does that mean i have to replace my line 2 with:
require '/home/username123/public_html/includes/common.php';
?
thanks!
With all do respect to all the current answers, they all went to "change the php code" approach.
I don't like to change my PHP files just to run it from a cron because it reduces the code portability and increases the chances to forget to change one or two relative paths and break the program.
Instead change the directory at the cron tab line, and leave all your relative paths and your PHP files untouched. For example
1 1 * * * cd /home/username/public_html/&& php -f script.php
check this answer
also check this article, I will quote the relative part
Depending on the code in your PHP script, it may only run correctly when called from a specific directory. For example, if the script uses relative paths to include files, it will only run if it is called from the correct directory. The following command shows how to call a PHP script from a specific directory:
cd /home/username/public_html/; php -q script.php
Technically seen the php script is run where cron is located; ex. If cron was in /bin/cron, then this statement would look for common.php in /bin/includes/common.php.
So yeah, you'll probably have to use fullpaths or use set_include_path
set_include_path('/home/username123/public_html/includes/');
require 'common.php';
nono. you need to use absolute paths on crons.
what I do is:
// supouse your cron is on app/cron and your lib is on app/lib
$base = dirname(dirname(__FILE__)); // now $base contains "app"
include_once $base . '/lib/db.inc';
// move on
An alternative to the solutions which recommend absolute path specification is using a chdir in your script. That way, your relative paths will work as expected.
For example, to change to the directory of the script:
$curr_dir = dirname(__FILE__);
chdir($curr_dir);
To change to the parent directory of the script:
$curr_dir = dirname(__FILE__);
chdir($curr_dir . "/..");
And so forth.
If the relative path doesn't work, then it means that the current directory set when the cron tasks are running is not /home/username123/public_html. In such cases, you can only use an absolute path.
It sounds as simple as just some script you are running is setting the include_path and you are including that script. use phpinfo() to check the include_path global vs local setting.

Why do I get a Warning and a Fatal error when I use ../?

When I use ../mysqlConnect.php I get the following messages.
Warning: require_once(../mysqlConnect.php) [function.require-once]:
failed to open stream: No such file or directory in /home/content/etc...
Fatal error: require_once() [function.require]: Failed opening required
'../mysqlConnect.php' (include_path='.:/usr/local/php5/lib/php') in /home/content/etc...
When I use the directory name - mydir/mysqlConnect.php - everything works fine.
require_once('../mysqlConnect.php') asks PHP to look in the directory above the one your script is currently in for mysqlConnect.php.
Since your connection file appears to be in a mydir directory, require_once('mydir/mysqlConnect.php') works because it looks in that directory, which is contained by the one it's currently in.
Visual representation (assuming script.php is your script including that file):
dir/
subdir/ # PHP looks here for ../mysqlConnect.php
script.php
mydir/ # PHP looks here for mydir/mysqlConnect.php
mysqlConnect.php
Require is relative to the invoced script, not the script you call require() in. Use something like this to have an absolute path:
require(dirname(__FILE__) . '/../mysqlConnect.php');
In PHP 5 you can also use DIR.
because it doesn't find your file then. to give a more specific answer I need to see you file-/folder-structure
That's because you are not specifying the correct include path. ../ refers to parent directory. ../../ goes two directories back, ../../../ goes three of them back. If the mysqlConnect.php file is present in the same folder as your script, you don't need to specify ../ in the include.
Make sure that you specify the correct path. You can easily check whether or not you are specifying correct path like:
if (file_exists('../mysqlConnect.php'))
{
echo 'Iam specifying the correct path !!';
}
else
{
echo 'Well, I am not :(';
}

How does include path resolution work in require_once?

I was writing an web app in PHP, when I encountered a strange situation. To illustrate my problem, consider a web app of this structure:
/
index.php
f1/
f1.php
f2/
f2.php
Contents of these files:
index.php:
<?php require_once("f1/f1.php"); ?>
f1.php:
<?php require_once("../f2/f2.php"); ?>
f2.php: blank
now when I try to open index.php in my browser I get this error:
Warning: require_once(../f2/f2.php) [function.require-once]:
failed to open stream: No such file or directory in /var/www/reqtest/f1/f1.php on line 2
Fatal error: require_once() [function.require]:
Failed opening required '../f2/f2.php' (include_path='.:/usr/share/php:/usr/share/pear') in /var/www/reqtest/f1/f1.php on line 2
Is there something obvious I'm missing? how do include paths work in PHP?
Before I asked this question, I attempted to experiment and find out. I set up another test, like so:
/
index.php
f1/
f1.php
f2.php
index.php:
<?php require_once("f1/f1.php"); ?>
f1.php:
<?php require_once("f2.php"); ?>
f2.php: blank
To my surprise (and utter confusion), this worked out fine!
So, what is the secret behind the path resolution?
PS I saw this question, but it still does not answer the second case that I've stated here.
If you include another file, the working directory remains where the including file is.
Your examples are working as intended.
Edit: The second example works because . (actual directory) is in your include path (see your error message).
Edit2:
In your second example, the key point of your interest is this line:
<?php require_once("f2.php"); ?>
At first it will look in the current working dir (/var/www/req_path_test), but does not find f2.php.
As fallback, it will try to find f2.php in your include_path ('.:/usr/share/php:/usr/share/pear'), starting with '.' (which is relative to the actual file, not the including one).
So './f2.php' works and the require does not fail.
When you open index.php, working dir is set to the folder this file resides in. And inside insluded f1.php this working dir does not change.
You can include files by using their absolute paths, relative to the current included file like this:
require_once(dirname(__FILE__).'/../../test/file.php')
But better consider using an autoloader if these files contain classes.
Normaly in you old structure
<?php require_once("f2/f2.php"); ?>
instead of
<?php require_once("../f2/f2.php"); ?>
should work. As far as i know php takes the paths from the initial script
It sounds like your server has the open_basedir setting enabled in the PHP configuration. This makes it impossible to include (and open) files in folders above your in the directory structur (i.e., you can't use ../ to go up in the folder structure).
From the PHP Docs PHP include
Files are included based on the file path given or, if none is given, the include_path specified. If the file isn't found in the include_path, include will finally check in the calling script's own directory and the current working directory before failing.
If the file path is not given then i.e require_once("f2.php");
1st. The include_path is checked
2nd. The calling scripts own directory is checked
3rd. Finally the current working directory is checked
If file not found then PHP throws warning on file include & fatal error on require
If a path is defined — whether absolute (starting with a drive letter or \ on Windows, or / on Unix/Linux systems) or relative to the current directory (starting with . or ..) — the include_path will be ignored altogether. For example, if a filename begins with ../, the parser will look in the parent directory to find the requested file.
If you include/require your file beginning with . or .. or ./ then PHP's parser will look in the parent directory which is the current working directory i.e require_once("../f2/f2.php"), php will check at the root directory as the calling script index.php is in that directory.
Now You have not defined any include path in your PHP script thus it always falls back to the calling script and then into the current working directory.
// Check your default include path, most likely to be C:\xampp\php\PEAR
echo get_include_path();
// To set include path
set_include_path ( string $new_include_path ) : string
The Current Working Directory is derived from your main calling script index.php.
// The Current Working Directory can be checked
echo getcwd();
In the first Example where the required file "../f2/f2.php" is from f1.php
You code does not work because -
The specified path is ignored by PHP as your filename begins with ../
f1/ the calling script's own directory is ignored as well.
The parser directory looks into the parent directory to find the requested file. The current working directory is root directory, this is from where you have initiated the working script index.php. The file is not located at this directory, wrong path given.
Thus you get the Fatal Error
In the Second example you have changed the directory & from f1.php you require_once("f2.php").
Your code works because -
This time you require("f2.php") no leading ../ or ./ This time PHP checks the include_path but does find it there, as you haven't defined it and the file does not reside in the default preset include_path.
This time the calling script f1.php's directory is f1/. and you require file ("f2.php") is located at this directory. PHP This time checks the file in this directory and finds it.
PHP does not have to check the working directory as the file was found.
Thus Your Code Works Fine!

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