script in cronjob doesn't work with: "Could not open input file"

Question

This is my code in scriptrun (name of my shell script):

php -f a1.php; php -f b2.php; sh -e c3.txt

This is my cronjob command: /home/telia/www/robot/scriptrun, created as root

When I run script I get error message

Could not open input file: a1.php      
Could not open input file: b2.php 

scriptrun file is already have +x and I already tried

/usr/bin/php -f a1.php; /usr/bin/php b2.php ;sh -e c3.txt

I tried giving 777 chmod to php files however doesn't change anything.

script runs perfectly if I try manually it is just not work with cronjob.

Answer 1

As was answered in a comment, the problem appears to be that your a1.php and b2.php scripts are not in your $HOME directory, which is where cron jobs will execute. Either add a cd /to/that/path command to your scriptrun script, or change the php commands to use the full path to those scripts.