Shell Cron错误 - 什么可能导致此错误？

Question

Description: I am not very familiar with using a lot of bash/shell. I currently have a cron tab set up on an Ubuntu server that runs a Shell script. The Shell script then is suppose to run a PHP script, however, instead I am getting the following error message:

Could not open input file: wscript.php

At the top of my shell script I have written #!/bin/bash

Then the shell script itself I am passing a bash variable to PHP script.

 while read bashvariable
 do
   php wscript.php "$bashvariable"
 done

Keep in mind when I run this this shell script manually the script executes and fires correctly.

At the top of wscript.php I have placed in #!/usr/local/bin/php.

wscript.php has an include file of wscript-add.php

I have attempted to change the permission of all files to 777 and I haven't had any luck on getting the cron tab to run correctly.

Below is what my cron tab looks like:

*/2 * * * * sh /var/www/website/wcron/wcron.sh

My Question: What could cause my PHP file to not fire correctly when used by cron? Do I need specific file permissions on each file to run correctly?

Answer 1

您需要指定php脚本的完整路径，因为当cron运行时，它使用不同的当前目录。