I'm trying to make a simple bash script to check for "php" process. If the spesific PHP script process (eg: cleanlog.php) does not run, it will then be executed from bash.
When I tried to execute the bash script, it throws error:
What is wrong with the above code, please help.
Thank you.
Last edited by Scrutinizer; 01-04-2013 at 03:11 AM..
Reason: Additional code tags for error message
Hi,
For my bash script, terminal with bash is generate an OK output and program works right.
already,
terminal with ash have "line 48: syntax error: Bad substitution" output and program don't work.
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LEARN ABOUT DEBIAN
php-config5
php-config(1) Scripting Language php-config(1)NAME
php-config - get information about PHP configuration and compile options
SYNOPSIS
php-config [options]
DESCRIPTION
php-config is a simple shell script for obtaining information about installed PHP configuration.
OPTIONS --prefix Directory prefix where PHP is installed, e.g. /usr/local
--includes List of -I options with all include files
--ldflags LD Flags which PHP was compiled with
--libs Extra libraries which PHP was compiled with
--man-dir The directory prefix where the manpages is installed
--extension-dir
Directory where extensions are searched by default
--include-dir Directory prefix where header files are installed by default
--php-binary Full path to php CLI or CGI binary
--php-sapis Show all SAPI modules installed on the Debian system
--configure-options
Configure options to recreate configuration of current PHP installation
--version PHP version
--vernum PHP version as integer
SEE ALSO
php(1)VERSION INFORMATION
This manpage describes php, version 5.4.4-14+deb7u14.
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This source file is subject to version 3.01 of the PHP license, that is bundled with this package in the file LICENSE, and is available
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http://www.php.net/license/3_01.txt
If you did not receive a copy of the PHP license and are unable to obtain it through the world-wide-web, please send a note to
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