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Editorial Team
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Asked: 2026-06-03T14:12:56+00:00

I am generating a shell script from within a script and it needs root

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I am generating a shell script from within a script and it needs root permissions so I use 

sudo bash -c "echo 'Hello There!' > '/var/www/cgi-bin/php-cgi-5.3.8'"

I get an error if I try to output the shebang line and I not sure how to escape it — like i do with other variables.

sudo bash -c "echo '#!/bin/bash

version=\"5.3.8\"
export PHPRC=/etc/php/phpfarm/inst/php-\${version}/lib/php.ini
export PHP_FCGI_CHILDREN=3
export PHP_FCGI_MAX_REQUESTS=5000
exec /etc/php/phpfarm/inst/php-\${version}/bin/php-cgi' > '/var/www/cgi-bin/php-cgi-5.3.8'"

Where am I going wrong here?

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1 Answer

  1. Editorial Team

    ! is looking for a command beginning with ' in your bash_history. You should have the ‘!’ unquoted and escape it with a backslash. You may as well take the # out of quotes and escape it too, since a \ is shorter than two quotes.

    sudo bash -c "echo \#\!'/bin/bash

version=\"5.3.8\"
export PHPRC=/etc/php/phpfarm/inst/php-\${version}/lib/php.ini
export PHP_FCGI_CHILDREN=3
export PHP_FCGI_MAX_REQUESTS=5000
exec /etc/php/phpfarm/inst/php-\${version}/bin/php-cgi' > '/var/www/cgi-bin/php-cgi-5.3.8'"
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