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Top Forums Shell Programming and Scripting [Bash] passing variables to executable doesn't work Post 303039704 by Neo on Saturday 12th of October 2019 06:36:42 PM
Old 10-12-2019
Quote:
Originally Posted by sushi2k7
You can't reproduce the error, maybe you can improve the code that it should work then.
What does this (above) mean exactly?
 

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BASHBUG(1)							      bashbug								BASHBUG(1)

NAME
bashbug - report a bug in bash SYNOPSIS
bashbug [--help] [--version] [bug-report-email-addresses] DESCRIPTION
bashbug is a utility for reporting bugs in Bash to the maintainers. bashbug will start up your preferred editor with a preformatted bug report template for you to fill in. Save the file and quit the editor once you have completed the missing fields. bashbug will notify you of any problems with the report and ask for confirmation before sending it. By default the bug report is mailed to both the GNU developers and the Debian Bash maintainers. The recipients can be changed by giving a comma separated list of bug-report-email-addresses. If you invoke bashbug by accident, just quit your editor. You will always be asked for confirmation before a bug report is sent. OPTIONS
--help Show a brief usage message and exit. --version Show the version of bashbug and exit. bug-report-email-addresses Comma separated list of recipients' email addresses. By default the report is mailed to both the GNU developers and the Debian Bash maintainers. ENVIRONMENT
DEFEDITOR Editor to use for editing the bug report. EDITOR Editor to use for editing the bug report (overridden by DEFEDITOR). SEE ALSO
bash(1), reportbug(1), update-alternatives(8) for preferred editor. AUTHOR
This manual page was written by Christer Andersson <klamm@comhem.se> for the Debian project (but may be used by others). GNU Bash 3.1 11 December 2007 BASHBUG(1)
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