Bash prints syntax error when command in $(...) contains case-esac

Configuration Information [Automatically generated, do not change]: Machine: i686 OS: cygwin Compiler: gcc Compilation CFLAGS: -DPROGRAM='bash.exe' -DCONF_HOSTTYPE='i686' -DCONF_OSTYPE='cygwin' -DCONF_MACHTYPE='i686-pc-cygwin' -DCONF_VENDOR='pc' -DLOCALEDIR='/usr/share/locale' -DPACKAGE='bash'

Re: Bash prints syntax error when command in $(...) contains case-esac

Juergen Gohlke schrieb: Configuration Information [Automatically generated, do not change]: Machine: i686 OS: cygwin Compiler: gcc Compilation CFLAGS: -DPROGRAM='bash.exe' -DCONF_HOSTTYPE='i686' -DCONF_OSTYPE='cygwin' -DCONF_MACHTYPE='i686-pc-cygwin' -DCONF_VENDOR='pc'

Re: Bash prints syntax error when command in $(...) contains case-esac

Juergen Gohlke wrote: Description: If a command in $(...) contains a case-esac construction, the bash prints a syntax error instead of executing the code: bash: syntax error near unexpected token `;;' In case you have trouble with code you need, use this workaround:

Re: Bash prints syntax error when command in $(...) contains case-esac

Machine Type: i686-pc-cygwin Bash Version: 3.2 Patch Level: 9 Release Status: release Description: If a command in $(...) contains a case-esac construction, the bash prints a syntax error instead of executing the code: bash: syntax error near unexpected token