Re: Turning off job control doesn't make async jobs ignore SIGINT

2007-09-25 Thread Chet Ramey
AnMaster wrote: > Configuration Information [Automatically generated, do not change]: > Machine: x86_64 > OS: linux-gnu > Compiler: x86_64-pc-linux-gnu-gcc > Compilation CFLAGS: -DPROGRAM='bash' -DCONF_HOSTTYPE='x86_64' > -DCONF_OSTYPE='linux-gnu' -DCONF_MACHTYPE='x86_64-pc-linux-gnu' > -DCONF_V

Re: scripting help

2007-09-25 Thread The Wanderer
Reply addresses set by hand to work around broken defaults. (Again.) suser wrote: I am trying to create a script to go through /etc/passwd and change anyone who's home directory is "/mnt/home" to "/mnt/". I have been playing around with loops and regular expressions but have not got anything cl

Re: scripting help

2007-09-25 Thread Andreas Schwab
suser <[EMAIL PROTECTED]> writes: > I am trying to create a script to go through /etc/passwd and change anyone > who's home directory is "/mnt/home" to "/mnt/". This does not really has anything to do with bash, let alone with bugs in bash, so it is rather off-topic here. > Original - user:x:100

Re: Turning off job control doesn't make async jobs ignore SIGINT

2007-09-25 Thread Andreas Schwab
AnMaster <[EMAIL PROTECTED]> writes: >>From man bash: >Non-builtin commands run by bash have signal handlers set to the > values inherited by the shell from its parent. >When job control is not in effect, asynchronous commands ignore SIGINT > and SIGQUIT in addition to these in

scripting help

2007-09-25 Thread suser
I am trying to create a script to go through /etc/passwd and change anyone who's home directory is "/mnt/home" to "/mnt/". I have been playing around with loops and regular expressions but have not got anything close to the results i want. I have roughly 1000 users so changing them all by hand is

Turning off job control doesn't make async jobs ignore SIGINT

2007-09-25 Thread AnMaster
Configuration Information [Automatically generated, do not change]: Machine: x86_64 OS: linux-gnu Compiler: x86_64-pc-linux-gnu-gcc Compilation CFLAGS: -DPROGRAM='bash' -DCONF_HOSTTYPE='x86_64' -DCONF_OSTYPE='linux-gnu' -DCONF_MACHTYPE='x86_64-pc-linux-gnu' -DCONF_VENDOR='pc' -DLOCALEDIR='/usr/s

Re: Linux: bash 'declare -f' adds trailing whitespaces to output

2007-09-25 Thread Jari Aalto
* Tue 2007-09-25 Chet Ramey INBOX > Jari Aalto wrote: > >> Tested in: >> >> GNU bash, version 2.05.0(1)-release (sparc-sun-solaris2.9) >> GNU bash, version 3.2.25(16)-release (i686-pc-cygwin) >> >> An example: >> >> function testfunc () >> { >> echo "this is test" >>

Re: Linux: bash 'declare -f' adds trailing whitespaces to output

2007-09-25 Thread Chet Ramey
Jari Aalto wrote: > Tested in: > > GNU bash, version 2.05.0(1)-release (sparc-sun-solaris2.9) > GNU bash, version 3.2.25(16)-release (i686-pc-cygwin) > > An example: > > function testfunc () > { > echo "this is test" > } > > From shell: > > $ declare -f | grep t

Re: Linux: bash 'declare -f' adds trailing whitespaces to output

2007-09-25 Thread Jan Schampera
Jari Aalto wrote: testfunc () $ { $ echo "this is test"$ }$ Just a thought: It looks like there is a system behind it. So I guess it's some parser thing that needs it like that. However, just guessing. Jan

Linux: bash 'declare -f' adds trailing whitespaces to output

2007-09-25 Thread Jari Aalto
Tested in: GNU bash, version 2.05.0(1)-release (sparc-sun-solaris2.9) GNU bash, version 3.2.25(16)-release (i686-pc-cygwin) An example: function testfunc () { echo "this is test" } >From shell: $ declare -f | grep testfunc -A 3 | cat -A Result; notice the trai