Re: Bash script

2003-04-03 Thread Bret Hughes
On Thu, 2003-04-03 at 07:50, santosh kumar wrote: > Hi guys, > > Have small doubt in bash scripting..(#!/bin/bash) > Need to write script to get all m/c info, If I do cat /proc/cpuinfo will > display CPU info. > My requirement is grep CPU model, speed and display the outputs like > For example m/c

Re: Bash script

2003-04-03 Thread Emmanuel Seyman
On Thu, Apr 03, 2003 at 10:57:31AM -0300, Rodrigo Nascimento wrote: > > What you want? I believe that the poster wants our advice on how to obtain the output he posted. [EMAIL PROTECTED] seyman]$ cat /proc/cpuinfo | grep -E "MHz|name" model name : Pentium III (Coppermine) cpu MHz :

Re: Bash script

2003-04-03 Thread Rodrigo Nascimento
Hi, I didn't understand. What is wrong or confused in output for more one CPU? 1st CPU model : pentium III 1st CPU speed : 1.73 GHz 2nd CPU model : pentium IV 2nd CPU speed : 2.4 GHz The 1st CPU is a PIII with 1.7 Ghz and 2nd CPU is a P4 with 2.4 Ghz. What you want? --- santosh kumar <[E

Re: Bash script .bash_logout

2003-03-20 Thread Todd A. Jacobs
On 19 Mar 2003, will wrote: > The .bash_logout script is not executing when my system is in run level If your xterm is not defined as a login shell, it won't execute .bashrc or .bash_profile, either. You may need to define the xterm as a login shell to get .bash_logout working. -- Guvf gntyva

Re: Bash script .bash_logout

2003-03-19 Thread will
TERM=xterm COLORTERM=gnome-terminal BTW -- I've read the man on xdm that recommends using the Xreset script (e.g., xdm-config: DisplayManager._0.reset=/etc/X11/xdm/TakeConsole) to run after the users session is terminated. It should contain commands such as unmounting directories from file servers

Re: Bash script .bash_logout

2003-03-19 Thread Ritesh Raj Sarraf
what terminal do you use on your X Window.. Quoting will <[EMAIL PROTECTED]>: > The .bash_logout script is not executing when my system is in run level > init 5 (Xsession); however, in init 3 it runs fine. Any one know how > this can be fixed? > > Red Hat Linux release 8.0 (Psyche) -- X

RE: Bash script help ?

2003-02-05 Thread David Simmons
[ test ] then Dave -Original Message- From: R P Herrold [mailto:[EMAIL PROTECTED]] Sent: Wednesday, February 05, 2003 12:11 PM To: [EMAIL PROTECTED] Subject: RE: Bash script help ? On Wed, 5 Feb 2003, David Simmons wrote: > I have noticed in this thread that everyone is puttin

RE: Bash script help ?

2003-02-05 Thread R P Herrold
On Wed, 5 Feb 2003, David Simmons wrote: > I have noticed in this thread that everyone is putting a ";" after their > tests: if [ test ] ; and for [ test ] ; > > When is the ";" required or is it always required after the test in a > conditional statement? Ehhh? Not all people do. I never

Re: Bash script help ?

2003-02-05 Thread Raymundo M. Vega
quot; required or is it always required after the test in a conditional statement? Thanks, Dave Simmons -Original Message- From: Raymundo M. Vega [mailto:[EMAIL PROTECTED]] Sent: Wednesday, February 05, 2003 11:00 AM To: [EMAIL PROTECTED] Subject: Re: Bash script help ? I think sev

RE: Bash script help ?

2003-02-05 Thread David Simmons
undo M. Vega [mailto:[EMAIL PROTECTED]] Sent: Wednesday, February 05, 2003 11:00 AM To: [EMAIL PROTECTED] Subject: Re: Bash script help ? I think several lines are not quite right: - regular expression in gawk should be inside the {}, but you will have to pass the argument to gawk. - next line aft

Re: Bash script help ?

2003-02-05 Thread Raymundo M. Vega
I think several lines are not quite right: - regular expression in gawk should be inside the {}, but you will have to pass the argument to gawk. - next line after if should be then, it is usually used like if [ -z "$processid" ] ; then the script will only echo first time it finds the string,

Re: Bash script help ?

2003-02-05 Thread Ryan Babchishin
[EMAIL PROTECTED] wrote: I'm trying to get a pidof a php script by capturing the path: like /home/somebody/my_script which will be the first arg. to the bash script Could anybody tell me what's wrong with this script ? I'm new to scripting in general so any help would be apreciated ! #!/bin/ba

Re: Bash script help ?

2003-02-05 Thread Lars
On Wednesday, February 5, 2003, at 02:24 PM, Jan wrote: In your awk stmt you search for '1st_arg' rather than the value of the variable 1st_arg, which would be $1st_arg - or perhaps ${1st_arg}, as $1 is something else... Ok ! Thanks ! Lars -- redhat-list mailing list unsubscribe mailto:[E

Re: Bash script help ?

2003-02-05 Thread lardi
>Jon Haugsand wrote: >> * [EMAIL PROTECTED] >> >>>#!/bin/bash >>>1st_arg=$1 >> >> number not allowed in front. >> >> >>>prosesses=`lsof -i` >>>prosessid=`echo $prosesses | gawk /1st_arg/'{print $2}'` >> >> Cannot understand this to work in gawk. In any case, you might do >> this a lot more e

Re: Bash script help ?

2003-02-05 Thread Jan
Jon Haugsand wrote: * [EMAIL PROTECTED] #!/bin/bash 1st_arg=$1 number not allowed in front. prosesses=`lsof -i` prosessid=`echo $prosesses | gawk /1st_arg/'{print $2}'` Cannot understand this to work in gawk. In any case, you might do this a lot more effective: echo $prosessid if [ -z

Re: Bash script help ?

2003-02-05 Thread Jon Haugsand
* [EMAIL PROTECTED] > #!/bin/bash > 1st_arg=$1 number not allowed in front. > prosesses=`lsof -i` > prosessid=`echo $prosesses | gawk /1st_arg/'{print $2}'` Cannot understand this to work in gawk. In any case, you might do this a lot more effective: > echo $prosessid > if [ -z "$prosessid" ] mis

Re: Bash script help ?

2003-02-05 Thread Jan
[EMAIL PROTECTED] wrote: I'm trying to get a pidof a php script by capturing the path: like /home/somebody/my_script which will be the first arg. to the bash script Could anybody tell me what's wrong with this script ? I'm new to scripting in general so any help would be apreciated ! #!/bin/ba

Re: Bash Script question

2003-01-10 Thread Robert P. J. Day
On Thu, 9 Jan 2003, David Busby wrote: > List, > I've got a bash script, that executes other scripts (wow!) and the sub > scripts return a value. > How can I get the parent script to capture that return value? I've been all > over the BASH manual (more than one hour) and Google, still no luck

Re: Bash Script question

2003-01-09 Thread Ashley M. Kirchner
David Busby wrote: Here's the BASH code that I currently have #!/bin/bash for YEAR in 2003 2004 2005 2006 do CMD="./import.php $YEAR" $CMD done #!/bin/bash for YEAR in $(seq 2003 1 2006) do ./import.php $YEAR RETVAL=$? echo $RETVAL; done You can create different RETVALs if you

Re: Bash Script question

2003-01-09 Thread David Busby
to the number of items it imported. I would like BASH to trap this value and at the end echo the total of the 4 script executions. /B - Original Message - From: "Todd A. Jacobs" <[EMAIL PROTECTED]> To: <[EMAIL PROTECTED]> Sent: Thursday, January 09, 2003 19:39 Subject:

Re: Bash Script question

2003-01-09 Thread Todd A. Jacobs
On Thu, 9 Jan 2003, David Busby wrote: > How can I get the parent script to capture that return value? I've been > all over the BASH manual (more than one hour) and Google, still no luck. Not sure what you're really trying to do. If you posted real code, that would help. The exit status of th

Re: Bash Script || Loop Through Parameters

2002-11-06 Thread Todd A. Jacobs
On Wed, 6 Nov 2002, MET wrote: > for i in $# > do echo Loop iteration - $i > done What shell? Have you read the appropriate manual? Have you bought the appropriate O'Reilly book? Is this a homework assignment? -- "Whenever I feel blue, I start breathing again."

RE: Bash Script || Loop Through Parameters

2002-11-06 Thread Rechenberg, Andrew
That is what you told it to do :) $# outputs the number of arguments input to a script. You should look into the shift built-in function of (ba)sh. Try something like this: #!/bin/sh while [ "$1" ] do echo $1 shift done Check out: http://www.cyberciti.biz/nixcraft/linux/docs

RE: Bash Script || Loop Through Parameters

2002-11-06 Thread Babu, Sreedhar (CAP, GCF, Satyam, Consultant)
Replace $# with $* in ur script.. -Original Message- From: MET [mailto:met@;uberstats.com] Sent: Wednesday, November 06, 2002 20:07 To: RedHat List Subject: Bash Script || Loop Through Parameters I'm trying to loop through all the parameters given to a script. My script is executed l

Re: Bash Script Question

2002-11-06 Thread Robert P. J. Day
On Wed, 6 Nov 2002, Hella wrote: > You want to use sed and awk for a truly robust solution. a sadly-overlooked but useful text processing command is "expr". $ man expr which can do simple things like matching, extraction, length, substr and stuff like that. check it out before going on to sed,

Re: Bash Script Question

2002-11-06 Thread Hella
You want to use sed and awk for a truly robust solution. If you need further help, I can dig up examples of doing this. Generally, when I need to do text processing, I turn to perl. However, this can be done in bash, but larger tasks will get ugly in bash. -Chuck John H Darrah wrote: On Mon,

Re: Bash Script Question

2002-11-05 Thread John H Darrah
On Mon, 4 Nov 2002, Chad Skinner wrote: > Is there a way in a bash script to trim the spaces from the front and end of > a variable > > I have a script that contains the following variable definition > > > BLOCKED_SERVICES="tcp,111,Sun RPC;\ > udp,111,Sun RPC;\ >

RE: Bash Script Question

2002-11-05 Thread Cowles, Steve
> -Original Message- > From: mark > Subject: Re: Bash Script Question > > > For example, with a datafile, you could then say > export BLOCKED_SERVICES=`cat myblocked` but where you > need the services broken out, youi would want to use awk: > > # start o

Re: Bash Script Question

2002-11-05 Thread mark
On Tuesday 05 November 2002 09:51 am, Chad is done writ: > Is there a way in a bash script to trim the spaces from the front and > end of a variable > I have a script that contains the following variable definition > BLOCKED_SERVICES="tcp,111,Sun RPC;\ >udp,111,Sun

RE: Bash Script Question

2002-11-05 Thread mail-lists
you can use xargs to get rid of that whitespace in the protocol declaration: BLOCKED_SERVICES="tcp,111,Sun RPC;\ udp,111,Sun RPC;\ tcp,443,Microsoft DS;\ udp,443,Microsoft DS" IFS=";" for SERVICE in $BLOCKED

Re: Bash Script Question

2002-11-04 Thread Todd A. Jacobs
On Mon, 4 Nov 2002, Chad Skinner wrote: > Is there a way in a bash script to trim the spaces from the front and end of > a variable Not the way you're doing it. The easiest thing to do is to change your data representation, rather than spending a lot of time trying to strip whitespace. Try:

RE: Bash Script Question

2002-11-04 Thread Cowles, Steve
> -Original Message- > From: Chad Skinner > Subject: Bash Script Question > > > Is there a way in a bash script to trim the spaces from the > front and end of a variable > > I have a script that contains the following variable definition > > > BLOCKED_SERVICES="tcp,111,Sun RPC;

Re: bash script to test smtp relay

2002-10-12 Thread jkinz
On Fri, Oct 11, 2002 at 08:05:49PM -0500, Freddy Chavez wrote: > I've read many examples about using a shell script to test a mail server > (sendmail, postfix, etc) such as: > #!/bin/bash > telnet 1.2.3.4 25 << _EOF_ > HELO abc.com > <- smtp conversation here (mail from, r

RE: Bash Script Questions

2001-01-26 Thread Brad Doster
>> Look at manipulating the IFS variable; you can do wondrous things with that. Yep, I've thought about it, but that's all so far. >> Next, we get into opening and using file descriptors... All in good time. :) bd ___ Redhat-list mailing list [EMA

RE: Bash Script Questions

2001-01-26 Thread Brad Doster
>> See my other replies for more thoughts on this, but it actually has to be >> "s/$//g", and must use "s rather than 's. Whoops! My bad. :) Both of these work... 's/\\$//g' # single quotes "s/$//g"# double quotes Thanks to Matthew for catching my goof

RE: Bash Script Questions

2001-01-26 Thread Brad Doster
>> That's making the big assumption that wrong sort of quotes is your problem. >> If not then umm.. err.. *shurg* ... Did i meantion in worked for me? ;) Yep, you nailed it. Thanks for taking time to post the echo examples! bd ___ Redhat-list maili

Re: Bash Script Questions

2001-01-26 Thread Dave Ihnat
On Fri, Jan 26, 2001 at 09:41:30AM -0800, Brad Doster wrote: > IOW, I can treat the list as space or line delimited. Using var3=`echo > "$var1\n$var2"` eliminates the ability to treat it as a space delimilted > list as 'echo $var3' and 'echo "$var3"' both produce 'a\nb'. Well, in one sense--in a

RE: Bash Script Questions

2001-01-26 Thread Brad Doster
>> Now what you should do is store the \n's in var3 like: >> var3=`echo "$var1\n$var2"` >> after which: >> echo -e $var3 >> would give the desired output Yep, that works too. But :), var3=`echo -e "$var1\n$var2"` seems to be a bit more flexible in that 'echo $var3' produces... a b ...a

RE: Bash Script Questions

2001-01-26 Thread Brad Doster
>> varlist=`echo "$varlist" | sed 's/\\$//'` See my other replies for more thoughts on this, but it actually has to be "s/$//g", and must use "s rather than 's. Thanks for your input! bd ___ Redhat-list mailing list [EMAIL PROTECTED] https://li

RE: Bash Script Questions

2001-01-26 Thread Brad Doster
= Brad Doster Insight Network Solutions www.InsightNetSolutions.net 925.335.9510 -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]]On Behalf Of Luke C Gavel Sent: Friday, January 26, 2001 2:18 AM To: [EMAIL PROTECTED] Subject: RE: Bash Script Questions

RE: Bash Script Questions

2001-01-26 Thread Matthew Melvin
On Fri, 26 Jan 2001 at 8:34am (-0800), Brad Doster wrote: > >> Ahhh.. we need another \ - one to protect $ from sed and > >> another to protect the first \ from the backticks. > > >>varlist=`echo "$varlist" | sed 's/\\$//g'` > > Apparently that's still not enough. I think '\\$' parses to '\$

RE: Bash Script Questions

2001-01-26 Thread Brad Doster
-Original Message- From: Matthew Melvin [mailto:[EMAIL PROTECTED]] Sent: Thursday, January 25, 2001 9:57 PM To: Brad Doster Cc: [EMAIL PROTECTED] Subject: RE: Bash Script Questions On Thu, 25 Jan 2001 at 7:56pm (-0800), Brad Doster wrote: > Hi Matthew, > > Perhaps what I'm do

RE: Bash Script Questions

2001-01-26 Thread Stephen_Reilly
>> var3=`echo -e "$var1\n$var2"` >>...does the trick. It's the -e option in echo that's doing this, it allows the \n to be read and used as a new line. e.g. echo -e "$var1\n$var2" should read: a b but echo var3 after above would still give: a b Now what you should do is store the \n's in v

RE: Bash Script Questions

2001-01-26 Thread Stephen_Reilly
probably a bit late with this but; > > varlist=`echo "$varlist" | sed 's/$//'` > varlist=`echo "$varlist" | sed 's/\\$//'` > > 2) Given two variables, say var1="a" and var2="b", I want to create var3 > > such that it is "ab", i.e 'echo "$var3"' produces: > > > > a > > b > > > You w

RE: Bash Script Questions

2001-01-26 Thread Luke C Gavel
On Thu, 25 Jan 2001, Brad Doster wrote: > But why? Why doesn't good ol' '\$' work right out of the > box? Because the '$' is used in a special by the shell (script) and sed too. sed uses the '$' to anchor a search pattern from the end of a line, and, of course the script uses it to expand vari

RE: Bash Script Questions

2001-01-25 Thread Matthew Melvin
On Thu, 25 Jan 2001 at 7:56pm (-0800), Brad Doster wrote: > Hi Matthew, > > Perhaps what I'm doing in the '$' case needs a bit more explanation. The > following is a script segment that gleans variable names from the script in > which it is run: > > for varname in case select until while

RE: Bash Script Questions

2001-01-25 Thread Brad Doster
>> varlist=`echo "$varlist" | sed -e "s/$//g" Yep, that did it! But, what did it do? Or, where can I find info that will help me make sense of it? FWIW, it looks to me like it's searching for '\\$' (???). Oooh it is! But the '\\$' gets reduced to '\$' which then works as desired.

RE: Bash Script Questions

2001-01-25 Thread Brad Doster
Hi Matthew, Perhaps what I'm doing in the '$' case needs a bit more explanation. The following is a script segment that gleans variable names from the script in which it is run: for varname in case select until while ; do varlist=`cat "$0" | grep "^[]*$varname " | se

Re: Bash Script Questions

2001-01-25 Thread Matthew Melvin
On Thu, 25 Jan 2001 at 10:21am (-0800), Brad Doster wrote: > 1) I have a variable with a list of variable names as its content, e.g. > 'varlist=$var1 $var2 $var3'. I want to manipulate $varlist such that it > ='var1 var2 var3', i.e get rid of the '$'s in front of each variable name. > A 'sed' ex

Re: Bash Script Questions

2001-01-25 Thread Dave Ihnat
On Thu, Jan 25, 2001 at 10:21:39AM -0800, Brad Doster wrote: > 1) I have a variable with a list of variable names as its content, e.g. > 'varlist=$var1 $var2 $var3'. I want to manipulate $varlist such that it > ='var1 var2 var3', i.e get rid of the '$'s in front of each variable name. > A 'sed' e

Re: bash script question (for loops)

2000-11-27 Thread David Talkington
-BEGIN PGP SIGNED MESSAGE- Matthew Melvin wrote: > >You can use $IFS to tell bash to split on just the line feeds instead of the >spaces >[mloe@riverbank (1) bashtest]$ IFS=' >> ' Brilliant, thank you -- this was driving me nuts. I finally figured out that the reason this works: LI

Re: bash script question (for loops)

2000-11-27 Thread David Talkington
-BEGIN PGP SIGNED MESSAGE- Mark - You're right, that's an odd inconsistency. I thought perhaps we could modify IFS, but it uses space/tab/newline by default, yet it doesn't seem to be splitting on the space in the first example. (I would think it should.) These two commands produce i

Re: bash script question (for loops)

2000-11-26 Thread Statux
Use quotes.. these things ---> " " :) I didn't get to read your whole message, but quotes protect text from default parameterization rules of the shell, etc. Therefore: cat "Filename with spaces.txt" will.. well.. you get the idea A common script I do do is like: for f in *.c do echo -e "$f"

Re: bash script question (for loops)

2000-11-26 Thread Todd A. Jacobs
Put double-quotes around your variables. :) -- Todd A. Jacobs Senior Network Consultant ___ Redhat-list mailing list [EMAIL PROTECTED] https://listman.redhat.com/mailman/listinfo/redhat-list

Re: bash script question (for loops)

2000-11-26 Thread Matthew Melvin
On Sat, 25 Nov 2000, Mark Ivey wrote: > Here are two more scripts, similar to the first 2: >LIST=*; for NAME in $LIST; do echo "-->" $NAME; done >LIST=$(\ls); for NAME in $LIST; do echo "-->" $NAME; done > > They give different output. The first one gives: >--> Astral Projection

Re: Bash script

2000-01-28 Thread John Darrah
On Thu, 27 Jan 2000, Jeff Smelser wrote: > Just trying to write a bash script and can't seem to figure out how to > do addition with variables. If someone could give me a quick example, I > would appreciate it. Thanks Here is a script with lots of math examples. -- cut -- #!/bin/ksh #

Re: Bash script

2000-01-28 Thread Steven W Orr
If no one minds, I'd just like to amplify on this a little. The solution below uses the primitive Bourne shell method which uses an external program (expr) to do the arithmetic. Bash has internal functionality to perform stuff like this. The hashpling at the beginning specifies which interpreter i

Re: Bash script

2000-01-28 Thread rpjday
On Thu, 27 Jan 2000, Jeff Smelser wrote: > this worked great, thanks a lot! > > > On Thu, 27 Jan 2000, Bret Hughes wrote: > > > Try some thing like this > > > > > > #! /bin/sh > > # addnums - adds two numbers input as args and outputs the result > > newval=`expr $1 + $2`

Re: Bash script

2000-01-27 Thread Rick L. Mantooth
You can also use bc #!/bin/sh var1=5 var2=6 echo "$var1 + $var2" | bc # end Have fun, -- Rick L. Mantooth [EMAIL PROTECTED] As I said before, I never repeat myself! On Thu, 27 Jan 2000, Jeff Smelser wrote: => Just trying to write a bash script and can't seem to figure out how to => do additio

Re: Bash script

2000-01-27 Thread Jeff Smelser
this worked great, thanks a lot! On Thu, 27 Jan 2000, Bret Hughes wrote: > Try some thing like this > > > #! /bin/sh > # addnums - adds two numbers input as args and outputs the result > newval=`expr $1 + $2` > echo 'the sum of ' $1 ' and ' $2 ' is ' $newval > > > Jeff

Re: Bash script

2000-01-27 Thread Bret Hughes
Try some thing like this #! /bin/sh # addnums - adds two numbers input as args and outputs the result newval=`expr $1 + $2` echo 'the sum of ' $1 ' and ' $2 ' is ' $newval Jeff Smelser wrote: > > Just trying to write a bash script and can't seem to figure out how to > do addition with variab