Re: [expert] bash scripting question - simple regular expression?
-BEGIN PGP SIGNED MESSAGE-
Hash: SHA1
Jim C wrote on Tue, Jan 14, 2003 at 07:28:07PM -0800 :
> >newldif="./file2"
> Herein lies part of the problem.
> I don't want to use a file for scaleability reasons and I can't think of
> any reason why it should be nescesary. In theory I should be able to
> store all of the text in a shell variable and then redirect the output
> of echo to ldapadd. So far I can get the text in but when I pass it to
You're right, this is part of the problem. Shell variables don't keep
the newline characters. Witness:
[todd@fiji ~/tmp]$ cat file1
#!/bin/bash
TEMP="line1
line2
line3"
echo $TEMP
[todd@fiji ~/tmp]$ ./file1
line1 line2 line3
I don't see why you just don't do this:
#!/bin/bash
if [ ! $1 ]; then
echo "Sorry, you must pass the userid to add to directory."
exit 1
fi
binddn="cn=root,dc=microverse,dc=net"
pw4binddn="[deleted for security]"
ldaphost="ldap://localhost";
base="ou=Computers,dc=microverse,dc=net"
basetest="ou=People,dc=microverse,dc=net"
minimumUID=501
groupnum=421
# 1. Search the LDAP database and return all uidNumber attributes in a
# given base
store=`ldapsearch -LLL -D $binddn -H $ldaphost -b$base -x "(cn=*)" uidNumber | \
grep uidNumber | \
sed -e 's/^uidNumber://' | sort -nr | head -n 1`
newtest=`ldapsearch -LLL -D $binddn -H $ldaphost -b$basetest -x "(cn=*)" uidNumber | \
grep uidNumber | \
sed -e 's/^uidNumber://' | sort -nr`
echo ${newtest[0]}
#It is best not to start at 0 or 1 as these could be privledged.
if [ "$store" = "" ]
then
store=$minimumUID
else
store=`expr $store + 1`
fi
LDAPADD="ldapadd -vx -D $binddn -W $pw4binddn"
$LDAPADD
Re: [expert] bash scripting question - simple regular expression?
Herein lies part of the problem. I don't want to use a file for scaleability reasons and I can't think of any reason why it should be nescesary. In theory I should be able to store all of the text in a shell variable and then redirect the output of echo to ldapadd. So far I can get the text in but when I pass it to ldapadd it executes, returns no error messages but the new record does not show up on the ldap server. Since it works from the command line using an ldif file it logically cannot be an issue of access to the server. Also, no encryption is currently enabled. Now Whoops, accidentally sent the message before finishing it. As I was going to say, the idea of not having the delimiters correct has some merit as I wouldn't know either way. How can I check or set them? Want to buy your Pack or Services from MandrakeSoft? Go to http://www.mandrakestore.com
Re: [expert] bash scripting question - simple regular expression?
echo $newldif
This produces:
[root@enigma scripts]# ./adduser adlfalj
./adduser: line 42: $newldif: ambiguous redirect
You've got something else going on cause it works on mine. (Are you
sure that you're using the bash shell?)
[root@enigma scripts]# echo $SHELL
/bin/bash
[root@enigma scripts]#
[todd@fiji ~/tmp]$ cat file1
#!/bin/bash
#
newldif="./file2"
Herein lies part of the problem.
I don't want to use a file for scaleability reasons and I can't think of
any reason why it should be nescesary. In theory I should be able to
store all of the text in a shell variable and then redirect the output
of echo to ldapadd. So far I can get the text in but when I pass it to
ldapadd it executes, returns no error messages but the new record does
not show up on the ldap server. Since it works from the command line
using an ldif file it logically cannot be an issue of access to the
server. Also, no encryption is currently enabled. Now
#
cat > $newldif <
...
description: Machine Account
...
Post the WHOLE script so we can see what you're doing. If you change
anything make sure you say exactly what you change.
It's a mess so remember, you asked for it. ;-)
I've been doing all kinds of tests to try and figure a way around the
problem. The algorithm works like this:
1. Get a list of uidNumbers
2. Sort them.
3. Take the one off the top (the largest)
4. Add one to it.
5. Create the text of a new record using the new uidNumber.
#!/bin/bash
binddn="cn=root,dc=microverse,dc=net"
pw4binddn="[deleted for security]"
ldaphost="ldap://localhost";
base="ou=Computers,dc=microverse,dc=net"
basetest="ou=People,dc=microverse,dc=net"
minimumUID=501
groupnum=421
#complete=`echo $line1 $line2 $line3 $line4 $line5 $line6 $line7 $line8
$line9 $line10 $line11 $line12`
#ldapsearch -LL -v -D "cn=proxyuser,dc=microverse,dc=net" -H
ldap://localhost -b"dc=microverse,dc=net" -x "(cn=proxyuser)"
#1. Search the LDAP database and return all uidNumber attributes in a
given base
store=`ldapsearch -LLL -D $binddn -H $ldaphost -b$base -x "(cn=*)"
uidNumber | \
grep uidNumber | \
sed -e 's/^uidNumber:
//' | sort -nr | head -n 1`
newtest=`ldapsearch -LLL -D $binddn -H $ldaphost -b$basetest -x "(cn=*)"
uidNumber | \
grep uidNumber | \
sed -e 's/^uidNumber:
//' | sort -nr`
echo ${newtest[0]}
#It is best not to start at 0 or 1 as these could be privledged.
if [ "$store" = "" ]
then
store=$minimumUID
else
store=`expr $store + 1`
fi
#ldapadd -x -D $binddn -w $pw4binddn
line1="dn: uid=$1,ou=Computers,dc=microverse,dc=net\n";
line2="objectClass: top\n"
line3="objectClass: account\n"
line4="objectClass: posixAccount\n"
line5="uidNumber: $store\n"
line6="uid: $1\n"
line7="cn: $1\n"
line8="gidNumber: $groupnum\n"
line9="homeDirectory: /dev/null\n"
line10="loginShell: /bin/false\n"
line11="gecos: Machine Account\n"
line12="description: Machine Account\n"
output=$line1$line2$line3$line4$line5$line6$line7$line8$line9$line10$line11$line12
echo -e "$output" | ldapadd -vx -D $binddn -W $pw4binddn
echo -e "$output" > ldapadd -vx -D $binddn -W $pw4binddn
echo -e $output
echo -e $output > test.ldif
#cat $output
#echo $output
#$output <
#dn: uid=$1,ou=Computers,dc=microverse,dc=net
#objectClass: top
#objectClass: account
#objectClass: posixAccount
#uidNumber: $store
#uid: $1
#cn: $1
#gidNumber: $groupnum
#homeDirectory: /dev/null
#loginShell: /bin/false
#gecos: Machine Account
#description: Machine Account
#hrdoc
#cat $line1 $line2 $line3 $line4 $line5 $line6 $line7 $line8 $line9
$line10 $line11 $line12 > ldapadd -x -D $binddn -w $pw4binddn
#complt=$line1$line2$line3$line4$line5$line6$line7$line8$line9$line10$line11$line12
#echo `expr $store + 1`
Want to buy your Pack or Services from MandrakeSoft?
Go to http://www.mandrakestore.com
Re: [expert] bash scripting question - simple regular expression?
jipe wrote: here is what says man: The entry information is read from standard input or from file through the use of the -f option. so why not to try this: echo -e "$output" | ldapadd -x -D "cn=root,dc=microverse,dc=net" This produces an error, i.e. I get the usage text for ldapadd if I try it. Want to buy your Pack or Services from MandrakeSoft? Go to http://www.mandrakestore.com
Re: [expert] bash scripting question - simple regular expression?
-BEGIN PGP SIGNED MESSAGE- Hash: SHA1 Jim C wrote on Tue, Jan 14, 2003 at 11:31:20AM -0800 : > I finally got around to giving your suggestion a try. :-) > > cat > $newldif < dn: uid=$1,ou=Computers,dc=microverse,dc=net > objectClass: top > objectClass: account > objectClass: posixAccount > uidNumber: $store > uid: $1 > cn: $1 > gidNumber: $groupnum > homeDirectory: /dev/null > loginShell: /bin/false > gecos: Machine Account > description: Machine Account > hrdoc > > echo $newldif > > This produces: > > [root@enigma scripts]# ./adduser adlfalj > ./adduser: line 42: $newldif: ambiguous redirect You've got something else going on cause it works on mine. (Are you sure that you're using the bash shell?) [todd@fiji ~/tmp]$ cat file1 #!/bin/bash # newldif="./file2" # cat > $newldif < This causes the script to halt at some point and gives no output if you > replace the 'cat' with 'echo'. Sounds like 1) You've got a misplaced backtick. 2) You're using different delimiters for the << hrdoc start point and hrdoc endpoint. Post the WHOLE script so we can see what you're doing. If you change anything make sure you say exactly what you change. > >All you gotta do is change the "command" portion of the above that I > >quoted: > > > >#!/bin/bash > >ldapadd -x blah blah blah < >dn: uid=$1,ou=Computers,dc=microverse,dc=net > >objectClass: top > >objectClass: account > >objectClass: posixAccount > >uidNumber: $store > >uid: $1 > >cn: $1 > >gidNumber: $groupnum > >homeDirectory: /dev/null > >loginShell: /bin/false > >gecos: Machine Account > >description: Machine Account > >hrdoc What I said originally should still work. Blue skies... Todd - -- MandrakeSoft USA http://www.mandrakesoft.com cat /boot/vmlinuz > /dev/dsp #for great justice Cooker Version mandrake-release-9.1-0.1mdk Kernel 2.4.20-2mdk -BEGIN PGP SIGNATURE- Version: GnuPG v1.2.1 (GNU/Linux) iD8DBQE+JLl8lp7v05cW2woRAuSQAKDOd7bKbEgO3kt6eL8SLrZHLcwwHACcCSWk r7X2YShaye9fOJwX7rcsyWQ= =559n -END PGP SIGNATURE- Want to buy your Pack or Services from MandrakeSoft? Go to http://www.mandrakestore.com
Re: [expert] bash scripting question - simple regular expression?
On Tue, 14 Jan 2003 12:58:30 -0800 Jim C <[EMAIL PROTECTED]> wrote: > Grrr!! > > What gets me is why this doesn't work: > > echo -e $output > ldapadd -x -D $binddn -W $pw4binddn > > where echo -e produces: > > [root@enigma scripts]# ./adduser alkjdfal > > dn: uid=alkjdfal,ou=Computers,dc=microverse,dc=net > objectClass: top > objectClass: account > objectClass: posixAccount > uidNumber: 501 > uid: alkjdfal > cn: alkjdfal > gidNumber: 421 > homeDirectory: /dev/null > loginShell: /bin/false > gecos: Machine Account > description: Machine Account > > 502 <--different echo statement > > If I do echo -e $output > test.ldif > and then > ldapadd -x -D "cn=root,dc=microverse,dc=net" -f test.ldif from the > command line it works fine! Do you think this is a bug in ldapadd? > > here is what says man: The entry information is read from standard input or from file through the use of the -f option. so why not to try this: echo -e "$output" | ldapadd -x -D "cn=root,dc=microverse,dc=net" bye jipe Want to buy your Pack or Services from MandrakeSoft? Go to http://www.mandrakestore.com
Re: [expert] bash scripting question - simple regular expression?
Grrr!! What gets me is why this doesn't work: echo -e $output > ldapadd -x -D $binddn -W $pw4binddn where echo -e produces: [root@enigma scripts]# ./adduser alkjdfal dn: uid=alkjdfal,ou=Computers,dc=microverse,dc=net objectClass: top objectClass: account objectClass: posixAccount uidNumber: 501 uid: alkjdfal cn: alkjdfal gidNumber: 421 homeDirectory: /dev/null loginShell: /bin/false gecos: Machine Account description: Machine Account 502 <--different echo statement If I do echo -e $output > test.ldif and then ldapadd -x -D "cn=root,dc=microverse,dc=net" -f test.ldif from the command line it works fine! Do you think this is a bug in ldapadd? Todd Lyons wrote: -BEGIN PGP SIGNED MESSAGE- Hash: SHA1 Jim C wrote on Mon, Jan 13, 2003 at 10:16:41AM -0800 : What I am trying to do is create a sort of virtual file. I've already tried using the here document and it doesn't seem to be working for me. What I am trying to do exactly is create an LDIF file for passing to an LDAP database. I have each line stored in an individual shell variable like so: There are two ways to do this: 1) Create an ldif file, then call ldap* with that ldif file. 2) Just pass it all to ldap* on stdin. The syntax I've been trying looks like this: ldapadd -x -D $binddn -w "placepasswordhere" < Look at this: #!/bin/bash cat > $1.ldif < In all cases the here document doesn't seem to make it to the ldapadd command which I know reads from standard-in. All you gotta do is change the "command" portion of the above that I quoted: #!/bin/bash ldapadd -x blah blah blah < dn: uid=$1,ou=Computers,dc=microverse,dc=net objectClass: top objectClass: account objectClass: posixAccount uidNumber: $store uid: $1 cn: $1 gidNumber: $groupnum homeDirectory: /dev/null loginShell: /bin/false gecos: Machine Account description: Machine Account hrdoc Blue skies... Todd - -- MandrakeSoft USA http://www.mandrakesoft.com Mandrake: An amalgam of good ideas from RedHat, Debian, and MandrakeSoft. All in all, IMHO, an unbeatable combination. --Levi Ramsey on Cooker ML Cooker Version mandrake-release-9.1-0.1mdk Kernel 2.4.20-2mdk -BEGIN PGP SIGNATURE- Version: GnuPG v1.2.1 (GNU/Linux) iD8DBQE+Iy/Ylp7v05cW2woRAhwMAJoDELaI+O82WNuz5SRnujaXG5y7hQCfUdlY dCj3ZneaqniZkaaADmAFVD4= =O07o -END PGP SIGNATURE- Want to buy your Pack or Services from MandrakeSoft? Go to http://www.mandrakestore.com Want to buy your Pack or Services from MandrakeSoft? Go to http://www.mandrakestore.com
Re: [expert] bash scripting question - simple regular expression?
I finally got around to giving your suggestion a try. :-) cat > $newldif < dn: uid=$1,ou=Computers,dc=microverse,dc=net objectClass: top objectClass: account objectClass: posixAccount uidNumber: $store uid: $1 cn: $1 gidNumber: $groupnum homeDirectory: /dev/null loginShell: /bin/false gecos: Machine Account description: Machine Account hrdoc echo $newldif This produces: [root@enigma scripts]# ./adduser adlfalj ./adduser: line 42: $newldif: ambiguous redirect $newldif < dn: uid=$1,ou=Computers,dc=microverse,dc=net objectClass: top objectClass: account objectClass: posixAccount uidNumber: $store uid: $1 cn: $1 gidNumber: $groupnum homeDirectory: /dev/null loginShell: /bin/false gecos: Machine Account description: Machine Account hrdoc cat $newldif This causes the script to halt at some point and gives no output if you replace the 'cat' with 'echo'. Todd Lyons wrote: -BEGIN PGP SIGNED MESSAGE- Hash: SHA1 Jim C wrote on Mon, Jan 13, 2003 at 10:16:41AM -0800 : What I am trying to do is create a sort of virtual file. I've already tried using the here document and it doesn't seem to be working for me. What I am trying to do exactly is create an LDIF file for passing to an LDAP database. I have each line stored in an individual shell variable like so: There are two ways to do this: 1) Create an ldif file, then call ldap* with that ldif file. 2) Just pass it all to ldap* on stdin. The syntax I've been trying looks like this: ldapadd -x -D $binddn -w "placepasswordhere" < Look at this: #!/bin/bash cat > $1.ldif < In all cases the here document doesn't seem to make it to the ldapadd command which I know reads from standard-in. All you gotta do is change the "command" portion of the above that I quoted: #!/bin/bash ldapadd -x blah blah blah < dn: uid=$1,ou=Computers,dc=microverse,dc=net objectClass: top objectClass: account objectClass: posixAccount uidNumber: $store uid: $1 cn: $1 gidNumber: $groupnum homeDirectory: /dev/null loginShell: /bin/false gecos: Machine Account description: Machine Account hrdoc Blue skies... Todd - -- MandrakeSoft USA http://www.mandrakesoft.com Mandrake: An amalgam of good ideas from RedHat, Debian, and MandrakeSoft. All in all, IMHO, an unbeatable combination. --Levi Ramsey on Cooker ML Cooker Version mandrake-release-9.1-0.1mdk Kernel 2.4.20-2mdk -BEGIN PGP SIGNATURE- Version: GnuPG v1.2.1 (GNU/Linux) iD8DBQE+Iy/Ylp7v05cW2woRAhwMAJoDELaI+O82WNuz5SRnujaXG5y7hQCfUdlY dCj3ZneaqniZkaaADmAFVD4= =O07o -END PGP SIGNATURE- Want to buy your Pack or Services from MandrakeSoft? Go to http://www.mandrakestore.com Want to buy your Pack or Services from MandrakeSoft? Go to http://www.mandrakestore.com
Re: [expert] bash scripting question - simple regular expression?
-BEGIN PGP SIGNED MESSAGE- Hash: SHA1 Jim C wrote on Mon, Jan 13, 2003 at 10:16:41AM -0800 : > > What I am trying to do is create a sort of virtual file. > I've already tried using the here document and it doesn't seem to be > working for me. What I am trying to do exactly is create an LDIF file > for passing to an LDAP database. I have each line stored in an > individual shell variable like so: There are two ways to do this: 1) Create an ldif file, then call ldap* with that ldif file. 2) Just pass it all to ldap* on stdin. > The syntax I've been trying looks like this: > ldapadd -x -D $binddn -w "placepasswordhere" < line1="dn: uid=$1,ou=Computers,dc=microverse,dc=net"; > line2="objectClass: top" Look at this: #!/bin/bash cat > $1.ldif < In all cases the here document doesn't seem to make it to the ldapadd > command which I know reads from standard-in. All you gotta do is change the "command" portion of the above that I quoted: #!/bin/bash ldapadd -x blah blah blah
Re: [expert] bash scripting question - simple regular expression?
Great but now I have another problem.
What I am trying to do is create a sort of virtual file.
I've already tried using the here document and it doesn't seem to be
working for me. What I am trying to do exactly is create an LDIF file
for passing to an LDAP database. I have each line stored in an
individual shell variable like so:
line1="dn: uid=$1,ou=Computers,dc=microverse,dc=net";
line2="objectClass: top"
line3="objectClass: account"
line4="objectClass: posixAccount"
line5="uidNumber: $store"
line6="uid: $1"
line7="cn: $1"
line8="gidNumber: $groupnum"
line9="homeDirectory: /dev/null"
line10="loginShell: /bin/false"
line11="gecos: Machine Account"
line12="description: Machine Account"
The syntax I've been trying looks like this:
ldapadd -x -D $binddn -w "placepasswordhere" <
line1="dn: uid=$1,ou=Computers,dc=microverse,dc=net";
line2="objectClass: top"
line3="objectClass: account"
line4="objectClass: posixAccount"
line5="uidNumber: $store"
line6="uid: $1"
line7="cn: $1"
line8="gidNumber: $groupnum"
line9="homeDirectory: /dev/null"
line10="loginShell: /bin/false"
line11="gecos: Machine Account"
line12="description: Machine Account"
hrdoc
I've also tried it with newlines at the end of each line like so:
ldapadd -x -D $binddn -w "placepasswordhere" <
line1="dn: uid=$1,ou=Computers,dc=microverse,dc=net\n";
line2="objectClass: top\n"
line3="objectClass: account\n"
line4="objectClass: posixAccount\n"
line5="uidNumber: $store\n"
line6="uid: $1\n"
line7="cn: $1\n"
line8="gidNumber: $groupnum\n"
line9="homeDirectory: /dev/null\n"
line10="loginShell: /bin/false\n"
line11="gecos: Machine Account\n"
line12="description: Machine Account\n"
hrdoc
In all cases the here document doesn't seem to make it to the ldapadd
command which I know reads from standard-in. I've also tried the -f
switch which is for specifiying a filename.
Anybody got a clue here?
jipe wrote:
On Sun, 12 Jan 2003 09:15:50 -0800
Mark Alexander <[EMAIL PROTECTED]> wrote:
On Sat, Jan 11, 2003 at 05:56:16PM -0800, Jim C wrote:
I have a list of positive integers of which I only want the first one.
They are of arbitrary size. How can I cut the rest of them off?
I've been trying to write a regular expression for this using sed or awk.
#!/bin/sh
LIST="42 666 1776 2001"
echo $LIST | sed -e 's/^\([0-9]*\).*/\1/'
LIST="42 666 1776 2001"
echo ${LIST%% *} --> 42
or
LIST=(42 666 1776 2001)
echo ${LIST[0]} --> 42
both are about 5 times faster than a pipe with sed.
bye
jipe
Want to buy your Pack or Services from MandrakeSoft?
Go to http://www.mandrakestore.com
Want to buy your Pack or Services from MandrakeSoft?
Go to http://www.mandrakestore.com
Re: [expert] bash scripting question - simple regular expression?
Wow. I didn't know bash had arrays. Kool. :-)
jipe wrote:
On Sat, 11 Jan 2003 17:56:16 -0800
Jim C <[EMAIL PROTECTED]> wrote:
I have a list of positive integers of which I only want the first one.
They are of arbitrary size. How can I cut the rest of them off?
I've been trying to write a regular expression for this using sed or awk.
bash you said, so use only bash.
array=($(your_command_to_create_the list_of_intergers))
echo ${array[0]}
this displays the 1rst integer from your list.
echo ${#array[*]} displays the number of integers
echo ${array[$((${#array[*]}-1))]} displays the last integer
etc...
maybe you'll have to set IFS. depends of the separator.
bye
jipe
Want to buy your Pack or Services from MandrakeSoft?
Go to http://www.mandrakestore.com
Want to buy your Pack or Services from MandrakeSoft?
Go to http://www.mandrakestore.com
Re: [expert] bash scripting question - simple regular expression?
On Sun, 12 Jan 2003 09:15:50 -0800
Mark Alexander <[EMAIL PROTECTED]> wrote:
> On Sat, Jan 11, 2003 at 05:56:16PM -0800, Jim C wrote:
> > I have a list of positive integers of which I only want the first one.
> > They are of arbitrary size. How can I cut the rest of them off?
> > I've been trying to write a regular expression for this using sed or awk.
>
> #!/bin/sh
> LIST="42 666 1776 2001"
> echo $LIST | sed -e 's/^\([0-9]*\).*/\1/'
>
>
LIST="42 666 1776 2001"
echo ${LIST%% *} --> 42
or
LIST=(42 666 1776 2001)
echo ${LIST[0]} --> 42
both are about 5 times faster than a pipe with sed.
bye
jipe
Want to buy your Pack or Services from MandrakeSoft?
Go to http://www.mandrakestore.com
Re: [expert] bash scripting question - simple regular expression?
On Sat, 11 Jan 2003 17:56:16 -0800
Jim C <[EMAIL PROTECTED]> wrote:
> I have a list of positive integers of which I only want the first one.
> They are of arbitrary size. How can I cut the rest of them off?
> I've been trying to write a regular expression for this using sed or awk.
>
>
>
bash you said, so use only bash.
array=($(your_command_to_create_the list_of_intergers))
echo ${array[0]}
this displays the 1rst integer from your list.
echo ${#array[*]} displays the number of integers
echo ${array[$((${#array[*]}-1))]} displays the last integer
etc...
maybe you'll have to set IFS. depends of the separator.
bye
jipe
Want to buy your Pack or Services from MandrakeSoft?
Go to http://www.mandrakestore.com
Re: [expert] bash scripting question - simple regular expression?
On Sat, Jan 11, 2003 at 05:56:16PM -0800, Jim C wrote: > I have a list of positive integers of which I only want the first one. > They are of arbitrary size. How can I cut the rest of them off? > I've been trying to write a regular expression for this using sed or awk. #!/bin/sh LIST="42 666 1776 2001" echo $LIST | sed -e 's/^\([0-9]*\).*/\1/' Want to buy your Pack or Services from MandrakeSoft? Go to http://www.mandrakestore.com
Re: [expert] bash scripting question - simple regular expression?
> >>>I have a list of positive integers of which I only want the first one.
> >>>They are of arbitrary size. How can I cut the rest of them off?
> >>>I've been trying to write a regular expression for this using sed or
> >>> awk.
> >
> > you don't specify the format of the integers. are they space separated
> > and all on one line? or are they line separated (each integer on
> > its own line)?
> >
> > are there blank spaces (if the first) or blank lines (if the second)
> > before the first integer?
>
> Blank spaces are between the numbers.
>
In awk is pretty simple, just do:
awk '{ print $1 }' filename
(where filename is the name of the file with the numbers)
I'm assuming you have only one line in that file, if you have more than one
line and you want to specify which line to print do:
awk '{ if ( NR==line_number ) print $1 } filename
--
Toshiro
Want to buy your Pack or Services from MandrakeSoft?
Go to http://www.mandrakestore.com
Re: [expert] bash scripting question - simple regular expression?
On Saturday 11 January 2003 08:56 pm, Jim C wrote: I have a list of positive integers of which I only want the first one. They are of arbitrary size. How can I cut the rest of them off? I've been trying to write a regular expression for this using sed or awk. you don't specify the format of the integers. are they space separated and all on one line? or are they line separated (each integer on its own line)? are there blank spaces (if the first) or blank lines (if the second) before the first integer? Blank spaces are between the numbers. or are they in some other format? you say they're arbitrary size, so does that mean they aren't in some fixed column format?\ No fixed columns. What they are is a line of userids with spaces between them. if they're one integer to a line, then head (for the first), tail (for the last), or some combination of head and tail (for anything in between) will get you the integer you want. There is only one line... but that is a good idea. I might be able to arrange for them to be in one column. Then I could use head. :-) I'll try that and let you know. if they're delimited by something (spaces, commas, colons, whatever), then cut -f -d "" might help. otherwise, well, sed, awk, or some custom program in your favorite language. sed and awk are driving me nuts. ;-) if there are blank lines or blank fields before the first number, then some sort of bash while loop would be needed to ignore the blank lines or blank fields, print the first non-blank field, and then exit the loop. Want to buy your Pack or Services from MandrakeSoft? Go to http://www.mandrakestore.com
Re: [expert] bash scripting question - simple regular expression?
The are already sorted. I just need the largest one which is the first one. Mark Weaver wrote: -BEGIN PGP SIGNED MESSAGE- Hash: SHA1 On Saturday 11 January 2003 08:56 pm, Jim C wrote: I have a list of positive integers of which I only want the first one. They are of arbitrary size. How can I cut the rest of them off? I've been trying to write a regular expression for this using sed or awk. without sorting them or anything why not just use a while loop with a counter? Mark -BEGIN PGP SIGNATURE- Version: GnuPG v1.2.1 (GNU/Linux) iD8DBQE+IM23JuZ1geTzHgERAmKPAKDSouh+GrtkZTPj9o5CDvTG7tIEtgCgjOZk buhxrkvmzeqXdrW7n0a8ACk= =Mw7P -END PGP SIGNATURE- Want to buy your Pack or Services from MandrakeSoft? Go to http://www.mandrakestore.com Want to buy your Pack or Services from MandrakeSoft? Go to http://www.mandrakestore.com
Re: [expert] bash scripting question - simple regular expression?
> On Saturday 11 January 2003 08:56 pm, Jim C wrote: > > I have a list of positive integers of which I only want the first one. > > They are of arbitrary size. How can I cut the rest of them off? > > I've been trying to write a regular expression for this using sed or awk. you don't specify the format of the integers. are they space separated and all on one line? or are they line separated (each integer on its own line)? are there blank spaces (if the first) or blank lines (if the second) before the first integer? or are they in some other format? you say they're arbitrary size, so does that mean they aren't in some fixed column format? if they're one integer to a line, then head (for the first), tail (for the last), or some combination of head and tail (for anything in between) will get you the integer you want. if they're delimited by something (spaces, commas, colons, whatever), then cut -f -d "" might help. otherwise, well, sed, awk, or some custom program in your favorite language. if there are blank lines or blank fields before the first number, then some sort of bash while loop would be needed to ignore the blank lines or blank fields, print the first non-blank field, and then exit the loop. tiger -- Gerald Timothy Quimpo tiger*quimpo*org gquimpo*sni-inc.com tiger*sni*ph Public Key: "gpg --keyserver pgp.mit.edu --recv-keys 672F4C78" Veritas liberabit vos. Doveryai no proveryai. Want to buy your Pack or Services from MandrakeSoft? Go to http://www.mandrakestore.com
Re: [expert] bash scripting question - simple regular expression?
-BEGIN PGP SIGNED MESSAGE- Hash: SHA1 On Saturday 11 January 2003 08:56 pm, Jim C wrote: > I have a list of positive integers of which I only want the first one. > They are of arbitrary size. How can I cut the rest of them off? > I've been trying to write a regular expression for this using sed or awk. without sorting them or anything why not just use a while loop with a counter? Mark -BEGIN PGP SIGNATURE- Version: GnuPG v1.2.1 (GNU/Linux) iD8DBQE+IM23JuZ1geTzHgERAmKPAKDSouh+GrtkZTPj9o5CDvTG7tIEtgCgjOZk buhxrkvmzeqXdrW7n0a8ACk= =Mw7P -END PGP SIGNATURE- Want to buy your Pack or Services from MandrakeSoft? Go to http://www.mandrakestore.com
[expert] bash scripting question - simple regular expression?
I have a list of positive integers of which I only want the first one. They are of arbitrary size. How can I cut the rest of them off? I've been trying to write a regular expression for this using sed or awk. Want to buy your Pack or Services from MandrakeSoft? Go to http://www.mandrakestore.com
Re: [expert] Bash Scripting
Necrotica wrote: > > I'm interested in learning a little about bash scripting. In particular, I'd > like to learn how to tell if the user logging on is logging in under X or from > a command prompt. Can anyone point me in the direction of where I can find this > information? Thanks... > > -Chris Just see if $DISPLAY is set. (Sorry: I only know csh; there it's $?DISPLAY, but probably not the same for bash.) -- "Brian, the man from babble-on" [EMAIL PROTECTED] Brian T. Schellenberger http://www.babbleon.org Support http://www.eff.org. Support decss defendents. Support http://www.programming-freedom.org. Boycott amazon.com.
Re: [expert] Bash Scripting
Thanks for the help, guys. That was exactly what I was looking for! -Chris On Sat, 10 Jun 2000, Civileme wrote: > Civileme wrote: > > > > Necrotica wrote: > > > > > > I'm interested in learning a little about bash scripting. In particular, I'd > > > like to learn how to tell if the user logging on is logging in under X or from > > > a command prompt. Can anyone point me in the direction of where I can find this > > > information? Thanks... > > > > > > -Chris > > > > printenv from Konsole when you are logged in in graphics mode > > > > printenv from a console (logged in on runlevel 3) > > > > Check the settings of the DISPLAY variable > > > > Civileme > > Ummm... Well under the console perhaps better to printenv | > less (blush). That fills more than a screen! > > > The point is DISPLAY has a value in graphics mode and does NOT > exist in console > > So > > if [ $DISPLAY ] then > # DISPLAY is set and is a nonzero length string so do GUI stuff > elif [ -z $DISPLAY ] > # do the console stuf here (length of string is zero) > else > # do any other stuff here > fi > > Civileme
Re: [expert] Bash Scripting
Civileme wrote: > > Necrotica wrote: > > > > I'm interested in learning a little about bash scripting. In particular, I'd > > like to learn how to tell if the user logging on is logging in under X or from > > a command prompt. Can anyone point me in the direction of where I can find this > > information? Thanks... > > > > -Chris > > printenv from Konsole when you are logged in in graphics mode > > printenv from a console (logged in on runlevel 3) > > Check the settings of the DISPLAY variable > > Civileme Ummm... Well under the console perhaps better to printenv | less (blush). That fills more than a screen! The point is DISPLAY has a value in graphics mode and does NOT exist in console So if [ $DISPLAY ] then # DISPLAY is set and is a nonzero length string so do GUI stuff elif [ -z $DISPLAY ] # do the console stuf here (length of string is zero) else # do any other stuff here fi Civileme
Re: [expert] Bash Scripting
Necrotica wrote: > > I'm interested in learning a little about bash scripting. In particular, I'd > like to learn how to tell if the user logging on is logging in under X or from > a command prompt. Can anyone point me in the direction of where I can find this > information? Thanks... > > -Chris printenv from Konsole when you are logged in in graphics mode printenv from a console (logged in on runlevel 3) Check the settings of the DISPLAY variable Civileme
[expert] Bash Scripting
I'm interested in learning a little about bash scripting. In particular, I'd like to learn how to tell if the user logging on is logging in under X or from a command prompt. Can anyone point me in the direction of where I can find this information? Thanks... -Chris
