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What RavinderSingh says is correct for regexes, but not for shell globbing. * itself is the wildcard character, matching any sequence of any chars.

So - *[^a]* will match anything followed by a...

Hello carloszhang,

While using * the preceding item will be matched zero or more times. Here you haven't mentioend any thing in last expressions so it is taking everything as a match. Also * to...

So try:

$ cat tmp.pl
#!/usr/local/bin/perl
use strict;
use warnings;
use DBI;
use CGI;

system("tmp.sh", "");

carloszhang,

Check it out : First n line and last n lines :

n=3;f=file;(head -n $n $f;tail -n $n $f)

with Classic ed:
n=3;printf "1,${n}p\n\$-${n}+,\$p\nq\n"|ed -s file

Just for the fun of it, using awk
awk '{a[NR]=$0} NR<=n {print} END{for (i=NR-n+1;i<=NR;i++) print a[i]}' n=3 file
This prints the first 3 and last 3 records of file

Or if you like it to the...

I know that you asked for a sed solution, but ed is well suited to something like this and only needs to be invoked once to do it. This was tested using a Korn shell, but will work with any shell...

Is your need to print them at one go using a single sed command ?

Can't you go with something like this ?


sed '3q' filename && sed -e :a -e '$q;N;4,$D;ba' filename

Check "selective printing of certain lines" section on this page

http://www.catonmat.net/blog/wp-content/uploads/2008/09/sed1line.txt

od -c after your perl command. perl -e blah...blah... | od -c

I cannot reproduce this behavior:

bash-2.03$ perl -e '@f=("aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa","1","911");...

Hi.

On my system:
OS, ker|rel, machine: Linux, 2.6.26-2-amd64, x86_64
Distribution : Debian GNU/Linux 5.0.8 (lenny)


The command zgrep is a shell script:
% file /bin/zgrep...

this very strange i don't what is reason for this....


zcat 1.zip
abc in csv
zcat: 1.zip has more than one entry--rest ignored
$

I believe this is an implementation limit, not a bug, i.e. these numbers are large for awk.

You may use Perl though:


perl -le'
$pat111 = 0x1000000002E3E02;
$snBegin = ...

Looks like a bug in awk
awk 'BEGIN{ pat111=sprintf ("%d", 0x1000000002E3E02); snBegin=sprintf("%d",0x1000000002E3E01);print pat111;print snBegin; if (pat111<=snBegin) printf "a\n"}'
...

echo "0000000011179E0A" | sed 's/^0*//'

Convert to base 10 using bash/ksh base conversion and back to hex with printf:

$ printf "%X" $((16#0000000011179E0A))
11179E0A

Edit: Beware - After more testing of the above I found it...

$ echo 0000000011179E0A | ruby -e 'print gets.sub(/^0*/,"")'
11179E0A

$ echo 0000000011179E0A | awk '{print gensub(/^0*/,"","")}'
11179E0A

From Man Page for less (Linux Section 1) - The UNIX and Linux Forums (https://www.unix.com/man-page/Linux/1/less/)

Greetings Franklin!
I understand now, that what is wanted is feeding find command with the names supplied from a file:
So its going to use a loop...
e.g.

cat filename | while read FILETOFIND...

Try something like this, but I'm not shure if this is what you're looking for:
find <your options> | grep -f namefile

By the way, your specific issue is that you do not have quotes around your variable:


echo "$list" # now see what happens...