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what is the relation b/w the input and output ?

I implemented line swapping as well, check if it is working properly: #!/usr/bin/perl
open I, "$ARGV[0]";
@x=<I>;
$i=0;
for (@x){
s/ +$//;
$t=$_;
@s=split / +/;
if ($#s==12){
...

Try this one: #!/usr/bin/perl
open I, "$ARGV[0]";
@x=<I>;
$i=0;
for (@x){
s/ +$//;
$t=$_;
@s=split / +/;
if ($#s==12){
$i=1;
s/([^ ]+ +EUR)1([SB]\d+ +)[^ ]+( +)1/$1$2 ...

Try this then: #!/usr/bin/perl
open I, "$ARGV[0]";
@x=<I>;
$i=0;
for (@x){
s/ +$//;
$t=$_;
@s=split / +/;
if ($#s==12){
$i=1;
s/([^ ]+ +EUR)1([SB]\d+ +)[^ ]+( +)1/$1$2 ...

No need to apologize ;) Try: #!/usr/bin/perl
open I, "$ARGV[0]";
@x=<I>;
$i=0;
for (@x){
s/ +$//;
$t=$_;
@s=split / +/;
if ($#s==11){
$i=1;
s/([^ ]+ +EUR)1([SB]\d+ +)[^ ]+(...

Try: #!/usr/bin/perl
open I, "$ARGV[0]";
@x=<I>;
chomp($d=`date +%Y%m%d`);
$i=0;
for (@x){
s/ +$//;
$t=$_;
@s=split / +/;
if ($#s==10){
$i=1;
s/([^ ]+ +EUR)1([SB]\d+ +)[^...

Are you sure "30416938966" is Unix time? Current Unix timestamp is "1307530654", so 23 times less than your number. Also using simple Perl converter doesn't seem to recognize your number as Unix...

30416938966 is not an epoch time, at least not this millennium it's not.

to convert from epoch to your format try

perl -e'
@time=localtime($epoch) ;
printf...

Hi,

Can you post the the code which gives the output as shown above.

EUR1B892 and EURB1892 are quite different... Try this:#!/usr/bin/perl
open I, "$ARGV[0]";
@x=<I>;
$i=0;
for (@x){
s/ +$//;
$t=$_;
@s=split / +/;
if ($#s==10){
$i=1;
s/([^ ]+...

[root@linux ~]# ./a.pl c
H: EURS891 10440000000000.003000000450 FR0000130007 2ALUp 20110607-08:34:25TRADETES PA TRADETES TESTCLRXX #
H: EURB891 ...

$ nawk '{if(NF>10) {print $0"-"NR"\n"$0} else {print $0}}' input.txt|nawk 'NF>11 {$2="EUR"$3;$3="";$NF=""}{print |"sort -n -k3 -r"}'


Thanks
Sha

Try this script: #!/usr/bin/perl
open I, "$ARGV[0]";
@x=<I>;
$i=0;
for (@x){
s/ +$//;
$t=$_;
@s=split / +/;
if ($#s==10){
$i=1;
s/([^ ]+ +EUR[SB])1(\d+ +)[^ ]+( +)1/$1$2 ...

Ok, so if three consecutive lines have 11 columns, then three new lines should be inserted above them? Example.. input:H: EURS1892 S892 10440000000000.003000000450 FR0000130007 2ALUp ...

Can the output look like this? H: EURS890 00440000000069.110100000963 DE0008032004 2CBKd 20110607-07:50:59BNABFRPP DE BNABFRPP PARBFRPP #
H: EURB890 ...

Try: perl -0ne '$x=$_;s/^([^ ]+ +AAA)1([^ ]+)( +)[^ ]+( +)1([^ ]+ +[^ ]+$)/$1$2$3 ${4}0$5/mg;print $_,$x' file

Try: perl -0ne '$x=$_;s/^([^ ]+ +AAA)1([^ ]+)( +)[^ ]+( +)1/$1$2$3 ${4}0/mg;print $_,$x' file

let me know the time taken, once the script is finished

use echo -e

Try this..


echo "hour\tminutes\tseconds\tmiliseconds" > output.txt
for hour in `seq -w 00 23`
do
for minutes in `seq -w 00 59`
do
for seconds in `seq -w 00 59`
do
for...

is that you looking for ?

i dont know how much time it will take to complete


echo "hour\tminutes\tseconds\tmiliseconds" > output.txt
for hour in `seq -w 00 23`
do
for minutes in `seq -w...

perl -0pe 's/(.*\n)(.*\n)/\2\1/g' file

awk 'getline x{print x}1' infile

sed -n 'h; n; G; p'

If you'd prefer to keep the last line when there are an odd number of lines:
sed -n '$p; h; n; G; p'

Regards,
Alister

One way would be..
awk '{s=$0;getline;print $0"\n" s}' inputfile > outfile