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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){
...

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 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 ...

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 ...

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...

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+ +)[^...

Hi,

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

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...

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

$ 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 ...

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

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/([^ ]+...

to swap columns 4 - adjust as needed.

nawk -v c=4 'BEGIN{ARGV[ARGC++]=ARGV[1]} FNR==NR{a[FNR]=$c;next}{$c=(FNR%2)?a[FNR+1]:a[FNR-1]}1' myFile

nawk 'BEGIN{ARGV[ARGC++]=ARGV[1]} FNR==NR{a[FNR]=$(NF-1);next}{$(NF-1)=(FNR%2)?a[FNR+1]:a[FNR-1]}1' myFile

awk -F'x' '{a=$2;printf FS $2 FS;getline;printf $3"\n" FS $3 FS a"\n";}' mydata.txt

Try: awk 'NR%2{x=$2}!NR%2{$2=x;print;$2=$4;$4=x;print}' mydata.txt

perl -anse 'BEGIN{$c--};if ($.%2){$x=$_;$y=$F[$c]};if ($.%2==0) {$x=~s/(([^\s]+\s+){$c})\w+/\1$F[$c]/;print $x;s/(([^\s]+\s+){$c})\w+/\1$y/;print}' -- -c=5 mydata.txt
c=5 means 5th column.

You can also try this: perl -ane 'if ($.%2){$x=$_;$y=$F[2]};if ($.%2==0) {$x=~s/(([^\s]+\s+){4})\w+/\1$F[4]/;print $x;s/(([^\s]+\s+){4})\w+/\1$y/;print}' mydata.txt

assuming all the fields are of the constant width....

nawk -v c=5 '
BEGIN {
ARGV[ARGC++]=ARGV[1]
}
FNR==NR && NR==1 {
copy0=$0
for(i=1;i<=NF;i++){
match(copy0, FS FS "*")
...

I don't follow your sample.
Given your latest sample and

nawk -v c=5 'BEGIN{ARGV[ARGC++]=ARGV[1]} FNR==NR{a[FNR]=$c;next}{$c=(FNR%2)?a[FNR+1]:a[FNR-1]}1' yourLatestSample

produces:

x 20 x y...