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While you can always fix the date format using a scripting language after you get it out of the database, you may want to explore the possibility of getting it in the correct format from the database...

A Perl alternative:

perl -pe 's/(\d{4})-(\d{2})-(\d{2})/$3\/$2\/$1/' roozo.file

Output:
SAP|23/03/2017
REEF-U|24/03/2017
FEB|12/06/2016
BEN-JEF|23/09/2017
TUJJI|30/03/2017

How about
awk 'BEGIN {FS = OFS = "|"} {split ($2, T, "-"); print $1, T[3] "/" T[2] "/" T[1]}' file
SAP|23/03/2017
REEF-U|24/03/2017
FEB|12/06/2016
BEN-JEF|23/09/2017
TUJJI|30/03/2017

When you say that the following is working from the command line:
echo "2014-09-10" | awk
-F '/'
'{print ( $1 >0 && length($1) == 4 && $2 <= 12 && $3 <= 31 ) ? "good" : "bad" }'
did you mean...

This might help you to start
awk 'function error(v){
print "invalid "v ; exit
}

BEGIN{
month="Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec"
split(month,MON,/ /)

...

Looks like you managed to print just the state of each server.
Now, try to push each state as the key in an associative array and keep incrementing the value each time you find the same state in the...

Here is my own little concoction I use for such things.
Produces output like:

=====================================
host total FIN_WAIT2 ESTABLISHED
192.168.1.115 | ...

ed -s file <<-EOF
g/Pattern to find/s//replacement string/g
w
q
EOF

There is always the straightforward ed or ex that you can use to edit a file:
ed -s file <<-EOF
g/Pattern to delete/d
w
q
EOF

Leave off the -s option if you want to...

Hi,
If you have perl:
perl -i -n -e 'if(!/Pattern to delete/) {print}' file
Regards.

OR

Try

if second file size is too big then you may have to face memory problem

$ awk 'FNR==NR{A[$1+0]; next}{if($3+0 in A){$5=replace;print}else{last[++i]=$0}}END{for(j=1;j<=i;j++)print...

Yes., this will avoid memory problem.

Try :

$ awk 'BEGIN{ print "Enter input "; getline input < "-"; print input >"output_file" }'
Enter input
Roozo

$ cat output_file
Roozo

By adding0 awk treats array index$1 as numeric value, if you do not add 0 then string

Example :

$ awk 'FNR==NR{A[$1]; next}END{for(i in A)print i }' input
55
066

$ awk 'FNR==NR{A[$1+0];...

Hi, try this fix:
awk 'FNR==NR{A[$1+0]; next} $3+0 in A{$5=replace}1' replace=77 FS=, OFS=, input.txt file

You are appending in every iteration below one does your job


Try :

$ awk -F, 'FNR==NR{A[$1]++;next}A[$3]{$5=replace}1' replace=77 OFS=\, input.txt file >output.txt

$ cat output.txt...

It is more efficient to use $3 in A instead of A[$3]:
awk 'FNR==NR{A[$1]; next} $3 in A{$5=replace}1' replace=77 FS=, OFS=, input.txt file

Then there is no need for A[$1]++ which saves an...

for input posted in #10

$ cat input1.txt
aster;PAGE;1233;4;1222

$ cat input2.txt
1111 A
1234 B
9999 B
2355 C

$ cat input3.txt
1111 2
1234 1
9999 3
2355 4

$ cat config.txt
A...

wow this is just getting more and more complex! This was fun. Would be more understandable if I knew what I was working with and could name the variables something..


mute@thedoctor:~/temp/roozo$...

mute@thedoctor:~/temp/roozo$ ./script
aster;PAGE;1234;4;KART
aster;PAGE;2355;4;HOME
aster;PAGE;1234;4;KAR2
aster;PAGE;1234;4;MART
aster;PAGE;1111;4;DART



#!/bin/sh
awk '
...

Are you trying to join three files?

If yes, something like this might work for the input that you posted:
awk -F'[; ]' '
NR == FNR {
i = $2
next
...

For input in post #6 and order is important then you may try

$ cat config.txt
A DART
B MART
B KART
B KAR2
C HOME

$ cat input2.txt
1111 A
1234 B
2355 C

$ cat input1.txt ...

If your ksh is ksh93 you can also use the builtin date formatting capability:

d=$(printf "%(%Y%m%d%H%M%S)T" "30 day ago")


To check for ksh93 just use

print $KSH_VERSION


If the output...

awk 'NR==FNR{A[$1];next}!($1 in A){print $1}' file2 file1

Hello,


Could you please try this script.
Where find_replace.ksh is the script name. Also lets say we have all entries in the file named test_file121.




$ cat find_replace.ksh
...