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wait? Will not help.
The problem is that ssh sucks from the input, here file1.
The simple solution is
ssh -n ... or </dev/null ssh ...

You forgot to reset j
j=1
while [[ $j -lt 3 ]]

j remains at value 3 once it reaches it. Reset to one just before the j loop.

With a classic for loop:
for f in *; do cp "$f" /other_dir/ABC_"$f"_`date +%m%d%y`; done

If you are looking for files that belong to a userID or groupID that is not current in your system try -nouser or -nogroup

For example where userID 57484 does not exist in the system:
$ chown...

Does ^K appear to be a single character on the screen (that is, when you move the cursor over it (in vi), does the cursor move only to the ^ or also to the K)?

catting the file will typically not...

That ^K in fact IS invisible when printed with normal methods.. You have to make it appear by special options like the -v to cat, or use commands built for that, like od.

Hi.

This looks like it does what you wish:
$ cat z8; tr '_-' '-_' < z8
abc-def_hij-klm_nop-qrs_tuv-wxy_z
abc_def-hij_klm-nop_qrs-tuv_wxy-z
and
$ echo "$s" ; echo "$s" | tr '_-' '-_'...

Try also
awk 'sub ("^001.","&##") {sub (/##../, " ")} 1' fileYou may want to choose a better placeholder pattern / value.

Setting FS and OFS to an empty string to treat character positions as fields is a GNU extension that is not supported by nawk, /usr/xpg4/bin/awk, nor required by the POSIX standards. Just using...

different tack:

nawk '/^001 / {$0="001 " substr($0, index($0," ")+3)}1' myFile
or try this - Solaris' awk/nawk are finicky...

nawk 'BEGIN {FS=SUBSEP}/^001 / {$0="001 " substr($0,...

OK, this is not a Julian date - this the cumulative number of day since the begining of the year....
But you can utilize the gre2jul conversion and substraction for Jan/1 - using the same command...

var=$(printf "%100s" " ") && echo "length of var variable = ${#var}"


The first part assigns spaces to var, the second part is just information to show it worked.

Firstly, I would add the -b flag to ignore leading blanks, just in case. You also need to beware that LDAP supported group names can contain spaces, so that throws things out of what as to what is...

The reason is the way the shell parses its input. We tend to implicitly take for granted that spaces (or other whitespace) separate parts of the shells input (the "command line"), but that is not...

a="abc"
b="def"
c="ghi"
printf "%-41s%-31s%s\n" $a $b $c

Read up on printf to see what is happening, but basically I am telling it to left-justify the strings in the fields of the given...

Try this:

$ cat infile
test one
12345678901234567890123456789012345678901234567890123456789012345678901234567890
two

$ awk '
BEGIN{
R[1]="abc"
R[42]="def"
R[73]="ghi"
...

Tried:

ssh <hostname> ls >$HOME/ssh_out.log

yet?

You have to write something that creates a file mtime in epoch seconds, then formats it usefully:

filetime()
{
perl -e '
use POSIX qw(strftime);
$mtime = (stat...

Yes.. that is the file Soham, last reads from that file /var/adm/wtmpx.

Since it does not have data of last one year. You cannot view it with last command.
It nullifies after every reboot.

If you did a little search in the forums, you would have found e.g.

https://www.unix.com/shell-programming-scripting/115495-need-year-value-users-last-login.html

How old is your oldest /var/log/wtmp* file, I doubt that this is the issue you think it is (you could test with a wtmp > 12months old of course.

last command reads from wtmp.. you sure that wtmp is since the last year?