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Slight change to Yoda's solution to replace ID within file.

Note this should also work find with /bin/bash (if you don't have ksh installed).

#!/bin/ksh

file="file_000"

while read id...

Try:awk 'NR==FNR{a[NR]=$0;n=NR;next}{for (i=1;i<=n;i++){x=a[i];gsub("000",$0,x);print x >> "file_"$0}}' file_000 file_id

#!/bin/ksh

file="file_000"

while read id
do
cp "$file" "${file%_*}_$id"
done < file_id

See Advanced Bash Scripting Guide -- String Operations (http://tldp.org/LDP/abs/html/refcards.html#AEN22664)

It matches the longest match of "*abc2" from the start of the string, and replaces it...

From man expr:

index STRING CHARS
index in STRING where any CHARS is found, or 0

I don't think expr can do that.

You can do that with replacement and some arithmetic:
...

That's a little tricky. I would suggest the following as the best simple solution:
$ var999=abc$'\n'def$'\n'defabc$'\n'def$'\n'def
$ echo -n "$var999" | wc -l
4

Reports the right number of...

The echo command writes its argument to standard output, followed by a <newline>. If there are no arguments, only the <newline> is written.
$ echo | od -c
0000000 \n
0000001

This is why you...

The variable does not have a newline.

The echo command adds the newline, which "wc -m" counts. That's why you get a different count with "echo -n", which does not add a newline.

Yes, you can use the second field alone, using cut. Here is how I would do it, which seems simple and clear.
n=0
cut -f 2 myfile > fld2.txt
while read fld_2; do
array_var[n]=$fld_2
n=`expr...

Don't know about other awk implementations but gawk does have these egrep like regexp characters. From man gawk:
\y matches the empty string at either the beginning or the end
...

In perl, the word anchor is \b

For any questions regarding special perl variables, the perlvar (http://perldoc.perl.org/perlvar.html) man page is the definitive source.

Regards,
Alister

Try $.

You can also do something like this:
awk '{A[$1,$2]=$3; I[$1]; J[$2]} END{ for(i in I)for(j in J) print i, j, A[i,j]}' file
or for example:
awk '{A[$1,$2]=$3; J[$2]} END{ for(j in J) print 5, j,...

awk does not actually have two-dimensional arrays; it handles two indexes by squeezing them into a single string. As such, your first loop will work but may look weird; the SUBSEP character which...

With perl you can specify a backup extension after the -i switch:
perl -i.bak 's/:\S*//g' file #will save the original as file.bakGNU sed can also edit in-place, here is sed solution:
sed -i 's/:[^...

Yes, perl -i ... does it.
Here perl seems to be preferable anyway.
perl -pe 's/:\S*//g' file

There is no default print if the main loop consists of only {}
You must add a print command. Lazy people do a 1, that's a true condition where the default action is again a print.
For demonstration...

Another method:
awk '{ gsub(/:[^ ]*/,x); $1=$1 }1' input_file > input_file.tmp

mv input_file.tmp input_file

Awk is a filter, it doesn't modify the input. You should redirect into a temporary file, then overwrite the original.
How about:

cp $input ${input}.copy
awk -F: '{print $1}' RS="[ \t]+" ORS=" "...

You cannot embed shell variables in an awk statement that way.
Generally

var=foo
awk -v myvar=$var '{.....}' somefile

-v [awkvariable name]=$variable

It looks like you are trying to get a...

You have to assign an external variable to awk variable to start working on it.

No need to use /dev/null , you can print inside BEGIN block.

Also starting character in substr function is...

That is because when you used $d you were actually assigning NULL to your array: myarray for 3 consecutive times.

This is the reason why it was printing NULL followed by number of occurrence: 3....

Please explain the logic that goes in to get the expected output. Merely looking at your input and output, I could suggest this:
ls -1 | awk 'BEGIN { i=1 } {print i, $0; i++}'

awk 'BEGIN{while("ls"|getline d) myarray[++j]=d; close("ls")}END{ for (i in myarray){print i, myarray[i]}}' /dev/null