SED replace specyific line with variable

zwierzq · 01-23-2010, 11:50 AM

Hi!

I have a little problem. i have file like this..

1 (130.216.1.125)
2 (210.7.32.2)
3 (210.7.36.67)
4 (210.7.47.22)
5 (207.231.240.8)
6 (64.57.28.27)
7 (64.57.28.25)
8 (64.57.28.37)
9 (64.57.28.13)
10 *
11 (62.40.112.37)
12 (62.40.112.5)
13 (62.40.124.2)
14 (193.171.15.38)
15 (193.171.15.41)
16 *
17 (143.205.170.1)
18 (143.205.172.11)
1 (130.216.1.125)
2 *
3 (210.7.36.67)
4 (210.7.47.22)
5 (207.231.240.8)
6 (64.57.28.27)

and i need to replace * to this this format (0.0.0.0) where last zero is a variable
output file mus be like that

1 (130.216.1.125)
2 (210.7.32.2)
3 (210.7.36.67)
4 (210.7.47.22)
5 (207.231.240.8)
6 (64.57.28.27)
7 (64.57.28.25)
8 (64.57.28.37)
9 (64.57.28.13)
10 (0.0.0.1)
11 (62.40.112.37)
12 (62.40.112.5)
13 (62.40.124.2)
14 (193.171.15.38)
15 (193.171.15.41)
16 (0.0.0.2)
17 (143.205.170.1)
18 (143.205.172.11)
1 (130.216.1.125)
2 (0.0.0.3)
3 (210.7.36.67)
4 (210.7.47.22)
5 (207.231.240.8)
6 (64.57.28.27)

i have created bash scrip like this:

amount_of_lines=`wc -l < zlit`

for((x=1;x<$amount_of_lines+1;x++))
do
sed -e "$x s/*/(0.0.0.$x)/g" zlit
done

but it does not work i think that he problem is in $x but i can't find the way to mask $x

jschiwal · 01-23-2010, 12:27 PM

One problem is that you need to escape the asterisk character. The other is that sed will read each line and all of the asterisks will be replaced the first time through the loop.

It would be better to use awk for this problem. You can initialize a variable in the BEGIN block and increment it if you have a match.
If you want the 4th octet to equal the line number in the zlit file, you can use the NR awk variable which tracks the record number.

awk '/\*/{ printf "%d (0.0.0.%s)\n",$1,NR };!/\*/{print}' zlt

zwierzq · 01-23-2010, 12:46 PM

Quote:

Originally Posted by jschiwal

One problem is that you need to escape the asterisk character. The other is that sed will read each line and all of the asterisks will be replaced the first time through the loop.

Yeach i know so i want to insert a variable
if i do like this: sed -e "1s/*/(0.0.0.$x)/g" zlit
it changes none because first line of file is address
if i do like this sed -e "10s/*/(0.0.0.$x)/g" zlit
it chnges * to (0.0.0.10) in 10th line.
So i want to insert at 1 and 10 position a vriable from loop.

I am a beginer so i do not know AWK well. It is too complicated to me at this level

colucix · 01-23-2010, 01:08 PM

The problem is what jschiwal already pointed out: sed will parse the entire file at every call and if you put the sed command inside a loop, you will get the entire file duplicated multiple times (with some substitution made at some point). The problem is how to increment a numeric variable in sed and use it as the replacement string in the substitution. Since there is not an easy way to do that in sed, the suggested awk code is more handy.

Anyway, to stick with the shell, you can do something like this:

Code:

#!/bin/bash
while read line
do
  if [[ $line =~ '*' ]]
  then
    ((count++))
    echo ${line/\*/(0.0.0.$count)}
  else
    echo $line
  fi
done < zlit

The if statement check if the line matches a literal asterisk and use substring replacement accordingly. Note that the =~ operator is available from bash version 3 and you may have compatibility issues if you use the code with an older version.

jschiwal · 01-23-2010, 09:23 PM

To the general question about inserting variables, you could do something like:
sed '...'"$variable"'...' filename

The variable is evaluated by the bash shell before the sed command is executed.

license=GPL
set sed -n '/'$license'/p' README.txt
> echo $*
sed -n /GPL/p README.txt

Sed is usually used by putting it's command in single quotes. There are fewer bash meta-characters you would need to escape.

You would have to escape the \$ in regular expressions otherwise. It precedes a bash variable. In a regular expression it is a meta-character representing the end of a line.