Shell Programming and Scripting

Hi all,

I have a file with multiple lines. I want to replace characters 7 through 14 of every line with 0000000

Input:
12345678901234567890
23456789012345678901

Output
12345600000004567890
23456700000005678901

Please help.

JaK

awk '{print substr($0,1,6)"0000000"substr($0,14,10)}' infile

Thanks rikxik, it works but when i run the same command on record which has lenght of 6656 bytes .. it says "record too long" .. is there some limitation on awk? How do i get around it?

Please help
JaK

in bash

Code:

# while read line; do echo ${line//${line:7:7}/0000000}; done < file

or

Code:

# for line in `cat file`; do echo ${line//${line:7:7}/0000000}; done

sed 's/\(.\{7\}\).\{7\}/\10000000/' oldfile > newfile

A lot faster than calling substr() two times in awk.

If your lines are too long to handle you could cut them before making the exchange and then put them together again:

Code:

typeset line=""
typeset start=""
typeset end=""

cat oldfile | while read line ; do
     start="(print - "$line" | cut -c1-14)"
     end="$(print - "$line" | cut -c15-)"
     start="${start%%???????}0000000"
     print - "${start}${end}" >> newfile
done

bakunin

Thanks guys. To get around "too long" issue, I just used nawk instead and it worked.

Great to see people helping others!
JaK

Consider this:


@bakunin

sed definitely takes the cake (as long as no too long line issues):

But wrongly produces lines like this - nothing which cannot be fixed:
Instead of this:
However, speaking of speed, the bash script you posted is as fast as a Snail:

And it produced incorrect output too.

@ghostdog74
This wasn't very fast either:



Mine wasn't too bad - not as fast as the sed version:
Nothing against anyone - just fair comparison.

HTH