[SOLVED] Delete lines in a file by piping search pattern into Sed

Issue

Is there anyway we can delete lines in a file like below.

grep {SearchPattern} {file} | cut -c 1-9 | Sed {file}

Whatever I am searching can actually occur fewer than lines I want to delete. I want to actually search and take first 9 char and search again and delete whatever lines come in output.

my input file would like as below .

0002206993022      Enrollment Status             Terminated                              08/01/201412/31/9999 
0003119343022      Enrollment Status             Terminated                              05/28/201512/31/9999
0003119343009999   Pay Status                    N/A                                     09/10/201405/28/2015

you can I would to search for all "Terminated" persons and take their IDs by taking first 9 digits in the first column. I would like to delete all of their records be it Pay Status or Enrollment Status or whatever

Solution

You want something like this:

awk '/Terminated/{$0=substr($0, 1, 9);print "^" $0}' file | grep -vf - file

Update:

Try (Non GNU grep):

awk '/Terminated/{$0=substr($0, 1, 9);print "^" $0}' file | grep -vf /dev/stdin file

Test:

$ cat file
0002206993022      Enrollment Status             Terminated                              08/01/201412/31/9999 
0003119343022      Enrollment Status             Terminated                              05/28/201512/31/9999
0003119343009999   Pay Status                    N/A                                     09/10/201405/28/2015
1003119343009999   Pay Status                    N/A                                     09/10/201405/28/2015

$ awk '/Terminated/{$0=substr($0, 1, 9);print "^" $0}' file | grep -vf - file
1003119343009999   Pay Status                    N/A                                     09/10/201405/28/2015

Answered By - sat