Search Results

This thread is a duplicate of the thread Perl removing line match with pattern in column...

Here is a slightly different approach that seems to work:
awk -F"'" '
BEGIN { OFS = ","
h[++hc] = "recordOpeningTime"
h[++hc] = "servedMSISDN"
h[++hc] = "ratingGroup"
h[++hc] =...

For highest precision do the mult before the div
$2*100/1024/1024/9787
or force to floating-point
$2/1024.0/1024/9787*100

printf("%.2f\n", $3/1024/.....)

---------- Post updated at 10:07 PM ---------- Previous update was at 10:05 PM ----------






No bc, no sed - just awk

Hello justbow,

Following may help you, it will work only if we have not more than 2 columns and if this has one column it has the AA-HOST-04 etc values in it.


awk 'NF==1{V=$0;next}...

awk '$1 !~ /^[0-9]*$/ {key=$0; next} {print key, $0}' file

To lose the EPC-Per=673,HYHGU as well, extend above: awk '!/EPC-IpAddress:/ && /EPC-SessionId/ {print $2}' RS= file

Try:
awk '!/EPC-IpAddress:/{print $2}' RS= file

Hello justbow,

Following may help you in same.


awk '{X[$1]+=$2;Y[$1]+=$3;} END{for(i in X){print i OFS X[i] OFS Y[i]}}' OFS="\t" Input_file


Output will be as follows.

A 2 ...

As long as you know that you're only working with numeric input, a simpler approach is:
grep '2014....23' file

$ grep -E '2014[[:digit:]]{4}23' file
2014112023
2014020123
2014020123
2014030223

grep 2014[0-9][0-9][0-9][0-9]23 file

The data you say you want in the results doesn't match the headings in the 2nd and 3rd fields of your output. If the output you really want is:
number,bidirVol1,subsGName1,bidirVol2,subsGName2...

Try something like:
awk '/Date:/{c=$2 FS $3} NR>3{print c, $1 $2, $NF}' file

Emergency UNIX and Linux Support !! Help Me! Forum (Request Urgent Help) (https://www.unix.com/emergency-unix-linux-support-help-me/)

README FIRST: How to Request Emergency or Urgent Help:


...