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Print the first word based on the last 5 words from the file

I have a file with the content below and I want to print the first word only if it starts with P

and if the last 5 characters12

File:

papa;1;2;3;4;5;12;12;12;12;12
Peter;12;12;12;12;12
alpha;2;2;2;4;5;6;7;8;9;3;3;3;3;3
Prod;1;2;3;4;1;1;1;1;12;12;12;12;12

Expected Result:

Peter  
Prod

My attempt:

cat filename | awk -F '[;]' '{print substr( $0,length($0)-5,length($0))}' 
cat filename | egrep -V 'p|a' | awk -F '[;]' '{print $1}'

But I couldn't join the exit.

Update: -

From the comments in the accepted answer, obviously the OP is saying that there may also be lines of less than 5 fields in the final file.

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3 answers


Rough way to do it in Awk

just for the first word

awk -F';' '$1 ~ /^P/ && NF > 5 {c=0; for(i=NF-4;i<=NF;i++) {if($i==12){c++}} if(c==5) print $1} ' file
Peter
Prod

and if the whole line is needed, remove $1

in print

statement



awk -F';' '$1 ~ /^P/ && NF > 5 {c=0; for(i=NF-4;i<=NF;i++) {if($i==12){c++}} if(c==5) print} ' file
Peter;12;12;12;12;12
Prod;1;2;3;4;1;1;1;1;12;12;12;12;12

How it works?

  • The part $1 ~ "^P"

    is a regular match for filtering lines starting withP

  • On the lines starting with above, parse the last 5 fields (loop with NF

    and maintaining a counter and print the line if 5 instances are found 12

    .
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Here's another one with awk 

$ cat ip.txt 
papa;1;2;3;4;5;12;12;12;12;12
Peter;12;12;12;12;12
alpha;2;2;2;4;5;6;7;8;9;3;3;3;3;3
Prod;1;2;3;4;1;1;1;1;12;12;12;12;12
Post;1;3;12;12;12

$ awk -F';' '/^P.*(;12){5}$/{print $1}' ip.txt 
Peter
Prod


  • /^P.*(;12){5}$/

    a line starting with P

    ends with;12;12;12;12;12

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another approach of recursive checks, with everything in variables. Let me know if this helps you.

awk -F";" -v num=12 -v count=5 -v cha="P" 'function check(count){if($(NF-(--count))==num && count){check(count)};if(!count && substr($0,1,1)==cha){print $1}} check(count)'  Input_file

Will soon add more than one liner form and explanation of the above code.

EDIT1: Adding more than one linear form of the solution too.

awk -F";" -v num=12 -v count=5 -v cha="P" '
          function check(count){
          if($(NF-(--count))==num && count){
                check(count)
                                           };
          if(!count && substr($0,1,1)==cha){
          print $1
                                           }
                               }
         check(count)
                                          '   Input_file
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