I am writing an awk script which looks like this inside: This is my file. This tries to print the else if method once, but instead, it prints for all of the lines that do not include the searched value. If I entered 20, my output looks like this: I have to use the awk command to print else if
Tag: awk
Remove eveything before including colon and replace comma to newline
I am looking just to accommodate my query into one-liner. I have a below file content where i want to remove all before last colon(:) including itself and then replace comma(,) into newline to get the desired results. these all comma separated values are linux user ID’s File content: Already tried: working solution: result: Is there a better trick, please
Filtering access logs by last 15 minutes with awk not working when there are no entries
I have an apache access log with the below format and I’m trying to use the awk command to filter out the requests by the last 15 minutes. It works fine when there are entries but returns everything when there are none found in the last 15 minutes. Access log format Answer It’s not possible to compare date directly in
Using awk to make changes to nth character in nth line in a file
I have written an awk command awk ‘NR==5 {sub(substr($1,14,1),(substr($1,14,1) + 1)); print “test.py”}’ > test.py This is trying to change the 14th character on the 5th line of a python file. For some reason this doesn’t stop executing and I have to break it. It also deletes the contents of the file. Sample input: ` Output: Answer Get away from
How to apply regex to specific column in awk
This will colorize the output of ls -lhAXF How to apply the color to specific type of item using regex, for example i want the folder to be green, the .config folder red, the regular file blue or something like that. folder – match $9 column containing ‘/’ in the end of word hidden folder – match $9 column containing
converting 4 digit year to 2 digit in shell script
I have file as: Here the 1st column represents dates as YYYYMMDDHH I would like to write the dates as YYMMDDHH. So the desire output is: My script: It is printing Any help please. Thank you. Answer Please don’t kill me for this simple answer, but what about this: You simply cut the first two digits by showing character 3
Awk average of column by moving difference of grouping column variable
I have a file that look like this: File is sorted by column 3. I want the average of 4th column grouped by column 3 every 0.5 unit apart. For example it should output like this: I can print all positions without average like this: But I am not able to figure out how to print average of 4th column.
How to replace newlines between brackets
I have log file similar to this format Here is the echo command to produce that output $ echo -e “test {nseq-cont {nttt0,nttt67,nttt266nttt},nttgrp-id 505nt}n}ntest{nttest1{nttvalnt}n}n” Question is how to remove all whitespace between seq-cont { and the next } that may be multiple in the file. I want the output to be like this. Preferably use sed to produce the output.
print specific field from specific line of csv file linux
I am trying to extract a specific field in a specific line of a CSV file. I’m able to do it according to the number row but sometimes the row number of the file will change so this is not that flexible. I wanted to try and do it to extract a specific name before the field I’m interested in.
Combine multiple files csv into one using awk
I want to combine two .csv files based on the unique id that exists in both files. First file consist of 17 columns and the second one in 2 columns where in both files the first column is the same unique id. In the to be created file 3 i would like 18 columns. I have been trying paste paste