UNIX for Dummies Questions & Answers

I am trying to find only those entries where 7018 and another number appear in the end of the line.

Code:

7018 2828 1423 2351
7018 2828 14887
2828 7018 1222
123 7018 1487

I am looking for a way to generate only the last two lines. I was trying to do just "grep '7018[\s\t][0-9]{1,5}" but it does not work. What I am looking for is a way to find those lines where there is 7018 and exactly one more number after it... Can someone suggest a better way please?

Code:

awk '$(NF-1)=="7018" && $NF+0==$NF' file

Thank You. That works perfectly. I wanted to do this task in a directory recursively on the files and note down the file in which this was found. It was possible with grep but is there a way to make awk to it? (Sorry... I am new to awk and am still learning it)

I am actually thinking of using "find ./ -name '*ext' | xargs awk". I'm still experimenting on how to print the filename along with the line that awk finds... I will get back if I get it...

I think I figured it out...

Code:

 find ./ -name "*ext" | xargs awk '$(NF-1)=="7018" && $NF+0==$NF {print FILENAME "::" $0}'

Thanks for your help...

x

with sed

Code:

sed -n '/.*[ \t]7018[ \t][0-9]\{1,5\}$/p' filename

i don't think the second check condition is required..
because if last but one filed is 7018 then it automatically implies there is one more field after that..

yes i had taken it into consideration as well, but my final solution takes care of the case where the last field the OP's data happens not to be a number but a word.