How to use AWK to print line with highest number? - linux

Question

I have a question. Assuming I dump a file and do a grep for foo and comes out the result like this:
Foo-bar-120:'foo name 1'
Foo-bar-130:'foo name 2'
Foo-bar-1222:'foo name 3'
Etc.
All I want is trying to extract the foo name with largest number. For instance in this case, largest number is 1222 and the result I expect is foo name 3
Is there a easy way using awk and sed to achieve this? Rather than pull the number out line by line and loop through to find the largest number?

Answer 1

Code for awk:
awk -F[-:] '$3>a {a=$3; b=$4} END {print b}' file
$ cat file
Foo-bar-120:'foo name 1'
Foo-bar-130:'foo name 2'
Foo-bar-1222:'foo name 3'
$ awk -F[-:] '$3>a {a=$3; b=$4} END {print b}' file
'foo name 3'

Answer 2

Here's how I would do it. I just tested this in Cygwin. Hopefully it works under Linux as well. Put this into a file, such as mycommand:
#!/usr/bin/awk -f
BEGIN {
FS="-";
max = 0;
maxString = "";
}
{
num = $3 + 0; # convert string to int
if (num > max) {
max = num;
split($3, arr, "'");
maxString = arr[2];
}
}
END {
print maxString;
}
Then make the file executable (chmod 755 mycommand). Now you can pipe whatever you want through it by typing, for example, cat somefile | ./mycommand.

Answer 3

Assuming the line format is as shown with 2 hyphens before "the number":
cut -d- -f3- | sort -rn | sed '1{s/^[0-9]\+://; q}'

Answer 4

is this ok for you?
awk -F'[:-]' '{n=$(NF-1);if(n>m){v=$NF;m=n}}END{print v}'
with your data:
kent$ echo "Foo-bar-120:’foo name 1’
Foo-bar-130:’foo name 2’
Foo-bar-1222:’foo name 3’"|awk -F'[:-]' '{n=$(NF-1);if(n>m){v=$NF;m=n}}END{print v}'
’foo name 3’
P.S. I like the Field separator [:-]

Answer 5

$ awk '{gsub(/.*:.|.$/,"")} (NR==1)||($NF>max){max=$NF; val=$0} END{print val}' file
foo name 3

Answer 6

You don't need to use grep. you can use awk directly on your file as:
awk -F"[-:]" '/Foo/ && $3>prev{val=$NF;prev=$3}END{print val}' file