Awk Find And Replace Fields Values

I have a data as follows :

foo bar 12,300.50
foo bar 2,300.50
abc xyz 1,22,300.50

How do I replace all , from 3rd field using awk and pass output to bc -l in the following format to get sum of all numbers:

You can use gsub() funcion as follows. The syntax is:

gsub("find", "replace")
gsub("find-regex", "replace")
gsub("find-regex", "replace", t)
gsub(r, s [, t])

From the awk man page:

For each substring matching the regular expression r in the string t, substitute the string s, and return the number of substitutions. If t is not supplied, use $0. An & in the replacement text is replaced with the text that was actually matched. Use & to get a literal &.

You can also use the following syntax:

gensub(r, s, h [, t])

From the awk man page:

Search the target string t for matches of the regular expression r. If h is a string beginning with g or G, then replace all matches of r with s. Otherwise, h is a number indicating which match of r to replace. If t is not supplied, $0 is used instead. Within the replacement text s, the sequence n, where n is a digit from 1 to 9, may be used to indicate just the text that matched the n’th parenthesized subexpression. The sequence represents the entire matched text, as does the character &. Unlike sub() and gsub(), the modified string is returned as the result of the function, and the original target string is not changed.


Create a data file cat /tmp/data.txt

foo 	bar	12,300.50
foo	bar	2,300.50
abc	xyz	1,22,300.50

Type the following awk command:

awk '{ gsub(",","",$3); print $3 }' /tmp/data.txt

Sample outputs:


You can pass the output to any command or calculate sum of the fields:

awk 'BEGIN{ sum=0} { gsub(",","",$3); sum += $3 } END{ printf "%.2fn", sum}' /tmp/data.txt

OR build the list and pass to the bc -l:

awk '{ x=gensub(",","","G",$3); printf x "+" } END{ print "0" }' /tmp/data.txt   | bc -l

Posted by: SXI ADMIN

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