Print the line only if the line exists in a specific column
Need help with awk (or sed) command to select a row only if the value / String exists in a specific column.
Example:
I only want to return a row if the word "true" exists in column 5, do not return a row if "true" does not exist in column 5
input:
john,jacob,jingle,schmidt,true
jason,jack,john,true,false
oscar,meir,true,weiner,false
tiffany,amber,false,theissen,true
jack,john,sally,true,true
Desired output:
john,jacob,jingle,schmidt,true
tiffany,amber,false,theissen,true
jack,john,sally,true,true
I know how to pull back a line that has a word and doesn't have a word like this:
awk '/true/ && !/false/'
but that won't work and I know how to delete a word only if it's in a specific column
awk 'BEGIN{FS=OFS=","} {sub(/true */, "", $5)} 1'
but I don't know how to combine the two.
My best guess:
awk 'BEGIN{FS=OFS=","} {"/true/",$5}
worst case scenario .... :)
awk 'BEGIN{FS=OFS=","} {sub(/true */, "Unmatchablestring", $5)} 1' | awk '/Unmatchablestring/'
You can use:
awk -F, '$5 == "true"' file
Your question was a bit vague as to whether you want to check if column 5 contains a literal value true
or if it should be an exact match. If you want to check if a column contains 5 true
, than use a match operator ~
like:
awk -F, '$5 ~ /true/' file
Otherwise follow anubhava's answer .