Reputation: 2789
I have a text file with comma separated values. A sample line can be something like
"Joga","Bonito",7,"Machine1","Admin"
The " seen are part of the text and are needed when this csv gets converted back to a java object.
I want to filter out some lines from this file based on some field in the csv.
The following statement doesnt work.
awk -F "," '($2== "Bonito") {print}' filename.csv
I am guessing that this has something to do with the " appearing in the text.
I saw an example like:
awk -F "\"*,\"*"
I am not sure how this works. It looks like a regex, but the use of the last * flummoxed me.
Is there a better option than the last awk statement I wrote? How does it work?
Upvotes: 2
Views: 1015
Reputation: 37464
First a bit more through test file:
$ cat file
"Joga","Bonito",7,"Machine1","Admin"
"Joga",Bonito,7,"Machine1","Admin"
Using regex ^\"?
ie. starts with or without a double quote:
$ awk -F, '$2~/^\"?Bonito\"?$/' file
"Joga","Bonito",7,"Machine1","Admin"
"Joga",Bonito,7,"Machine1","Admin"
Upvotes: 1
Reputation: 13259
Since some parameters have double quotes and other not, you can filter with a quoted parameter:
awk -F, '$2 == "\"Bonito\""' filename.csv
To filter on parameter that do not have double quote, just do:
awk -F, '$3 == 7' filename.csv
Another way is to use the double quote in the regex (the command ?
that make the double quote optional):
awk -F '"?,"?' '$2 == "Bonito"' filename.csv
But this has a drawback of also matching the following line:
"Joga",Bonito",7,"Machine1","Admin"
Upvotes: 2