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regexgrep
Rebecca Abriam
Rebecca Abriam

Reputation: 560

Regex to find string that exactly has one pipe character

I'm new in using regex, hope someone can help me. I'm using the regex below to grep a csv file for string that exactly has one pipe character (i.e. |)

grep "^([^\\|]+\\|){1}[^\\|]+$" myfile.csv

Unfortunately, the above yields no result when used with grep. Any ideas?

A sample csv file content is as below, where I expect the 2nd line to be found.

"foo"|"foo"|"foo"

"bar"|"bar"

Solutions to this question:

grep -E "^([^|]+\|){1}[^|]+$" myfile.csv

and 

egrep "^[^|]+\|[^|]+$" myfile.csv

Upvotes: 2

Views: 2477

Answers (5)

jscs
jscs

Reputation: 64002

Grep and regexes are the wrong tool for this task. Use something that is intended for counting:

# Use a split function with the pipe as delimiter
awk 'split($0, _, "|") == 2 {print}' the_file

# Set awk's field separator to the pipe character
# and check the number of fields on each line
awk -F'|' 'NF == 2 {print}' the_file

Upvotes: 0

Rebecca Abriam
Rebecca Abriam

Reputation: 560

Here are the solutions to my question. Thanks to the comments that led me to solving this.

grep -E "^([^|]+\|){1}[^|]+$" myfile.csv

and

egrep "^[^|]+\|[^|]+$" myfile.csv

Upvotes: 0

Jotne
Jotne

Reputation: 41446

Solution using awk

awk 'gsub(/\|/,"|")==1' file

gsub(/\|/,"|") this counts number of | replaced, if this equal 1, then do default action, print $0

Edit:Another awk:

awk 'split($0,a,"|")==2' file

Count how many parts text is dived into by |, if 2 print.

Upvotes: 1

arshajii
arshajii

Reputation: 129497

You can try:

^[^|]*\|[^|]*$

You don't need to escape | in a character class. Also you presumably want * instead of + here to allow for strings like |abc, xyz|, and just | on its own.

Upvotes: 5

plalx
plalx

Reputation: 43718

Try the following:

^[^|]+\|[^|]+$

Upvotes: 1

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