How to get consolidated count of delimiter occurrence

user_8757495 0
Viewed 97

I have a requirement to fetch the count the occurrence of '|' in each line of a file then match the count with given inputcount, needs to throw exception when the count is wrong.

Say if the inputcount=3 and the file has following content

s01|test|aaa|hh
S02|test|bbb
so3|test|ccc|oo

then exception should get thrown on executing the line 2 and it should exit the file.

Tried below Awk command to fetch the count for each lines, but I was not sure how to compare and throw the exception, when it not matches

awk ' {print (split($0,a,"\|")-1) }' test.dat

Can anyone please help me with it?

awk shell linux python
4 Answers

You may use this awk:

awk -v inputcount=3 -F '\\|' 'NF && NF != inputcount+1 {exit 1}' file &&
 echo "good" || echo "bad"

Details:

  • -F '\\|' sets | as input field separator
  • NF != inputcount+1 will return true if any line doesn't have inputcount pipe delimiters.
user_5482250 
$ inputcount=3
$ awk -v c="$inputcount" 'gsub(/\|/,"&")  !=  c{exit 1}' file
$ echo $?
1
user_17450010

Maybe try

awk -F '[|]' -v cols="$inputcount" 'NF != cols+1 {
    print FILENAME ":" FNR ":" $0 >"/dev/stderr"; exit 1 }' test.dat

The -F argument says to split on this delimiter; the number of resulting fields NF will be one more than there are delimiters, so we scream and die when that number is wrong.

user_8741880

As you also tagged the post with python I will write a python answer that could be a simple script.

The core is:

with open(filename) as f:
    for n, line in enumerate(f):
        if line.count("|") != 3:
            print(f"Not valid file at line {n + 1}")

Than you can add some boilerplate:

import fileinput
import sys

with fileinput.input() as f:
    for n, line in enumerate(f):
        if line.count("|") != 3:
            print(f"Not valid file at line {n + 1}")
            sys.exit(1)

And with fileinput you can accept almost any sort of input: see Piping to a python script from cmd shell

user_12220260
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