Question 112678 in Ubuntu Question posted in
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Ubuntu – Print filenames & line number with number of fields greater than 'x'

I am running Ubuntu Linux. I am in need to print filenames & line numbers containing more than 7 columns. There are several hundred thousand files.

I am able to print the number of columns per file using awk. However the output I am after is something like

file1.csv-463 which is to suggest file1.csv has more than 7 records on line 463. I am using awk command awk -F"," '{print NF}' * to print the number of fields across all files.

Please could I request help?

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2

Answers


  1. 0

    If you have GNU awk with you, try following code then. This will simply check condition if NF is greater than 7 then it will print that particular file’s file name along with line number and nextfile will take program to next Input_file which will save our time because we need not to read whole Input_file then.

    awk -F',' 'NF>7{print FILENAME,FNR;nextfile}' *.csv

    Above will print only very first match of condition to get/print all matched lines try following then:

    awk -F',' 'NF>7{print FILENAME,FNR}' *.csv
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  2. 0

    This might work for you (GNU sed):

    sed -Ens 's/S+/&/8;T;F;=;p' *.csv | paste - - -

    If there is no eighth column, break.

    Output the file name F, the line number = and print the current line p.

    Feed the output into a paste command which prints three lines as one.

    N.B. The -s option resets the line numbers for each file, without it, it will number each line for the entire input.

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