bash script to extract entries from csv file whose values are greater than specified in the argument

Question

I am newbie to BASH scripting and was wondering if you could point me in the right direction. I have CSV file with the following field names separated by ","

no,name,score,city

I am trying to develop a script that will take 1 argument (in numeric format), search "score"column in csv file and extract entries greater than the given argument.

I am not allowed to use awk to achieve this, following is my draft code. I have tried different logic and approaches with cut and grep but don't get the desired results such as below. Is there any switch that I can use with grep to specify the field and apply condition ?

cut -f1,2,3,4 -d, sample.csv | grep 108

Additionally - I have prepared below script but having issues with the syntax and was wondering if anyone could help me or give me a hint. Thank you

EDITED

    #!/bin/bash
    file="sample.csv"
    arg="$1"
    while IFS=',' read -r f1 f2 f3 f4
    do
        if [ "$f3" -gt "$arg" ]
        then
             echo "$f1,$f2,$f3,$f4"
        else
            echo "No score greater than "$arg""
        fi
   done < $file

Also read the documentation for test(1) after you fix the typo. — Shawn
– Shawn, Commented Mar 1, 2020 at 7:54
A general comment about CSV: Fields can contain linefeeds (new line characters) and commas. So processing a CSV with something like IFS=',' read f1 f2 f3 f4 only works if you can ensure, that no member contains a newline and/or a comma! A counterexample for a city with comma is the german city Wetter, Ruhr. — Wiimm
– Wiimm, Commented Mar 1, 2020 at 11:02
As bash always call other tools/programs, I can provide a bash/php solution, where the php call looks like php -r 'some commands;'. — Wiimm
– Wiimm, Commented Mar 1, 2020 at 11:03
@Cyrus thank you - the site is really helpful. I have edited my script, I do get the entries greater than the argument passed on with the script. But it is also displaying all entries less than the passed argument value with the echo message. I don't want to show these values and echo the message only when there is no value greater than the argument. for example when giving pass 30 - it displays "1,john,40,NY" and also echos " No score greater than 30" for all entries less than 30. — N997
– N997, Commented Mar 1, 2020 at 16:54

Pierre François · Accepted Answer · 2020-03-01 17:10:41Z

1

Try this:

#!/bin/bash
    file="sample.csv"
    arg="$1"
    empty=1
    while IFS=',' read -r f1 f2 f3 f4
    do
        if [ "$f3" -gt "$arg" ]
        then
             echo "$f1,$f2,$f3,$f4"
             empty=0
        fi
    done < $file
    if [ "$empty" -eq 1 ]
    then
            echo "No score greater than $arg"
    fi

answered Mar 1, 2020 at 17:10

Pierre François

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