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I need to parse a csv file and store the value in a variable. Below is the sample csv file

Sample CSV file

And below is the ./script.sh

#!/bin/bash

D="";

P="";

./xyz --project "$P" --displayname "$D"

For example if we consider the first line:

When i run the script like

./script.sh /home/project.csv

it should store the values: 10V nmos and pmos for MA into $D and 10v_nmos_and_pmos_for_ma into $P.

How can this be done ?

Another scenario is, from the csv file if i call first three projects then it should run in a loop row by row

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2 Answers 2

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bash has no built-in support for parsing CSVs. You could use ksh93 instead which supports parsing CSV (at least some form of CSV, the one where a literal " is entered as "" within double quotes) with read -S:

#! /bin/ksh93 -
while IFS=, read -rSu3 P D ignore; do
  ./xyz --project "$P" --displayname "${D#*.}"
done 3< file.csv

Or use perl/python with a proper CSV parsing library that you could tune for the exact format of your csv. Example with perl:

perl -C -MText::CSV -e '
  $c = Text::CSV->new;
  while (($p, $d) = @{$c->getline(STDIN)}) {
    $d =~ s/.*?\.//;
    system "./xyz", "--project", $p, "--displayname", $d;
  }' < file.csv

If you can guarantee that the content of the CSV fields won't contain double-quote, comas or newline characters, with POSIX shells like bash, you could do:

tr -d \" < file.csv | while IFS=, read -r p d ignore; do
  ./xyz --project "$p" --displayname "${d#*.}"
done
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  • I am completly new to scripting, i did something like this #!/bin/bash export IFS="," cat /tmp/test/test_parse/parse.csv | while read a b; #do echo "$a:$b:$c:$d"; do echo $a echo $b done But in $a = it saves with 10V nmos and pmos for MA" but i need only the 10V nmos and pmos for MA in $b = it gives with projects.10v_nmos_and_pmos_for_ma which is correct but i need only 10v_nmos_and_pmos_for_ma please guide
    – Siddharth Sahoo
    Nov 11, 2016 at 13:14
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This assumes very reliable occurrences of ',' and '.' characters in your CSV:

#!/bin/bash
cat "$1" | while IFS='' read -r line; do
    D="$(echo "$line" | cut -d ',' -f 1 | tr -d '"')"
    P="$(echo "$line" | cut -d '.' -f 2)"
    echo 'D: '"$D"
    echo 'P: '"$P"
    ./xyz --project "$P" --displayname "$D"
done
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  • This is not working properly, when i run the script with inputs nothing is getting saved in $D and $P i mean i cant see value getting stored the Variable
    – Siddharth Sahoo
    Nov 11, 2016 at 12:48
  • You'll need to make sure the path to your 'xyz' executable is correct ('./xyz' obviously assumes it's in the same directory as the code (here) being run, and of course that your 'xyz' executable is consuming its arguments as expected. You can also substitute foo.csv with "$1" to utilize it exactly as you mentioned (./script.sh /home/project.csv).
    – Jan Kyu Peblik
    Nov 11, 2016 at 12:54
  • remove the last ./xyz for now If i run the below i dont see any value stroring to the variable #!/bin/bash cat foo.csv | while IFS='' read -r line; do D="$(echo "$line" | cut -d ',' -f 1 | tr -d '"')" P="$(echo "$line" | cut -d '.' -f 2)" echo $D echo $P
    – Siddharth Sahoo
    Nov 11, 2016 at 12:56
  • foo.csv is of course a placeholder. You can substitute foo.csv with "$1" to utilize it exactly as you mentioned (./script.sh /home/project.csv).
    – Jan Kyu Peblik
    Nov 11, 2016 at 12:59
  • 1
    A few bad practices here could justify a downvote: UUOC, use of echo for arbitrary data. But also running 9 commands, 4 of which not builtin and 5 pipes for each line of the file, while read can do the cutting itself, and bash expansions can remove the quotes by themselves. At least, you've quoted your variables. kudos for that.
    – Stéphane Chazelas
    Nov 11, 2016 at 15:20

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