I'm working on a bash script that reads a csv file with fields and field values and creates a json string out of the input.

For simplicity I have a bash script that takes in two arguments simulating above script.



jsonString='{"field1":"'$fieldValue1'", "field2":"'$fieldValue2'"}'
echo $jsonString

When I call above script with two arguments, I correctly get below output:

 ./test.sh "abc" "def"
{"field1":"abc", "field2":"def"}

What I am trying to achieve now, is to be able to assign and print a variable in the first argument, then reuse the variable from the first argument in the second argument.

For example when I call the script with below sample arguments (the syntax is flexible):

 ./test.sh "VAR=abc;echo $VAR" "$VAR"

then the output should look like this:

{"field1":"abc", "field2":"abc"}

The practical use of this is that for example for the real script inside the large input file, I only have to maintain a date once:

var startDate=1/1/2020;echo $startDate,someValue,$startDate

Does anybody have an idea on how to best achieve above output?

closed as unclear what you're asking by Kusalananda, RalfFriedl, G-Man, Thomas, roaima Sep 7 '18 at 19:42

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  • You realise that none of your variables in your script is double quoted? That's poor programming style. Try fieldValue1="$1" and jsonString="{\"field1\":\"$fieldValue1\", \"field2\":\"$fieldValue2\"}' or jsonString='{"field1":"'"$fieldValue1"'", "field2":"'"$fieldValue2"'"}' if you must, and finally echo "$jsonString". – roaima Sep 7 '18 at 19:40

Why not var=abc; ./test.sh "$var" "$var" ?

Or, call your script with one argument, and the script handles the missing value:


then: ./test.sh foo outputs {"field1"="foo", "field2"="foo"}

Note that's not JSON: use a colon not equal: {"field1":"foo", "field2":"foo"}

Also you're leaving the variables unquoted when you assign the jsonString variable. Do this

jsonString='{"field1":"'"$fieldValue1"'", "field2":"'"$fieldValue2"'"}'
# ......................^............^...............^............^
# or
jsonString="{'field1':'$fieldValue1', 'field2':'$fieldValue2'}"
# or
printf -v jsonString '{"field1":"%s","field2":"%s"}' "$fieldValue1" "$fieldValue2"

Also, you should protect against the values containing quote characters:

printf -v jsonString '{"field1":"%s","field2":"%s"}' \
    "${fieldValue1//\"/\\\"}" \


$ ./test.sh 'he said "foo"'
{"field1":"he said \"foo\"","field2":"he said \"foo\""}

Based on updated requirements. Assigning the variable must be a separate step. You're also missing the $(...) command substitution syntax

First an updated script:

for ((i=1; i<=$#; i++)); do
    pairs+=( "$(printf '"%s":"%s"' "field$i" "${value//\"/\\\"}")" )
echo "{${pairs[*]}}"

Then invoke it like this

bash test.sh "$startDate" someOtherValue "$startDate" "$(date -u -d "$startDate 1 hour" "+%FT%TZ")" "anotherValue"

and get this output:

  • I fixed the json string by using : instead of = I also added an explanation or the "real" use case to this post, which is having an large input file with values and declaring and using a variable only once in the input file instead of having to maintain a value in many places inside the input file... – Bernie Lenz Sep 6 '18 at 17:15

It seems as if you are trying to reinvent csvjson from CSVkit:

$ cat file.csv

$ csvjson file.csv
[{"A": 1.0, "B": 2.0, "C": 3.0}, {"A": 4.0, "B": 5.0, "C": 6.0}]

$ csvjson -i 4 file.csv
        "A": 1.0,
        "B": 2.0,
        "C": 3.0
        "A": 4.0,
        "B": 5.0,
        "C": 6.0
  • Similar but with variables: e.g. A,B,C\nVAR=1/1/20;echo $VAR,$VAR,3 – Bernie Lenz Sep 6 '18 at 17:18
  • @BernieLenz I'm not sure I understand. Could you give an example CSV input file that you are working on and what the JSON should look like after having processed it? – Kusalananda Sep 6 '18 at 17:52
  • Updated the post with an example – Bernie Lenz Sep 6 '18 at 18:06
  • 1
    @BernieLenz You made it very unclear. You have shell code in your CSV file? – Kusalananda Sep 6 '18 at 18:12
  • Correct. That's the reason I have my own shell script instead of using any tools. – Bernie Lenz Sep 6 '18 at 18:25

If you enclose your first argument within $(), you'll get the output you want.

./test.sh "$(VAR=abc; echo $VAR)" "def"

In this case, bash will use a subshell to execute setting VAR=abc, followed by echo $VAR, and then STDOUT from that subshell is being used as test.sh argument $1.


To get both variables assigned from a single setting of the variable, you'll need to get both from the subshell.

./test.sh $(VAR='abc'; echo $VAR $VAR)

By removing the quotation marks around $(), we know catch the first word in the STDOUT of the subshell as $1, and the second word as $2.

However, since $VAR isn't being set in the parent shell, you can't use $VAR directly.

You could however, use the subshell to set $VAR.

VAR=$(VAR=abc; echo $VAR) ./test.sh $VAR $VAR

or, for readability

VAR=$(FOO=abc; echo $FOO) ./test.sh $VAR $VAR
  • 1
    Does not answer the question: he wants to use the variable for the 2nd parameter as well. And since you're assigning it in a subshell, the variable is not available there. – glenn jackman Sep 6 '18 at 17:05
  • This seems to be on the right track, but when I try to reference VAR in the second argument it turns out to be empty: ./variableAssignmentTest.sh "$(VAR=abc; echo $VAR)" "$VAR" – Bernie Lenz Sep 6 '18 at 17:10
  • 1
    Yeah. That's because VAR isn't being set in the current shell. It's being set in the sub shell, and then we're catching it as STDOUT. – Tim Kennedy Sep 6 '18 at 17:55
  • If you assign $VAR in the parent shell, then you could use it to pass arguments to test.sh. – Tim Kennedy Sep 6 '18 at 18:01
  • Is there a way to export the variable from the sub to the parent? – Bernie Lenz Sep 6 '18 at 18:07

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