CODEWITHSUNDEEP
jsonbash
jm33_m0
jm33_m0

Reputation: 615

How to strip info from every JSON file under the same folder using BASH?

Let's say I have firstJSON.json secondJSON.json ... under the same folder, and each file looks like 

{
    "obj1":"item1",
    "obj2":"item2"
}

Now I want to strip item1 and item2, then put them into one file outFile, I have written the following code to do that:

function strip {
    item1=$(cat $1 | grep 'obj1' | cut -d '"' -f4)
    item2=$(cat $1 | grep 'obj2' | cut -d '"' -f4)
    name=$(echo $1 | cut -d '.' -f1)
    echo $name':'$item1':'$item2 >> ./outFile 
}

but how do I use this piece of code to strip info from every JSON file and then put them all into outFile?

Upvotes: 0

Views: 84

Answers (2)

pakistanprogrammerclub
pakistanprogrammerclub

Reputation: 827

A single grep on all files is enough piped into an awk munger

files=$*
grep -HPo '^ +"obj\d+" *: *"\K[^"]+' $files|
awk -F: '
src == $1 {
  printf(":%s", $2)
  next
}
{
  if(src) printf("\n")
  src=$1
  f1=gensub(/\.[^.]*$/,"",1,$1)
  printf("%s:%s", f1, $2)
}
END {printf("\n")}
'

Upvotes: 1

Mustafa DOGRU
Mustafa DOGRU

Reputation: 4112

you can use with for loop as below; 

for j in *.json; do
item1=$(cat $j | grep 'obj1' | cut -d '"' -f4)
item2=$(cat $j | grep 'obj2' | cut -d '"' -f4)
name=$(echo $j | cut -d '.' -f1)
echo $name':'$item1':'$item2 >> ./outFile 
done

Upvotes: 0

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