Removing curly brackets after using "to.String()" across a group of fields
- 4 years ago
Hey @darshthakkar,
I suppose you can remove the curly brackets from your final string, but if you want to go and make this solution totally dynamic, than I suggest to remove the brackets on an object level with the following expression:
$group.map(x=>x.toString().substring(1,x.toString().length-1)).toString()
What we’re doing here is first we’re mapping/iterating through the array and turning every object in it into a string:
[{ "name": "John", "age": 33 }, { "name": "Jack", "age": 23 }, { "name": "Jill", "age": 43 }]
will become:
[ "{name=John, age=33}" "{name=Jack, age=23}" "{name=Jill, age=43}" ]
In the same iteration we’re removing the first and last character which are the object curly brackets. The rest is clear, just concatenate the strings in the array:
[ "name=John, age=33" "name=Jack, age=23" "name=Jill, age=43" ]
End result:
name=John, age=33,name=Jack, age=23,name=Jill, age=43
With this we’re excluding any possibility of removing a curly brackets which are in the values of the JSON objects and should not be excluded.
Sample:
[ { "name": "John{test}", "age": 33 }, { "name": "Jack{test}", "age": 23 }, { "name": "Jill{test}", "age": 43 } ]
Result while replacing curly brackets on final string:
name=Johntest, age=33,name=Jacktest, age=23,name=Jilltest, age=43
Result while removing curly brackets on object level with mapping:
name=John{test}, age=33,name=Jack{test}, age=23,name=Jill{test}, age=43
Regards,
Bojan