nlohmann json insert value array partially to already existing data - c++

Question

I have a particular case which I am trying to solve with minimal changes if possible.
one of the data is
js["key1"]["subkey2"]["subsubkey3"].push_back({1,2,3,{4,5}});
[ 1,2,3,[[4,5]] ]
Later at some stage I want to insert
{1,2,3,{4,6}}
Then it should become
[ 1,2,3,[[4,5],[4,6]] ]
How can I make this possible without making 1,2,3 value as key?

Answer 1

I did some playing. I didn't get the results you were looking for. Here's my code and results so far.
#include <iostream>
#include <json.hpp>
using namespace std;
using JSON = nlohmann::json;
int main() {
JSON json = JSON::object();
JSON key1JSON = JSON::object();
JSON key2JSON = JSON::object();
JSON key3JSON = JSON::array();
key3JSON.push_back( {1,2,3, {4,5} } );
key3JSON.push_back( {6} );
key2JSON["subsubkey3"] = key3JSON;
key1JSON["subkey2"] = key2JSON;
json["key1"] = key1JSON;
cout << json.dump(2) << endl;
}
Output:
{
"key1": {
"subkey2": {
"subsubkey3": [
[
1,
2,
3,
[
4,
5
]
],
[
6
]
]
}
}
}
You'll see that the first push_back pushed an array inside an array, which is probably one level deeper than you wanted, and the second one added a second array, which is also not what you want.
Which means you're probably going to have to write your own method, especially as you want to also handle uniqueness. I personally never free-format data that way you have in your example. But maybe your method would look something like:
bool contains(const JSON &json, const JSON &value) {
... this looks like fun to write.
}
void appendUnique(JSON &json, const JSON &array) {
for (JSON & thisJson: array) {
if (!contains(json, thisJson)) {
json.push_back(thisJson);
}
}
}
I modified my code like this:
void appendUnique(JSON &json, const JSON & array) {
for (const JSON & thisJSON: array) {
json.push_back(thisJSON);
}
}
...
appendUnique(key3JSON, {1,2,3, {4,5} } );
appendUnique(key3JSON, {6} );
And got this:
{
"key1": {
"subkey2": {
"subsubkey3": [
1,
2,
3,
[
4,
5
],
6
]
}
}
}
I'm not going to write the isUnique method. But I think you may have to take this to conclusion.