4

My json object has a list in it and I would like to iterate through the elements of the list I saw the following post Iterating through objects in JsonCpp , but this did not work for me

My json object:

{
         "name": ["str1",str2" ... ]
}

The code that I have and is not working

Json::Value names= (*json)["name"];
for( Json::ValueIterator itr = names.begin() ; itr != names.end() ; itr++ ) {
    string name =  *itr.asString();}

I am getting the following error 

cannot convert from 'Json::Value' to 'std::basic_string<_Elem,_Traits,_Ax>

I am sure the elements are string because calling string name = names= (*json)["name"][0].asString() is working

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Quantico
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2 Answers2

9

In c++11 its even simpler now:

for (auto itr : json["name"]) {
    string name = itr.asString();
    // ...
}
Chris Sattinger
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7

I don't see how you can get a cannot convert error from your code, but I do think I see the problem.

Json::ValueIterator is an iterator; in other words, similar to a pointer, it needs to be dereferenced to access the value it points to.

  • itr.asString() tries to access the asString method of the iterator, and the iterator doesn't have an asString method.
  • *itr.asString() tries to access the asString method of the iterator then dereference the result (because of order of operations; . has a higher precedence than *), and the iterator doesn't have an asString method.
  • (*itr).asString() (using parentheses to clarify precedence) or itr->asString() (using the -> operator, the preferred approach) should work.
Josh Kelley
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    You're welcome. However, you should use `itr->` instead of `(*itr).`; they mean the same thing, but `->` is cleaner and is the universally accepted style. – Josh Kelley Nov 13 '14 at 17:11
  • Are you sure about itr->asString()? I am getting type 'Json::ValueIterator' does not have an overloaded member 'operator ->' – Quantico Nov 13 '14 at 17:22
  • You're right. I'm sorry; I hadn't tested it myself. So `->` is cleaner and is the almost universally accepted style, and I'd consider it a mistake or bug in JsonCpp to not support it, but since it doesn't, use `(*itr).`. – Josh Kelley Nov 13 '14 at 17:34