Question

有没有办法阻止Boost json解析器输出＆＃34;没有这样的节点＆＃34;每当传入无效的子位置时到控制台？

jsonNode.get_child("invalid.node.location");

我尝试更改代码，为默认值添加第二个参数，编译器说它是无效参数

Code Snippet（siteTree是另一个ptree），位置节点是JSON中可用的数组：

const ptree& test = siteTree.get_child("location", empty_ptree<ptree>());

错误1：

Function 'empty_ptree' could not be resolved

错误2：

Invalid arguments '
Candidates are:
boost::property_tree::basic_ptree<std::basic_string<char,std::char_traits<char>,std::allocator<char>>,std::basic_string<char,std::char_traits<char>,std::allocator<char>>,std::less<std::basic_string<char,std::char_traits<char>,std::allocator<char>>>> & get_child(const boost::property_tree::string_path<std::basic_string<char,std::char_traits<char>,std::allocator<char>>,boost::property_tree::id_translator<std::basic_string<char,std::char_traits<char>,std::allocator<char>>>> &)
const boost::property_tree::basic_ptree<std::basic_string<char,std::char_traits<char>,std::allocator<char>>,std::basic_string<char,std::char_traits<char>,std::allocator<char>>,std::less<std::basic_string<char,std::char_traits<char>,std::allocator<char>>>> & get_child(const boost::property_tree::string_path<std::basic_string<char,std::char_traits<char>,std::allocator<char>>,boost::property_tree::id_translator<std::basic_string<char,std::char_traits<char>,std::allocator<char>>>> &)
boost::property_tree::basic_ptree<std::basic_string<char,std::char_traits<char>,std::allocator<char>>,std::basic_string<char,std::char_traits<char>,std::allocator<char>>,std::less<std::basic_string<char,std::char_traits<char>,std::allocator<char>>>> & get_child(const boost::property_tree::string_path<std::basic_string<char,std::char_traits<char>,std::allocator<char>>,boost::property_tree::id_translator<std::basic_string<char,std::char_traits<char>,std::allocator<char>>>> &, boost::property_tree::basic_ptree<std::basic_string<char,std::char_traits<char>,std::allocator<char>>,std::basic_string<char,std::char_traits<char>,std::allocator<char>>,std::less<std::basic_string<char,std::char_traits<char>,std::allocator<char>>>> &)
const boost::property_tree::basic_ptree<std::basic_string<char,std::char_traits<char>,std::allocator<char>>,std::basic_string<char,std::char_traits<char>,std::allocator<char>>,std::less<std::basic_string<char,std::char_traits<char>,std::allocator<char>>>> & get_child(const boost::property_tree::string_path<std::basic_string<char,std::char_traits<char>,std::allocator<char>>,boost::property_tree::id_translator<std::basic_string<char,std::char_traits<char>,std::allocator<char>>>> &, const boost::property_tree::basic_ptree<std::basic_string<char,std::char_traits<char>,std::allocator<char>>,std::basic_string<char,std::char_traits<char>,std::allocator<char>>,std::less<std::basic_string<char,std::char_traits<char>,std::allocator<char>>>> &)'

Answer 1

一元ptree::get_child抛出传递无效节点。但是，如果传递的路径无效，则使用两个参数的重载将返回默认值。你可以使用这个重载。或者，ptree::get_child_optional<T>返回boost::optional<T>，如果传递的路径无效，则返回boost::null。

C ++ boost json解析器抑制“No such node”输出

1 个答案: