rapidjson 不正确的网络服务字符串格式

Question

我的网络服务中有一个 return 字符串：

std::string json_str = "{"Count":18,"Result":[{"Description":"DATABASE TASK MODE VALUE","Name":"TASK","Val":24,"addID":0,"CreatedAt":"2015-08-12T17:57:03.954Z","ModifiedAt":"2015-08-13T20:41:30.900Z","CreatedBy":"00000000-0000-0000-0000-000000000000","ModifiedBy":"00000000-0000-0000-0000-000000000000","Owner":"00000000-0000-0000-0000-000000000000","Id":"89fb5320-411b-11e5-a0dc-8babbc75a180","Meta":{"Permissions":{"CanRead":true,"CanUpdate":true,"CanDelete":true}}}, ...etc ]}"

此代码有效

document.Parse(json_str.c_str()).HasParseError())
assert(document.IsObject());
rapidjson::Value::MemberIterator localdata_0 = document.FindMember("89fb5320-411b-11e5-a0dc-8babbc75a180");

但随后在这些行上失败了

assert(localdata_0 != document.MemberEnd());
assert(localdata_0->value.IsDouble());

有没有一种简单的方法可以将我的字符串格式化为正确的 json 格式？

const char json[] = " { \"hello\" : \"world\", \"t\" : true , \"f\" : false, \"n\": null, \"i\":123, \"pi\": 3.1416, \"a\":[1, 2, 3, 4] } ";

Answer 1

解决方案是将字符串分割成行并解析它。将每个字符串行放入一个向量中。

std::vector <std::string> json_str_rows;
std::vector <float> json_float_ret;

// get values from rows
try
{
    for (size_t i = 0; i < json_str_rows.size(); i++)
    {
        if (document.Parse((json_str_rows.at(i)).c_str()).HasParseError())
        {
            std::cerr << "system_rapidjson::read_json_str >> document.Parse >> ERROR at(" << i << ") = " << json_str_rows.at(i) << std::endl;
            throw -3;
        }

        json_float_ret.push_back (document["Val"].GetDouble());    // get value from row
    }
}
catch (const std::exception &e)
{
    std::cerr << "system_rapidjson::read_json_str >> get values from rows >> ERROR = " << e.what() << std::endl;
    throw -2;
}

rapidjson 不正确的网络服务字符串格式

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