Writing rapidjson document to a file using PrettyWriter











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I have been unable to find a direct answer to this question. After searching for some time, I've written the following code but I'm sure that there exists a simpler way of doing the same task.



int persistJSONChanges(rapidjson::Document& fa_cloneDoc, string jsonFilePath)
{
FILE* lp_file = fopen(jsonFilePath.c_str(), "w");
rapidjson::StringBuffer buffer;
rapidjson::PrettyWriter<rapidjson::StringBuffer> writer(buffer);
fa_cloneDoc.Accept(writer);

string temp=buffer.GetString();
unique_ptr<char>l_writeBuffer(new char[temp.size()]);
rapidjson::FileWriteStream l_writeStream(lp_file, l_writeBuffer.get(), temp.size());
rapidjson::PrettyWriter<rapidjson::FileWriteStream> l_writer(l_writeStream);
bool l_returnStatus=fa_cloneDoc.Accept(l_writer);
if(l_returnStatus==false)
{
cout<<endl<<"file update failed"<<endl;
return -1;
}
fclose(lp_file);
return 0;
}









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    up vote
    0
    down vote

    favorite












    I have been unable to find a direct answer to this question. After searching for some time, I've written the following code but I'm sure that there exists a simpler way of doing the same task.



    int persistJSONChanges(rapidjson::Document& fa_cloneDoc, string jsonFilePath)
    {
    FILE* lp_file = fopen(jsonFilePath.c_str(), "w");
    rapidjson::StringBuffer buffer;
    rapidjson::PrettyWriter<rapidjson::StringBuffer> writer(buffer);
    fa_cloneDoc.Accept(writer);

    string temp=buffer.GetString();
    unique_ptr<char>l_writeBuffer(new char[temp.size()]);
    rapidjson::FileWriteStream l_writeStream(lp_file, l_writeBuffer.get(), temp.size());
    rapidjson::PrettyWriter<rapidjson::FileWriteStream> l_writer(l_writeStream);
    bool l_returnStatus=fa_cloneDoc.Accept(l_writer);
    if(l_returnStatus==false)
    {
    cout<<endl<<"file update failed"<<endl;
    return -1;
    }
    fclose(lp_file);
    return 0;
    }









    share|improve this question
























      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      I have been unable to find a direct answer to this question. After searching for some time, I've written the following code but I'm sure that there exists a simpler way of doing the same task.



      int persistJSONChanges(rapidjson::Document& fa_cloneDoc, string jsonFilePath)
      {
      FILE* lp_file = fopen(jsonFilePath.c_str(), "w");
      rapidjson::StringBuffer buffer;
      rapidjson::PrettyWriter<rapidjson::StringBuffer> writer(buffer);
      fa_cloneDoc.Accept(writer);

      string temp=buffer.GetString();
      unique_ptr<char>l_writeBuffer(new char[temp.size()]);
      rapidjson::FileWriteStream l_writeStream(lp_file, l_writeBuffer.get(), temp.size());
      rapidjson::PrettyWriter<rapidjson::FileWriteStream> l_writer(l_writeStream);
      bool l_returnStatus=fa_cloneDoc.Accept(l_writer);
      if(l_returnStatus==false)
      {
      cout<<endl<<"file update failed"<<endl;
      return -1;
      }
      fclose(lp_file);
      return 0;
      }









      share|improve this question













      I have been unable to find a direct answer to this question. After searching for some time, I've written the following code but I'm sure that there exists a simpler way of doing the same task.



      int persistJSONChanges(rapidjson::Document& fa_cloneDoc, string jsonFilePath)
      {
      FILE* lp_file = fopen(jsonFilePath.c_str(), "w");
      rapidjson::StringBuffer buffer;
      rapidjson::PrettyWriter<rapidjson::StringBuffer> writer(buffer);
      fa_cloneDoc.Accept(writer);

      string temp=buffer.GetString();
      unique_ptr<char>l_writeBuffer(new char[temp.size()]);
      rapidjson::FileWriteStream l_writeStream(lp_file, l_writeBuffer.get(), temp.size());
      rapidjson::PrettyWriter<rapidjson::FileWriteStream> l_writer(l_writeStream);
      bool l_returnStatus=fa_cloneDoc.Accept(l_writer);
      if(l_returnStatus==false)
      {
      cout<<endl<<"file update failed"<<endl;
      return -1;
      }
      fclose(lp_file);
      return 0;
      }






      rapidjson






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      share|improve this question











      share|improve this question




      share|improve this question










      asked Nov 20 at 8:22









      Vishal Sharma

      353320




      353320
























          2 Answers
          2






          active

          oldest

          votes

















          up vote
          1
          down vote



          accepted










          I think you misused the FileWriteStream. It just needs a buffer of arbitrary size.



          You simply needs:



          FILE* fp = fopen(...);
          char buffer[1024];
          FileWriteStream fs(fp, buffer, sizeof(buffer));
          PrettyWriter<FileWriteStream> writer(fs);
          document.Accept(writer);
          fclose(fp);





          share|improve this answer





















          • Is the size of the buffer really arbitrary? Can I use a really small value, regardless of the size of the document?
            – Vishal Sharma
            Nov 30 at 7:23






          • 1




            @VishalSharma Yes. This design is due to performance consideration. The size of buffer can be smaller than the JSON. You may check the source code of FileWriteStream and it is very simple.
            – Milo Yip
            Dec 1 at 15:54


















          up vote
          0
          down vote













          I have tried the following which worked for me:



          int persistJSONChanges(rapidjson::Document& fa_cloneDoc, string jsonFilePath)
          {

          rapidjson::StringBuffer buffer;
          rapidjson::PrettyWriter<rapidjson::StringBuffer> writer(buffer);
          bool l_returnStatus=fa_cloneDoc.Accept(writer);
          if(l_returnStatus==false)
          {
          fprintf(stdout,"n[%s::%d] JSON File update failedn",__FILE__,__LINE__);

          return -1;
          }

          string temp=buffer.GetString();
          ofstream out(jsonFilePath.c_str(),std::ofstream::trunc);
          out<<temp;
          return 0;
          }





          share|improve this answer





















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            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes








            up vote
            1
            down vote



            accepted










            I think you misused the FileWriteStream. It just needs a buffer of arbitrary size.



            You simply needs:



            FILE* fp = fopen(...);
            char buffer[1024];
            FileWriteStream fs(fp, buffer, sizeof(buffer));
            PrettyWriter<FileWriteStream> writer(fs);
            document.Accept(writer);
            fclose(fp);





            share|improve this answer





















            • Is the size of the buffer really arbitrary? Can I use a really small value, regardless of the size of the document?
              – Vishal Sharma
              Nov 30 at 7:23






            • 1




              @VishalSharma Yes. This design is due to performance consideration. The size of buffer can be smaller than the JSON. You may check the source code of FileWriteStream and it is very simple.
              – Milo Yip
              Dec 1 at 15:54















            up vote
            1
            down vote



            accepted










            I think you misused the FileWriteStream. It just needs a buffer of arbitrary size.



            You simply needs:



            FILE* fp = fopen(...);
            char buffer[1024];
            FileWriteStream fs(fp, buffer, sizeof(buffer));
            PrettyWriter<FileWriteStream> writer(fs);
            document.Accept(writer);
            fclose(fp);





            share|improve this answer





















            • Is the size of the buffer really arbitrary? Can I use a really small value, regardless of the size of the document?
              – Vishal Sharma
              Nov 30 at 7:23






            • 1




              @VishalSharma Yes. This design is due to performance consideration. The size of buffer can be smaller than the JSON. You may check the source code of FileWriteStream and it is very simple.
              – Milo Yip
              Dec 1 at 15:54













            up vote
            1
            down vote



            accepted







            up vote
            1
            down vote



            accepted






            I think you misused the FileWriteStream. It just needs a buffer of arbitrary size.



            You simply needs:



            FILE* fp = fopen(...);
            char buffer[1024];
            FileWriteStream fs(fp, buffer, sizeof(buffer));
            PrettyWriter<FileWriteStream> writer(fs);
            document.Accept(writer);
            fclose(fp);





            share|improve this answer












            I think you misused the FileWriteStream. It just needs a buffer of arbitrary size.



            You simply needs:



            FILE* fp = fopen(...);
            char buffer[1024];
            FileWriteStream fs(fp, buffer, sizeof(buffer));
            PrettyWriter<FileWriteStream> writer(fs);
            document.Accept(writer);
            fclose(fp);






            share|improve this answer












            share|improve this answer



            share|improve this answer










            answered Nov 30 at 5:12









            Milo Yip

            3,3791620




            3,3791620












            • Is the size of the buffer really arbitrary? Can I use a really small value, regardless of the size of the document?
              – Vishal Sharma
              Nov 30 at 7:23






            • 1




              @VishalSharma Yes. This design is due to performance consideration. The size of buffer can be smaller than the JSON. You may check the source code of FileWriteStream and it is very simple.
              – Milo Yip
              Dec 1 at 15:54


















            • Is the size of the buffer really arbitrary? Can I use a really small value, regardless of the size of the document?
              – Vishal Sharma
              Nov 30 at 7:23






            • 1




              @VishalSharma Yes. This design is due to performance consideration. The size of buffer can be smaller than the JSON. You may check the source code of FileWriteStream and it is very simple.
              – Milo Yip
              Dec 1 at 15:54
















            Is the size of the buffer really arbitrary? Can I use a really small value, regardless of the size of the document?
            – Vishal Sharma
            Nov 30 at 7:23




            Is the size of the buffer really arbitrary? Can I use a really small value, regardless of the size of the document?
            – Vishal Sharma
            Nov 30 at 7:23




            1




            1




            @VishalSharma Yes. This design is due to performance consideration. The size of buffer can be smaller than the JSON. You may check the source code of FileWriteStream and it is very simple.
            – Milo Yip
            Dec 1 at 15:54




            @VishalSharma Yes. This design is due to performance consideration. The size of buffer can be smaller than the JSON. You may check the source code of FileWriteStream and it is very simple.
            – Milo Yip
            Dec 1 at 15:54












            up vote
            0
            down vote













            I have tried the following which worked for me:



            int persistJSONChanges(rapidjson::Document& fa_cloneDoc, string jsonFilePath)
            {

            rapidjson::StringBuffer buffer;
            rapidjson::PrettyWriter<rapidjson::StringBuffer> writer(buffer);
            bool l_returnStatus=fa_cloneDoc.Accept(writer);
            if(l_returnStatus==false)
            {
            fprintf(stdout,"n[%s::%d] JSON File update failedn",__FILE__,__LINE__);

            return -1;
            }

            string temp=buffer.GetString();
            ofstream out(jsonFilePath.c_str(),std::ofstream::trunc);
            out<<temp;
            return 0;
            }





            share|improve this answer

























              up vote
              0
              down vote













              I have tried the following which worked for me:



              int persistJSONChanges(rapidjson::Document& fa_cloneDoc, string jsonFilePath)
              {

              rapidjson::StringBuffer buffer;
              rapidjson::PrettyWriter<rapidjson::StringBuffer> writer(buffer);
              bool l_returnStatus=fa_cloneDoc.Accept(writer);
              if(l_returnStatus==false)
              {
              fprintf(stdout,"n[%s::%d] JSON File update failedn",__FILE__,__LINE__);

              return -1;
              }

              string temp=buffer.GetString();
              ofstream out(jsonFilePath.c_str(),std::ofstream::trunc);
              out<<temp;
              return 0;
              }





              share|improve this answer























                up vote
                0
                down vote










                up vote
                0
                down vote









                I have tried the following which worked for me:



                int persistJSONChanges(rapidjson::Document& fa_cloneDoc, string jsonFilePath)
                {

                rapidjson::StringBuffer buffer;
                rapidjson::PrettyWriter<rapidjson::StringBuffer> writer(buffer);
                bool l_returnStatus=fa_cloneDoc.Accept(writer);
                if(l_returnStatus==false)
                {
                fprintf(stdout,"n[%s::%d] JSON File update failedn",__FILE__,__LINE__);

                return -1;
                }

                string temp=buffer.GetString();
                ofstream out(jsonFilePath.c_str(),std::ofstream::trunc);
                out<<temp;
                return 0;
                }





                share|improve this answer












                I have tried the following which worked for me:



                int persistJSONChanges(rapidjson::Document& fa_cloneDoc, string jsonFilePath)
                {

                rapidjson::StringBuffer buffer;
                rapidjson::PrettyWriter<rapidjson::StringBuffer> writer(buffer);
                bool l_returnStatus=fa_cloneDoc.Accept(writer);
                if(l_returnStatus==false)
                {
                fprintf(stdout,"n[%s::%d] JSON File update failedn",__FILE__,__LINE__);

                return -1;
                }

                string temp=buffer.GetString();
                ofstream out(jsonFilePath.c_str(),std::ofstream::trunc);
                out<<temp;
                return 0;
                }






                share|improve this answer












                share|improve this answer



                share|improve this answer










                answered Nov 30 at 7:18









                Vishal Sharma

                353320




                353320






























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