Writing rapidjson document to a file using PrettyWriter
0
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
asked Nov 20 '18 at 8:22
Vishal SharmaVishal Sharma
323322
add a comment |
0
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
asked Nov 20 '18 at 8:22
Vishal SharmaVishal Sharma
323322
add a comment |
0
0
0
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
asked Nov 20 '18 at 8:22
Vishal SharmaVishal Sharma
323322
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
rapidjson
asked Nov 20 '18 at 8:22
Vishal SharmaVishal Sharma
323322
asked Nov 20 '18 at 8:22
Vishal SharmaVishal Sharma
323322
asked Nov 20 '18 at 8:22
Vishal SharmaVishal Sharma
323322
asked Nov 20 '18 at 8:22
Vishal SharmaVishal Sharma
323322
asked Nov 20 '18 at 8:22
Vishal SharmaVishal Sharma
323322
323322
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
1
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);
answered Nov 30 '18 at 5:12
Milo YipMilo Yip
3,4891620
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 '18 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 '18 at 15:54
add a comment |
0
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;
}
answered Nov 30 '18 at 7:18
Vishal SharmaVishal Sharma
323322
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
1
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);
answered Nov 30 '18 at 5:12
Milo YipMilo Yip
3,4891620
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 '18 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 '18 at 15:54
add a comment |
1
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);
answered Nov 30 '18 at 5:12
Milo YipMilo Yip
3,4891620
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 '18 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 '18 at 15:54
add a comment |
1
1
1
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);
answered Nov 30 '18 at 5:12
Milo YipMilo Yip
3,4891620
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);
answered Nov 30 '18 at 5:12
Milo YipMilo Yip
3,4891620
answered Nov 30 '18 at 5:12
Milo YipMilo Yip
3,4891620
answered Nov 30 '18 at 5:12
Milo YipMilo Yip
3,4891620
answered Nov 30 '18 at 5:12
Milo YipMilo Yip
3,4891620
3,4891620
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 '18 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 '18 at 15:54
add a comment |
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 '18 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 '18 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 '18 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 '18 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 '18 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 '18 at 15:54
add a comment |
0
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;
}
answered Nov 30 '18 at 7:18
Vishal SharmaVishal Sharma
323322
add a comment |
0
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;
}
answered Nov 30 '18 at 7:18
Vishal SharmaVishal Sharma
323322
add a comment |
0
0
0
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;
}
answered Nov 30 '18 at 7:18
Vishal SharmaVishal Sharma
323322
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;
}
answered Nov 30 '18 at 7:18
Vishal SharmaVishal Sharma
323322
answered Nov 30 '18 at 7:18
Vishal SharmaVishal Sharma
323322
answered Nov 30 '18 at 7:18
Vishal SharmaVishal Sharma
323322
answered Nov 30 '18 at 7:18
Vishal SharmaVishal Sharma
323322
323322
add a comment |
add a comment |
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