oat++ 解决用户上传文件及获取文件名的解决方法

        这里先给出oat++ 中获取用户上传文件的官方文档 官方网址

        官方最简单的是以下这个

#include "oatpp/core/data/stream/FileStream.hpp"
 
...
 
ENDPOINT("POST", "/upload", upload, REQUEST(std::shared_ptr<IncomingRequest>, request)) {
    oatpp::data::stream::FileOutputStream fileOutputStream("/path/to/file");
    request->transferBodyToStream(&fileOutputStream); // transfer body chunk by chunk
    return createResponse(Status::CODE_200, "OK");
}

        用户只需要提供文件的位置即可。

        如果想获取用户上传的文件名，可以用以下代码

    ENDPOINT("POST", "datasets/upload", upload, REQUEST(std::shared_ptr<IncomingRequest>, request)) {
          /* Prepare multipart container. */
    auto multipart = std::make_shared<multipart::PartList>(request->getHeaders());
        /* Create multipart reader. */
    multipart::Reader multipartReader(multipart.get());
 
    /* Configure to read part with name "part1" into memory */
 
    multipartReader.setPartReader("mimiz", multipart::createInMemoryPartReader(-1));
 
    /* Read multipart body */
    request->transferBody(&multipartReader);
        /* Print value of "part1" */
    auto part1 = multipart->getNamedPart("mimiz");
 
    
    /* Assert part is not null */
 
    OATPP_ASSERT_HTTP(part1, Status::CODE_400, "part1 is null");
 
    auto filename  = part1->getFilename();
    OATPP_LOGD("test", "%s", filename.getValue("").c_str());
 
    oatpp::data::stream::FileOutputStream fileOutputStream(std::string("/root/pro_bz/data/" + filename.getValue("")).c_str());
    fileOutputStream.writeSimple(part1->getPayload()->getInMemoryData());
 
    return createResponse(Status::CODE_200, "OK");
  }

        其中filename就是用户上传的文件名，当用户使用postman测试的时候一定要保证key为mimiz

        成功！