如何用Delphi的TNetHttpClient/TRESTClient实现合规的CSV文件PUT上传？

以下分别用TNetHttpClient和TRESTClient两种组件实现需求，完全匹配你提供的curl操作逻辑：

一、使用TNetHttpClient实现 #

需要引用单元：System.Net.HttpClient, System.Net.HttpClientComponent, System.IOUtils

procedure UploadCSVWithNetHttpClient;
var
  HttpClient: TNetHttpClient;
  Request: TNetHttpRequest;
  Response: IHTTPResponse;
  LocalFilePath: string;
  ClientID, RemoteFileName: string;
  ApiKey: string;
  UploadURL: string;
  FileContent: TBytes;
begin
  // 配置参数
  ClientID := 'd32krpq';
  RemoteFileName := '20181023_update.csv';
  ApiKey := 'lXViWTzFic9sM8qe9Ew7JME8xTdBAOMJHdIjK7XkjQ00OWr';
  LocalFilePath := '20181023_update.csv'; // 本地CSV文件路径

  // 构造上传URL
  UploadURL := Format('https://example.com/upload/%s/%s', [ClientID, RemoteFileName]);

  // 读取本地文件内容
  if not TFile.Exists(LocalFilePath) then
    raise Exception.Create('本地CSV文件不存在');
  FileContent := TFile.ReadAllBytes(LocalFilePath);

  // 初始化HTTP客户端
  HttpClient := TNetHttpClient.Create(nil);
  try
    Request := TNetHttpRequest.Create(nil);
    try
      // 设置请求方法为PUT
      Request.Method := THTTPMethod.mPUT;
      // 添加API Key请求头
      Request.CustomHeaders['x-api-key'] := ApiKey;
      // 设置请求体为文件内容
      Request.ContentStream := TBytesStream.Create(FileContent);

      // 发送请求并获取响应
      Response := HttpClient.Execute(Request, UploadURL);
      try
        // 检查响应状态，200-299为成功
        if Response.StatusCode >= 200 and Response.StatusCode < 300 then
          ShowMessage('文件上传成功')
        else
          ShowMessage(Format('上传失败，状态码：%d，响应内容：%s', [Response.StatusCode, Response.ContentAsString]));
      finally
        Response := nil;
      end;
    finally
      Request.Free;
    end;
  finally
    HttpClient.Free;
  end;
end;

二、使用TRESTClient组件实现 #

需要引用单元：REST.Client, REST.Types, System.IOUtils

procedure UploadCSVWithRESTClient;
var
  RESTClient: TRESTClient;
  RESTRequest: TRESTRequest;
  RESTResponse: TRESTResponse;
  LocalFilePath: string;
  ClientID, RemoteFileName: string;
  ApiKey: string;
  UploadURL: string;
begin
  // 配置参数
  ClientID := 'd32krpq';
  RemoteFileName := '20181023_update.csv';
  ApiKey := 'lXViWTzFic9sM8qe9Ew7JME8xTdBAOMJHdIjK7XkjQ00OWr';
  LocalFilePath := '20181023_update.csv'; // 本地CSV文件路径

  // 构造上传URL
  UploadURL := Format('https://example.com/upload/%s/%s', [ClientID, RemoteFileName]);

  // 初始化REST组件
  RESTClient := TRESTClient.Create(nil);
  RESTRequest := TRESTRequest.Create(nil);
  RESTResponse := TRESTResponse.Create(nil);
  try
    RESTClient.BaseURL := UploadURL;
    RESTRequest.Client := RESTClient;
    RESTRequest.Response := RESTResponse;

    // 设置请求方法为PUT
    RESTRequest.Method := rmPUT;

    // 添加API Key请求头
    RESTRequest.Params.AddHeader('x-api-key', ApiKey);

    // 设置请求体为本地文件内容
    RESTRequest.AddBody(TFile.ReadAllBytes(LocalFilePath), ctTEXT_CSV);

    // 执行请求
    try
      RESTRequest.Execute;
      // 检查响应状态
      if RESTResponse.StatusCode >= 200 and RESTResponse.StatusCode < 300 then
        ShowMessage('文件上传成功')
      else
        ShowMessage(Format('上传失败，状态码：%d，响应内容：%s', [RESTResponse.StatusCode, RESTResponse.Content]));
    except
      on E: Exception do
        ShowMessage('请求执行出错：' + E.Message);
    end;
  finally
    RESTResponse.Free;
    RESTRequest.Free;
    RESTClient.Free;
  end;
end;

注意事项 #

• 确保本地文件路径正确，建议使用绝对路径避免找不到文件的问题
• 实际部署时需补充更完善的异常处理（如网络中断、权限不足等场景）
• 若服务器对请求体的Content-Type有特殊要求，可调整代码中对应的类型参数

内容的提问来源于stack exchange，提问作者Joe Wright