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var postObj = new { FolderName = folderName, FileName = fileName, FileContent = fileContent }; var postJson = JsonConvert.SerializeObject(postObj); va 阅读全文
posted @ 2023-09-19 15:21
hofmann
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摘要:
var fileStream = new MemoryStream(fileContent); var parametersList = new MultipartFormDataContent { { new StringContent(folderName), "FolderName" }, { 阅读全文
posted @ 2023-09-19 10:47
hofmann
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