Need to pass files from a folder and their details to an API dynamically

Question

User-1660589204 posted

I am working on a program which passes file details to an API and gets an ID. I am confused on how can I collect the values and set them as parameters to my API. API takes one file details at a time where as the source folder could contain one or many media files. 

public static void GetFilesID(HttpClient httpclient, string result)
        {
            HttpClient client = new HttpClient();
            DirectoryInfo info = new DirectoryInfo(path);
            info.GetFiles("*.trm");

            DataTable FileDetailsTable = new DataTable();
            System.Net.Http.HttpResponseMessage response = client
                                        .PostAsync(".../api/v2/file/uploader", fileDetails);
            response.EnsureSuccessStatusCode();
        } 

All the files I get are media files. 

My API takes four parameters as: FilePath, FileName, FileType, FileReferenceID(generated from DB). I have a model fileDetails which contains all the above parameters and some more columns. How can I pass these file details to DB save them and pass the saved values into API. I would like to use Linq for DB communication. Please help me with this.

 

Answer 1

User-939850651 posted

Hi ddesarajubyc,

According to your description, I think you may need the FileInfo Class to accept the file information that meets the filter, and then you could encapsulate this information into a custom fileDetails model object, and then serialize it as a parameter to your API.

Something like this:

DirectoryInfo info = new DirectoryInfo(path);
            List<FileDetails> details = new List<FileDetails>();
            FileInfo[] fileInfos = info.GetFiles("*.trm");
            if (fileInfos != null) { 
            foreach (FileInfo fileInfo in fileInfos) {
                    FileDetails detail = new FileDetails();
                    detail.FileName = fileInfo.Name;
                    detail.FilePath = fileInfo.FullName;
                    detail.FileType = fileInfo.Extension;
                    details.Add(detail);
                }
            }
            var fileDetails = JsonConvert.SerializeObject(details);
            HttpClient client = new HttpClient();
            string url = "your api address";
            HttpContent httpContent = new StringContent(fileDetails, Encoding.UTF8, "application/json");
            var response = client.PostAsync(url, httpContent);

Hope this can help you.

Best regards,

Xudong Peng

Answer 2

User-1660589204 posted

This is good code for my implementation XuDong Peng,

 I need one modification though. My FileDetails Model has following properties in it:

public partial class FileDetails
    {
        public string FileUrl { get; set; }
        public string Filename { get; set; }
        public string Case_ID { get; set; }
        public string FileType { get; set; }
        public Int64 File_Reference_Id { get; set; }
        public DateTime Created_Date { get; set; }
        public DateTime Updated_Date { get; set; }
        public string Status { get; set; }
        public string File_Id { get; set; }
        public string Preset { get; set; }
        public string Transcode_callbackUrl { get; set; }
        public string FileTranscodeJob_Id { get; set; }
        public string Transcription_callbackUrl { get; set; }
        public string jobDefinition { get; set; }
        public string TranscriptionJob_Id { get; set; }
    }

I am unable to set these values via fileInfo type. I need to set all these values in the foreach loop. My implementation as suggested by you shows compile time error:

public Task<FileDetails> UploadFiles(string filename)
        {
            DirectoryInfo info = new DirectoryInfo(path);
            List<FileDetails> details = new List<FileDetails>();
            FileInfo[] fileInfos = info.GetFiles("*.trm");
            if (fileInfos != null)
            {
                foreach (FileInfo fileInfo in fileInfos)
                {
                    FileDetails detail = new FileDetails();
                    detail.FileUrl = fileInfo.Name;
                    detail.Filename = fileInfo.FullName;
                    detail.FileType = fileInfo.Extension;
                    details.Add(detail);
                }
            }

I need to assign the exact values for my coding. Is there any operation for this. Please let me know.