massimopasquali Posted April 16, 2022 Share Posted April 16, 2022 Hello, someone have developed a class for uploading product's photo in c# calling the webservice? My c# code string filedto = FotoDto.UrlFolder; string webAddress = FotoDto.UrlPrestashop.Replace("languages", "images/products") + FotoDto.Id_Product; var credentials = new NetworkCredential(key, ""); var handler = new HttpClientHandler { Credentials = credentials }; var client = new HttpClient(handler); var pairs = new List<KeyValuePair<string, string>> { new KeyValuePair<string, string>("image", filedto) }; var content = new FormUrlEncodedContent(pairs); content.Headers.ContentType.MediaType = @"multipart/form-data"; var response = client.PostAsync(webAddress, content).Result; ResponseWebService.HttpStatus = response.StatusCode; ResponseWebService.Xml = response.ReasonPhrase; return ResponseWebService; But prestashop give me an error "bad request". tnx Link to comment Share on other sites More sharing options...
massimopasquali Posted April 16, 2022 Author Share Posted April 16, 2022 if i try to sending the request by postman <?xml version="1.0" encoding="UTF-8"?> <prestashop xmlns:xlink="http://www.w3.org/1999/xlink"> <errors> <error> <code> <![CDATA[76]]> </code> <message> <![CDATA[Please set an "image" parameter with image data for value]]> </message> </error> </errors> </prestashop> Link to comment Share on other sites More sharing options...
massimopasquali Posted April 17, 2022 Author Share Posted April 17, 2022 [SOLVED] byte[] ImageData; string filedto = FotoDto.UrlFolder; //FotoDto.Id_Product ResponseWebService = new ResponseWebServiceDto(); string webAddress = FotoDto.UrlPrestashop.Replace("languages", "images/products") + FotoDto.Id_Product + "?ws_key=" + key; ImageData = File.ReadAllBytes(filedto); using (var client = new HttpClient()) { var requestContent = new MultipartFormDataContent(); var imageContent = new ByteArrayContent(ImageData); imageContent.Headers.ContentType = MediaTypeHeaderValue.Parse("image/jpeg"); requestContent.Add(imageContent, "image", FotoDto.Name + ".jpg"); var response = client.PostAsync(new Uri(webAddress), requestContent).Result; ResponseWebService.HttpStatus = response.StatusCode; ResponseWebService.Xml = response.ReasonPhrase; } return ResponseWebService; Link to comment Share on other sites More sharing options...
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