我使用下面的代码发送图像以发布方法并将其另存为BLOB在数据库中,并且工作正常:
角度代码:
public postUploadedFile(file:any){
this.formData = new FormData();
this.formData.append('file',file,file.name);
this.Url='http://localhost:38300/api/site/PostUploadFiles';
console.log("url passed from here",this.Url)
return this.http.post(this.Url , this.img).subscribe()
}
API代码:
public IHttpActionResult PostUploadFiles()
{
int i = 0;
var uploadedFileNames = new List<string>();
string result = string.Empty;
HttpResponseMessage response = new HttpResponseMessage();
var httpRequest = HttpContext.Current.Request;
if (httpRequest.Files.Count > 0)
{
while(i < httpRequest.Files.Count && result != "Failed")
{
br = new BinaryReader(httpRequest.Files[i].InputStream);
ImageData = br.ReadBytes(httpRequest.Files[i].ContentLength);
br.Close();
if (DB_Operation_Obj.Upload_Image(ImageData) > 0)
{
result = "success";
}
else
{
result = "Failed";
}
i++;
}
}
else
{
result = "can't find images";
}
return Json(result);
}
但是现在我需要发送有关图像(类型ID,名称)的更多信息,而不仅仅是图像,因此角度代码将类似于:
public postUploadedFile(file:any, type_id:number,site_id:number){
this.img = new Image_List();
this.img.images = new Array<PreviewURL>();
this.img.type_id= type_id;
this.img.Reference_id = site_id;
this.img.images.push(file);
this.formData = new FormData();
this.formData.append('file',file,file.name);
this.Url='http://localhost:38300/api/site/PostUploadFiles';
console.log("url passed from here",this.Url)
return this.http.post(this.Url , this.img).subscribe()
}
任何在数据库中发送和插入的帮助。
我认为您可以只制作一个上传文件方法,并使用文件名制作另一个用于数据插入的方法,因此它将类似于:公共postUploadedFile(file:any){this.formData = new FormData(); this.formData.append('file',file,file.name); this.Url ='http://localhost:38300/api/site/PostUploadFiles';this.newMethod(filename); //在这里您上传其他数据console.log(“从此处传递的网址”,this.Url)返回this.http.post(this.Url,this.img).subscribe()}
使用FormData将其他信息附加到api调用。
const formData = new FormData();
formData.append(file.name, file,'some-data');
您可以使用相同名称的多个值。