How to receive file upload from Angular Reactive F

问题:

I have module to upload file in Angular 7 Reactive Forms ( I need reactive form because I need to upload files and some other information together)

I follow this article: https://medium.com/@amcdnl/file-uploads-with-angular-reactive-forms-960fd0b34cb5

the code as follow: https://pastebin.com/qsbPiJEX

onFileChange(event) {
    const reader = new FileReader();

    if(event.target.files && event.target.files.length) {
      const [file] = event.target.files;
      reader.readAsDataURL(file);

      reader.onload = () => {
        this.formGroup.patchValue({
          file: reader.result
       });

        // need to run CD since file load runs outside of zone
        this.cd.markForCheck();
      };
    }
}

as far I know, I will receive the file as text inside the json data. But I have no idea how to receive this as a file or convert the json data to file. The file could be images, pdf or other docs (excel, docs, or other formats)

I am using Dotnet Core 2.2 and Angular 7 Any idea how to receive the file ?

回答1:

I'm having a form in which i want to post image through formData, i got fileobject in my FormControl just by putting this attribute writeFile="true". This can write the FileList object in your FormControl as value. To Achieve this you need to install the package of '@rxweb/reactive-form-validators' and register the 'RxReactiveFormsModule' module. That's it.

here is my html code :

<form  [formGroup]="userFormGroup">       

        <label>Profile Photo</label>
        <input type="file" [writeFile]="true" formControlName="profilePhoto" multiple />        

      <button [disabled]="!userFormGroup.valid" class="btn btn-primary">Submit</button>
    </form>

Please refer this example : Stackblitz

回答2:

we are sending the file in request-body of the request so we can get the file in request using the below code. So we can access file in our request using below code

using System.IO;  
var filelocation = Path.GetTempFileName();
foreach (var FileData in Request.Form.Files)  
{ 
 if (FileData.Length > 0)  
  {  
    using (var inputStream = new FileStream(filelocation , FileMode.Create))  
    { 

          // read file to stream  
            await FileData.CopyToAsync(inputStream);  
             // stream to byte array  
              byte[] array = new byte[inputStream.Length];  
               inputStream.Seek(0, SeekOrigin.Begin);  

                inputStream.Read(array, 0, array.Length);  
     // get file name  
     string fName = formFile.FileName;  
  }  

}
}