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Here is the code. This code show the error that there is uri that is null.

public void uploadImage() {

       final StorageReference reference = storage.getReference()
               .child("Images");

The line below reference.putFile(uri).addOnSuccessListener(new OnSuccessListener<UploadTask.TaskSnapshot>() shows an error. In this line uri is null. So how to remove this error.

reference.putFile(uri).addOnSuccessListener(new OnSuccessListener<UploadTask.TaskSnapshot>() {
           @Override
           public void onSuccess(UploadTask.TaskSnapshot taskSnapshot) {

               reference.getDownloadUrl().addOnSuccessListener(new OnSuccessListener<Uri>() {
                   @Override
                   public void onSuccess(Uri uri) {

                       database.getReference().child("Image")
                               .setValue(uri.toString()).addOnSuccessListener(new OnSuccessListener<Void>() {
                           @Override
                           public void onSuccess(Void unused) {

                               Toast.makeText(getApplicationContext(), "Image Uploaded", Toast.LENGTH_SHORT).show();
                               uploadRecipeCredential();
                           }
                       });
                   }
               });
           }
       });
   }

And this is the error showing in log cat

Process: com.example.universityofloralaicafeteria, PID: 12956
    java.lang.IllegalArgumentException: uri cannot be null
        at com.google.android.gms.common.internal.Preconditions.checkArgument(com.google.android.gms:play-services-basement@@17.1.0:35)
        at com.google.firebase.storage.StorageReference.putFile(StorageReference.java:239)
        at com.example.universityofloralaicafeteria.Upload_Recipe.uploadImage(Upload_Recipe.java:95)
        at com.example.universityofloralaicafeteria.Upload_Recipe$3.onClick(Upload_Recipe.java:75)
        at android.view.View.performClick(View.java:6897)
Jabbar Kakar
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  • The error message clearly indicates that the `uri` object on which you are calling `.toString()` is null. – Alex Mamo Mar 08 '22 at 08:32

0 Answers0