Question

我正在调用函数validate()如果条件为真，它应该进入return并保持在该页面上，但是当我调试时即使条件为真，它也不会返回。任何人都可以帮助我。我已经尝试了很多，并决定发布。请帮忙

 if(balance>0) {

                   final AlertDialog.Builder alertDialog = new AlertDialog.Builder(DetailActivity.this);
                   alertDialog.setTitle("Confirmation");
                   alertDialog.setMessage("Order has been removed successfully.\n\nRepayable Amount is Rs " + balance);

                   final LinearLayout diagLayout = new LinearLayout(OrderDetailsActivity.this);
                   diagLayout.setOrientation(LinearLayout.VERTICAL);
                   final EditText  text = new EditText(OrderDetailsActivity.this);
                   text.setPadding(10, 10, 10, 10);
                   text.setHint("User Pin");
                   text.setGravity(Gravity.CENTER);
                   text.setTextSize(20);
                   diagLayout.addView(text);
                   alertDialog.setView(diagLayout);

                   alertDialog.setPositiveButton("REFUND", new DialogInterface.OnClickListener() {
                       public void onClick(DialogInterface dialog, int which) {

                         validate(text);
                    Intent intent = new Intent(getApplicationContext(), ScreenActivity.class);
                               startActivity(intent);

                       }
                   });

                   alertDialog.setNegativeButton("TO ACCOUNT", new DialogInterface.OnClickListener() {
                       public void onClick(DialogInterface dialog, int which) {

                          removeOrder();
                           dialog.cancel();
                       }
                   });


                   alertDialog.show();

               }
               else {
                   AlertDialog.Builder builder = new AlertDialog.Builder(OrderDetailsActivity.this);

                   builder.setMessage("Order has been removed successfully")
                           .setTitle("Removed");

                   builder.setPositiveButton(R.string.ok, new DialogInterface.OnClickListener() {
                       @Override
                       public void onClick(DialogInterface dialog, int i) {
                           dialog.dismiss();
                           finish();
                       }
                   });

                   AlertDialog dialog = builder.create();
                   dialog.show();
                   hideProgressWheel(true);
               }
            }

            @Override
            public void onErrorResponse(RPCRequest request, ResponseError error) {
                showAlert(error.getErrorSummary(), error.getErrorTitle());
                hideProgressWheel(true);
            }
        };



        private void validate(EditText text) {
            String pin = text.getText().toString();
            if (pin == null || pin.isEmpty()) {
                showToast("enter user pin");
                return;
            }

        }

Answer 1

如果我找到你的话，你将从validate()方法返回，而不是从正回调中返回。

Answer 2

只返回布尔值，否则不会在void中使用return，因为它只会从validate中返回而不是从你调用的地方返回

 private boolean validate(EditText text) {
        String pin = text.getText().toString();
        if (pin == null || pin.isEmpty()) {
            showToast("enter user pin");
            return false;
        }
        return true;
    }

并在调用网站检查这样

if(!validate(text)) return;

然后它不会再进一步了。

不给我回报价值

2 个答案: