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我想要下面的代码razrada.show();被执行后,我按保存在我的AlertDialog,但它显示并关闭 这是我的代码警报对话框在不按下的情况下关闭anything_android
AlertDialog.Builder razradaPlacanja = new AlertDialog.Builder(NoviRacun.this);
razradaPlacanja.setTitle("Način plaćanja");
LayoutInflater inflater = getLayoutInflater();
View vieww = inflater.inflate(R.layout.razrada, null);
razradaPlacanja.setView(vieww);
final EditText gotovinaEdit = (EditText) vieww.findViewById(R.id.gotovinaEdit);
final EditText karticeEdit = (EditText) vieww.findViewById(R.id.karticeEdit);
razradaPlacanja.setPositiveButton("Save",
new DialogInterface.OnClickListener() {
public void onClick(DialogInterface dialog, int which) {
try {
json.put("TotalCash", gotovinaEdit.getText().toString());
json.put("TotalCreditCards", karticeEdit.getText().toString());
} catch (JSONException e) {
e.printStackTrace();
}
}
});
AlertDialog razrada = razradaPlacanja.create();
razrada.getWindow().setSoftInputMode(WindowManager.LayoutParams.SOFT_INPUT_STATE_VISIBLE);
razrada.show();
//this i want to execute after pressing Save
Racuni racun = getInsertResponse(requestInsert(base64EncodedCredentials, json, httpclient1));
try {
...
我不能把 Racuni racun = getInsertResponse(requestInsert(base64EncodedCredentials,JSON,httpclient1)); 里面的onClick块导致然后racun是无法访问(如果宣布外,需要是最终的,然后我不能asign值)
谢谢你的任何帮助!
但我将如何获得变量racun? – sparrkli
你应该在新方法中处理它。你不能这样做吗? – Szymon
我只是把整个代码放在外面的方法中。感谢您的回复,我会接受答案。 – sparrkli