【发布时间】:2014-04-26 16:43:16
【问题描述】:
我正在尝试编写代码,该代码将弹出一个输入框并将一个按钮添加到 LinearLayout 中,并将用户输入的任何内容作为按钮的文本。这是我的弹出框代码
public void submitPublicQuestion(View view) {
AlertDialog.Builder question = new AlertDialog.Builder(this);
question.setTitle("Submit Question");
question.setMessage("Enter question: ");
// Set an EditText view to get user input
final EditText input = new EditText(this);
final View queue = this.findViewById(R.layout.activity_main_class);
final Button newBtn = new Button(this);
newBtn.setId((int) (Math.random() * 0xFFFFFFFF));
question.setView(input);
question.setPositiveButton("Submit", new DialogInterface.OnClickListener() {
public void onClick(DialogInterface dialog, int whichButton) {
String value = input.getText().toString();
newBtn.setText(value);
newBtn.setMinWidth(550);
newBtn.setTextSize(24);
newBtn.setVisibility(View.VISIBLE);
LinearLayout ll = (LinearLayout)findViewById(R.id.CSQuestionsLayout);
LayoutParams lp = new LayoutParams(LayoutParams.MATCH_PARENT, LayoutParams.WRAP_CONTENT);
newBtn.setLayoutParams(lp);
ll.addView(newBtn);
goToQueue(queue);
}
});
question.setNegativeButton("Cancel", new DialogInterface.OnClickListener() {
public void onClick(DialogInterface dialog, int whichButton) {
}
});
question.show();
}
当我调用ll.addView(newBtn); 时,此代码会导致 NullPointerException。这里有什么问题?
【问题讨论】:
标签: android android-layout nullpointerexception android-view android-button