Android无法得到的EditText的getText()的toString()在对话框 [英] Android can't get EditText getText().toString() in a Dialog

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问题描述

我想创建一个自定义对话框中的一个类。开始在主要活动的对话框:

  DialogLogin登录=新DialogLogin();
        login.show(getFragmentManager(),DISPLAY_SERVICE);
 

在启动主活动开始在后台应用程序,然后在对话框开始。 在对话框有一个EditText场来接收用户输入。在pressing保存按钮的的EditText场应该读出并显示输入,但它总是空的。此外还有在LogCat中没有错误......我想同样的问题的许多解决方案,但毫无效果。我希望任何人有一个有效的解决方案! =)

 进口android.app.AlertDialog;
进口android.app.Dialog;
进口android.app.DialogFragment;
进口android.content.DialogInterface;
进口android.os.Bundle;
进口android.view.LayoutInflater;
进口android.view.View;
进口android.widget.EditText;

公共类DialogLogin扩展DialogFragment {
字符串androidID;
LayoutInflater充气;

公共对话onCreateDialog(包savedInstanceState){

    AlertDialog.Builder建设者=新AlertDialog.Builder(getActivity());
    充气= getActivity()getLayoutInflater()。

    builder.setMessage(R.string.loginMessage)
            .setTitle(R.string.login)
            .setView(inflater.inflate(R.layout.loginlayout,NULL))
            .setPositiveButton(R.string.speichern,
                    新DialogInterface.OnClickListener(){
                        公共无效的onClick(DialogInterface对话框,INT ID){
                            的System.out.println(登录);
                            视图V = inflater.inflate(R.layout.loginlayout,NULL);
                            文字的EditText =(EditText上)v.findViewById(R.id.loginEdit);

                            的System.out.println(text.getText()的toString()); //显示什么
                            的System.out.println(text.length()); //为0

                        }
                    });

    返回builder.create();
}
 

}

而loginlayout.xml:

 < XML版本=1.0编码=UTF-8&GT?;
< LinearLayout中的xmlns:机器人=htt​​p://schemas.android.com/apk/res/android
机器人:layout_width =match_parent
机器人:layout_height =match_parent
机器人:方向=垂直>

<的EditText
    机器人:ID =@ + ID / loginEdit
    机器人:inputType =文本
    机器人:layout_width =match_parent
    机器人:layout_height =WRAP_CONTENT
    机器人:EMS =10
    机器人:提示=@字符串/ EDITTEXT>

< /的EditText>

< / LinearLayout中>
 

解决方案

正在膨胀的一个新的布局,其中的EditText 中没有任何文字。你只需要一次吹你的布局,并保持对它的引用。

 最终视图中查看= inflater.inflate(R.layout.loginlayout,NULL);

/ * ... * /
.setView(视图)
/ * ... * /
文字的EditText =(EditText上)view.findViewById(R.id.loginEdit);
 

I'm trying to create a custom Dialog in an individual class. The dialog is started in the main activity:

        DialogLogin login = new DialogLogin();
        login.show(getFragmentManager(), DISPLAY_SERVICE);

On starting the application the main activity starts in the background and then the dialog starts. In the dialog there is an EditText-field to receive user-input. On pressing the save-button the EditText-field should be read-out and the input displayed, but it's always empty. Moreover there are no errors in the LogCat... I tried many solutions of the same problem, but nothing worked. I hope anyone has a working solution ! =)

import android.app.AlertDialog;
import android.app.Dialog;
import android.app.DialogFragment;
import android.content.DialogInterface;
import android.os.Bundle;
import android.view.LayoutInflater;
import android.view.View;
import android.widget.EditText;

public class DialogLogin extends DialogFragment {
String androidID;
LayoutInflater inflater;

public Dialog onCreateDialog(Bundle savedInstanceState) {

    AlertDialog.Builder builder = new AlertDialog.Builder(getActivity());
    inflater = getActivity().getLayoutInflater();

    builder.setMessage(R.string.loginMessage)
            .setTitle(R.string.login)
            .setView(inflater.inflate(R.layout.loginlayout, null))
            .setPositiveButton(R.string.speichern,
                    new DialogInterface.OnClickListener() {
                        public void onClick(DialogInterface dialog, int id) {
                            System.out.println("LOGIN");
                            View v = inflater.inflate(R.layout.loginlayout, null);
                            EditText text = (EditText) v.findViewById(R.id.loginEdit);

                            System.out.println(text.getText().toString()); //Displays nothing
                            System.out.println(text.length());              //is 0

                        }
                    });

    return builder.create();
}

}

And the loginlayout.xml:

<?xml version="1.0" encoding="utf-8"?>
<LinearLayout xmlns:android="http://schemas.android.com/apk/res/android"
android:layout_width="match_parent"
android:layout_height="match_parent"
android:orientation="vertical" >

<EditText
    android:id="@+id/loginEdit"
    android:inputType="text"
    android:layout_width="match_parent"
    android:layout_height="wrap_content"
    android:ems="10"
    android:hint="@string/editText" >

</EditText>

</LinearLayout>

解决方案

You are inflating a new layout, where the EditText has no text in it. You'll need to only once inflate your layout, and keep a reference to it.

final View view = inflater.inflate(R.layout.loginlayout, null);

/* ... */
.setView(view)
/* ... */
EditText text = (EditText) view.findViewById(R.id.loginEdit);

这篇关于Android无法得到的EditText的getText()的toString()在对话框的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!

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