java - 获取文本无法正常工作

Question

目前，我正在对“提交”按钮进行编程以从 EditText 获取文本。我已经实现了检测空文本的必要条件。但是，当我运行模拟器时，我插入了一个随机字符串并单击提交，然后我收到了 Toast 消息 - 请在框中写点什么？

请看代码和截图

Diary_Entry:

package com.example.ali.googleandroid;

import android.content.Intent;
import android.support.v7.app.AppCompatActivity;
import android.os.Bundle;
import android.view.View;
import android.widget.Button;
import android.widget.EditText;
import android.widget.Toast;

public class Diary_Entry extends AppCompatActivity {

    @Override
    protected void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.diary_entry);

        EditText entryBox = (EditText) findViewById(R.id.diaryentry_box);
        Button submitEntry = (Button) findViewById(R.id.submit);

        final String writtenText = entryBox.getText().toString().trim();

        submitEntry.setOnClickListener(new View.OnClickListener() {
            @Override
            public void onClick(View v) {

                // check for empty text if submit button is pressed
                if ( writtenText.isEmpty() || writtenText.length() == 0 || writtenText.equals("") || writtenText == null ) {

                    Toast.makeText(getApplicationContext(), "Please write something in the box!",
                            Toast.LENGTH_LONG).show();
                } else {
                    // pass the text to display entry box and close activity
                    Intent i = new Intent(Diary_Entry.this, Entry_Dislplay.class);
                    i.putExtra("Submitted Entry", writtenText);
                    startActivity(i);
                    finish();
                }
            }
        });

    }


    public void GoBack(View v) {
        // close current activity and go back to previous activity
        onBackPressed();
        finish();
    }


}

Answer 1

您必须在按下按钮时检索 Edittext 的文本，不能在此之前。

移动这条线

final String writtenText = entryBox.getText().toString().trim();

在 onClick 回调中:

public void onClick(View v) {
    String writtenText = entryBox.getText().toString().trim();
    ....
}