Android：EditText.getText（）在Android 1.6上返回“”（空）字符串

时间：2011-04-21 08:05:42

标签： android

我想点击一个按钮来获取EditText值。

String ETValue = ((EditText)findViewById(R.id.ETID)).getText().toString().trim();

在其他Android版本上，每个东西都能正常工作，但在1.6上我得到了“”（空）字符串。

Android 1.6上出了什么问题，这是怎么回事？

由于

屏幕截图：

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代码参考：

main.xml中

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

        <TextView 
              android:layout_marginTop="10dip" 
              android:layout_marginLeft="5dip" 
              android:text="Type you text here"  
              android:id="@+id/TextView02" 
              android:layout_height="wrap_content" 
              android:textColor="#333333" 
              android:layout_width="fill_parent" 
              android:textSize="16dip">
              </TextView>

            <LinearLayout android:orientation="horizontal"
             android:layout_width="fill_parent"
             android:layout_height="wrap_content"
             android:id="@+id/ID1">   
                <EditText android:layout_marginLeft="5dip" 
                      android:id="@+id/keyword"
                      android:hint="e.g. Text here" 
                      android:textSize="17dip"
                      android:singleLine="true" 
                      android:layout_height="wrap_content" 
                      android:layout_width="250dip"/>
                <Button android:id="@+id/btnID"
                      android:textStyle="bold" 
                      android:layout_width="wrap_content" 
                      android:layout_height="wrap_content"
                      android:textColor="#FFFFFF"
                      android:background="@drawable/icon"
                      android:gravity="bottom"
                      android:paddingBottom="9dip"
                      android:layout_marginLeft="3dip"/>      
            </LinearLayout>

            <LinearLayout android:orientation="horizontal"
             android:layout_width="fill_parent"
             android:layout_height="wrap_content"
             android:id="@+id/ID2"
             android:visibility="gone">   
                <EditText android:layout_marginLeft="5dip" 
                      android:id="@+id/keyword"
                      android:hint="e.g. Text here"
                      android:textSize="17dip" 
                      android:singleLine="true" 
                      android:layout_height="wrap_content" 
                      android:layout_width="fill_parent"
                      android:layout_marginRight="5dip" />
            </LinearLayout>      

<Button android:text="Click" android:id="@+id/Button01" android:layout_width="fill_parent" android:layout_height="wrap_content"></Button>
</LinearLayout>

Etext.Java 

public class EText extends Activity {
/** Called when the activity is first created. */
@Override
public void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.main);

    ((LinearLayout)findViewById(R.id.ID1)).setVisibility(View.GONE);
    ((LinearLayout)findViewById(R.id.ID2)).setVisibility(View.VISIBLE);

    Button but = (Button) findViewById(R.id.Button01);
    but.setOnClickListener(new OnClickListener() {

        public void onClick(View arg0) {
            String ETValue = ((EditText) findViewById(R.id.keyword)).getText().toString().trim();
            Toast.makeText(EText.this, ETValue, Toast.LENGTH_SHORT).show();
        }
    });
} }
  

即使这样也行不通

public class EText extends Activity {
    /** Called when the activity is first created. */
    EditText ETextt1 = null;
    EditText ETextt2 = null;
    @Override
    public void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.main);

        if(Integer.parseInt(Build.VERSION.SDK) < 7){
            ((LinearLayout)findViewById(R.id.ID1)).setVisibility(View.GONE);
            ((LinearLayout)findViewById(R.id.ID2)).setVisibility(View.VISIBLE);
            ETextt1 = ((EditText) findViewById(R.id.keyword1));
        }else{
            ETextt2 = ((EditText) findViewById(R.id.keyword2));
        }

        Button but = (Button) findViewById(R.id.Button01);
        but.setOnClickListener(new OnClickListener() {

            public void onClick(View arg0) {
                String ETValue = null;
                if(null == ETextt1){
                    ETValue = ETextt2.getText().toString().trim();
                }else if(null == ETextt2){
                    ETValue = ETextt1.getText().toString().trim();
                }

                Toast.makeText(EText.this, ETValue, Toast.LENGTH_SHORT).show();
            }
        });
    } }
  

这完美无瑕： - ）

package com.test.et;

import android.app.Activity;
import android.os.Build;
import android.os.Bundle;
import android.view.View;
import android.view.View.OnClickListener;
import android.widget.Button;
import android.widget.EditText;
import android.widget.LinearLayout;
import android.widget.Toast;

public class EText extends Activity {
    /** Called when the activity is first created. */
    EditText ETextt1 = null;
    EditText ETextt2 = null;
    @Override
    public void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.main);

        if(Integer.parseInt(Build.VERSION.SDK) < 7){
            ((LinearLayout)findViewById(R.id.ID1)).setVisibility(View.GONE);
            ((LinearLayout)findViewById(R.id.ID2)).setVisibility(View.VISIBLE);
        }

        //ETextt = ((EditText) findViewById(R.id.keyword1));

        Button but = (Button) findViewById(R.id.Button01);
        but.setOnClickListener(new OnClickListener() {

            public void onClick(View arg0) {
                String ETValue = null;
                if(Integer.parseInt(Build.VERSION.SDK) < 7){
                    ETValue = ((EditText) findViewById(R.id.keyword2)).getText().toString().trim();
                }else{
                    ETValue = ((EditText) findViewById(R.id.keyword1)).getText().toString().trim();
                }

                Toast.makeText(EText.this, ETValue, Toast.LENGTH_SHORT).show();
            }
        });
    } }

4 个答案:

答案 0 :(得分：7)

我发现了问题来源。您声明了两个具有相同ID的EditTexts。要解决问题，只需将EditText重命名为keyword1和keyword2。然后，从第二个EditText获取文本。

public class EdText extends Activity {
    @Override
    public void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.main);

    ((LinearLayout)findViewById(R.id.ID1)).setVisibility(View.GONE);
    ((LinearLayout)findViewById(R.id.ID2)).setVisibility(View.VISIBLE);

    Button but = (Button) findViewById(R.id.Button01);
    but.setOnClickListener(new OnClickListener() {

        public void onClick(View arg0) {
            String ETValue = ((EditText) findViewById(R.id.keyword2)).getText().toString().trim();
            Toast.makeText(EdText.this, ETValue, Toast.LENGTH_SHORT).show();
        }
    });
} }

布局：

` 

    <TextView 
          android:layout_marginTop="10dip" 
          android:layout_marginLeft="5dip" 
          android:text="Type you text here"  
          android:id="@+id/TextView02" 
          android:layout_height="wrap_content" 
          android:textColor="#333333" 
          android:layout_width="fill_parent" 
          android:textSize="16dip">
          </TextView>

        <LinearLayout android:orientation="horizontal"
         android:layout_width="fill_parent"
         android:layout_height="wrap_content"
         android:id="@+id/ID1">   
            <EditText android:layout_marginLeft="5dip" 
                  android:id="@+id/keyword1"
                  android:hint="e.g. Text here" 
                  android:textSize="17dip"
                  android:singleLine="true" 
                  android:layout_height="wrap_content" 
                  android:layout_width="250dip"/>
            <Button android:id="@+id/btnID"
                  android:textStyle="bold" 
                  android:layout_width="wrap_content" 
                  android:layout_height="wrap_content"
                  android:textColor="#FFFFFF"
                  android:background="@drawable/icon"
                  android:gravity="bottom"
                  android:paddingBottom="9dip"
                  android:layout_marginLeft="3dip"/>      
        </LinearLayout>

        <LinearLayout android:orientation="horizontal"
         android:layout_width="fill_parent"
         android:layout_height="wrap_content"
         android:id="@+id/ID2"
         android:visibility="gone">   
            <EditText android:layout_marginLeft="5dip" 
                  android:id="@+id/keyword2"
                  android:hint="e.g. Text here"
                  android:textSize="17dip" 
                  android:singleLine="true" 
                  android:layout_height="wrap_content" 
                  android:layout_width="fill_parent"
                  android:layout_marginRight="5dip" />
        </LinearLayout>

`

答案 1 :(得分：1)

我刚刚使用EditText挣扎，返回一个空字符串（“”）。 原因是setContentView(R.layout.main);被调用了两次：

private EditText editText1;
private EditText editText2;

public void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.main);  /// !!!
    editText1=(EditText)findViewById(R.id.editText1);
    editText2=(EditText)findViewById(R.id.editText2);

    // some really long and untrivial initialization stuff

    setContentView(R.layout.main);  /// !!!
}

看起来editText1和editText2指向 R.layout.main的前一个版本的部分......

答案 2 :(得分：0)

以下是我的案例测试示例：

public class EdText extends Activity {
/** Called when the activity is first created. */
@Override
public void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.main);

    // string initialised here will always have initial value and never changes
    final String ETValue = ((EditText) findViewById(R.id.EditText01)).getText().toString().trim();

    Button but = (Button) findViewById(R.id.Button01);
    but.setOnClickListener(new OnClickListener() {

        @Override
        public void onClick(View arg0) {
            Toast.makeText(EdText.this, ((EditText) findViewById(R.id.EditText01)).getText().toString().trim(), Toast.LENGTH_SHORT).show();
        }
    });
} }

API 4的一切正常。也许你的问题与读取字符串值时的情况有关。看我的评论。

答案 3 :(得分：0)

public class EditTextDemo extends Activity {     /** 在第一次创建活动时调用。 * / 

Button btnClick;
 String ETValue;
 @Override
    public void onCreate(Bundle savedInstanceState) 
 {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.main);
ETValue = ((EditText)findViewById(R.id.editText1)).getText().toString().trim();
btnClick.setOnClickListener(new OnClickListener() 
{
    @Override
        public void onClick(View view) 
        {
        //assign a default value
        ETValue.equals("Welcome!!");
        Toast.makeText(getBaseContext(), "Value is:"+ETValue, Toast.LENGTH_SHORT).show();
        Log.i("EditTextDemo","-----Value of EditText is:----------"+ETValue);
        }
});

}}

您还可以在运行时给出值。此值也可用于进一步编码。 希望这会有所帮助!!

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