我想知道是否有一种方法来处理用户在输入EditText时按下Enter,就像onSubmit HTML事件。

还想知道是否有一种方法来操纵虚拟键盘,以这样的方式,“完成”按钮被标记为其他的东西(例如“Go”),并在单击时执行特定的动作(再次,像onSubmit)。


当前回答

作为Chad响应的补充(对我来说几乎完美),我发现我需要在KeyEvent操作类型上添加一个检查,以防止代码执行两次(一次在key-up事件上执行,一次在key-down事件上执行)。

if (actionId == EditorInfo.IME_NULL && event.getAction() == KeyEvent.ACTION_DOWN)
{
    // your code here
}

有关重复动作事件(按住enter键)的信息,请参见http://developer.android.com/reference/android/view/KeyEvent.html。

其他回答

你应该这样做。它也隐藏在Android开发者的样例代码“蓝牙聊天”中。用你自己的变量和方法替换“example”的粗体部分。

首先,把你需要的东西导入到主Activity中,你想让返回按钮做一些特殊的事情:

import android.view.inputmethod.EditorInfo;
import android.widget.TextView;
import android.view.KeyEvent;

现在,创建一个类型为TextView的变量。OnEditorActionListener作为返回键(这里我使用exampleListener);

TextView.OnEditorActionListener exampleListener = new TextView.OnEditorActionListener(){

然后,您需要告诉侦听器当按下返回按钮时该做什么。它需要知道我们谈论的EditText是什么(这里我使用exampleView),然后它需要知道当按下Enter键时要做什么(这里是example_confirm())。如果这是活动中最后一个或唯一一个EditText,它应该做与提交(或确定,确认,发送,保存等)按钮的onClick方法相同的事情。

public boolean onEditorAction(TextView exampleView, int actionId, KeyEvent event) {
   if (actionId == EditorInfo.IME_NULL  
      && event.getAction() == KeyEvent.ACTION_DOWN) { 
      example_confirm();//match this behavior to your 'Send' (or Confirm) button
   }
   return true;
}

最后,设置监听器(很可能在你的onCreate方法中);

exampleView.setOnEditorActionListener(exampleListener);

这个问题还没有被Butterknife回答

布局的XML

<android.support.design.widget.TextInputLayout
        android:layout_width="match_parent"
        android:layout_height="wrap_content"
        android:hint="@string/some_input_hint">

        <android.support.design.widget.TextInputEditText
            android:id="@+id/textinput"
            android:layout_width="match_parent"
            android:layout_height="wrap_content"
            android:imeOptions="actionSend"
            android:inputType="text|textCapSentences|textAutoComplete|textAutoCorrect"/>
    </android.support.design.widget.TextInputLayout>

JAVA应用程序

@OnEditorAction(R.id.textinput)
boolean onEditorAction(int actionId, KeyEvent key){
    boolean handled = false;
    if (actionId == EditorInfo.IME_ACTION_SEND || (key.getKeyCode() == KeyEvent.KEYCODE_ENTER)) {
        //do whatever you want
        handled = true;
    }
    return handled;
}

下面是一个简单的静态函数,您可以将它扔到Utils或keyboard类中,当用户敲击硬件或软件键盘上的返回键时,它将执行代码。这是@earlcasper精彩回答的修改版

 /**
 * Return a TextView.OnEditorActionListener that will execute code when an enter is pressed on
 * the keyboard.<br>
 * <code>
 *     myTextView.setOnEditorActionListener(Keyboards.onEnterEditorActionListener(new Runnable()->{
 *         Toast.makeText(context,"Enter Pressed",Toast.LENGTH_SHORT).show();
 *     }));
 * </code>
 * @param doOnEnter A Runnable for what to do when the user hits enter
 * @return the TextView.OnEditorActionListener
 */
public static TextView.OnEditorActionListener onEnterEditorActionListener(final Runnable doOnEnter){
    return (__, actionId, event) -> {
        if (event==null) {
            if (actionId == EditorInfo.IME_ACTION_DONE) {
                // Capture soft enters in a singleLine EditText that is the last EditText.
                doOnEnter.run();
                return true;
            } else if (actionId==EditorInfo.IME_ACTION_NEXT) {
                // Capture soft enters in other singleLine EditTexts
                doOnEnter.run();
                return true;
            } else {
                return false;  // Let system handle all other null KeyEvents
            }
        } else if (actionId==EditorInfo.IME_NULL) {
            // Capture most soft enters in multi-line EditTexts and all hard enters.
            // They supply a zero actionId and a valid KeyEvent rather than
            // a non-zero actionId and a null event like the previous cases.
            if (event.getAction()==KeyEvent.ACTION_DOWN) {
                // We capture the event when key is first pressed.
                return true;
            } else {
                doOnEnter.run();
                return true;   // We consume the event when the key is released.
            }
        } else {
            // We let the system handle it when the listener
            // is triggered by something that wasn't an enter.
            return false;
        }
    };
}

I had a similar purpose. I wanted to resolve pressing the "Enter" key on the keyboard (which I wanted to customize) in an AutoCompleteTextView which extends TextView. I tried different solutions from above and they seemed to work. BUT I experienced some problems when I switched the input type on my device (Nexus 4 with AOKP ROM) from SwiftKey 3 (where it worked perfectly) to the standard Android keyboard (where instead of handling my code from the listener, a new line was entered after pressing the "Enter" key. It took me a while to handle this problem, but I don't know if it will work under all circumstances no matter which input type you use.

这是我的解决方案:

在xml中设置TextView的输入类型属性为"text":

android:inputType="text"

自定义键盘上“Enter”键的标签:

myTextView.setImeActionLabel("Custom text", KeyEvent.KEYCODE_ENTER);

将OnEditorActionListener设置为TextView:

myTextView.setOnEditorActionListener(new OnEditorActionListener()
{
    @Override
    public boolean onEditorAction(TextView v, int actionId,
        KeyEvent event)
    {
    boolean handled = false;
    if (event.getAction() == KeyEvent.KEYCODE_ENTER)
    {
        // Handle pressing "Enter" key here

        handled = true;
    }
    return handled;
    }
});

我希望这能帮助其他人避免我遇到的问题,因为它们几乎把我逼疯了。

在xml中,将imeOptions属性添加到editText中

<EditText
    android:id="@+id/edittext_additem"
    ...
    android:imeOptions="actionDone"
    />

然后,在Java代码中,将OnEditorActionListener添加到相同的EditText中

mAddItemEditText.setOnEditorActionListener(new TextView.OnEditorActionListener() {
        @Override
        public boolean onEditorAction(TextView v, int actionId, KeyEvent event) {
            if(actionId == EditorInfo.IME_ACTION_DONE){
                //do stuff
                return true;
            }
            return false;
        }
    });

下面是解释 imeOptions=actionDone将“actionDone”分配给EnterKey。键盘上的EnterKey将从“Enter”变为“Done”。因此,当按下回车键时,它将触发这个动作,因此您将处理它。