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

首先,您必须设置EditText监听按键

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

    // Set the EditText listens to key press
    EditText edittextproductnumber = (EditText) findViewById(R.id.editTextproductnumber);
    edittextproductnumber.setOnKeyListener(this);

}

其次,定义按键时的事件,例如,设置TextView的文本的event:

@Override
public boolean onKey(View v, int keyCode, KeyEvent event) {
    // TODO Auto-generated method stub

 // Listen to "Enter" key press
 if ((event.getAction() == KeyEvent.ACTION_DOWN) && (keyCode == KeyEvent.KEYCODE_ENTER))
 {
     TextView textviewmessage = (TextView) findViewById(R.id.textViewmessage);
     textviewmessage.setText("You hit 'Enter' key");
     return true;
 }

return false;   

}

最后,不要忘记在顶部导入EditText,TextView,OnKeyListener,KeyEvent:

import android.view.KeyEvent;
import android.view.View.OnKeyListener;
import android.widget.EditText;
import android.widget.TextView;

其他回答

这将在用户按下返回键时为您提供一个可调用的函数。

fun EditText.setLineBreakListener(onLineBreak: () -> Unit) {
    val lineBreak = "\n"
    doOnTextChanged { text, _, _, _ ->
        val currentText = text.toString()

        // Check if text contains a line break
        if (currentText.contains(lineBreak)) {

            // Uncommenting the lines below will remove the line break from the string
            // and set the cursor back to the end of the line

            // val cleanedString = currentText.replace(lineBreak, "")
            // setText(cleanedString)
            // setSelection(cleanedString.length)

            onLineBreak()
        }
    }
}

使用

editText.setLineBreakListener {
    doSomething()
}

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;
}

我想知道是否有办法 处理用户按Enter键时 输入EditText之类的 onSubmit HTML事件。

是的。

也在想有没有办法 操作虚拟键盘 就像“完成”按钮一样 标记其他东西(例如 “Go”)并执行某个动作 当点击(再次,像onSubmit)。

也没错。

你会想看看android:imeActionId和android:imeOptions属性,加上setOnEditorActionListener()方法,所有的TextView。

要将“Done”按钮的文本更改为自定义字符串,请使用:

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

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。

Kotlin解决方案的反应进入按使用Lambda表达式:

        editText.setOnKeyListener { _, keyCode, event ->
            if(keyCode == KeyEvent.KEYCODE_ENTER && event.action==KeyEvent.ACTION_DOWN){
            //react to enter press here
            }
            true
        }

不做额外的检查类型的事件将导致这个监听器被调用两次时按一次(一次为ACTION_DOWN,一次为ACTION_UP)