Question

我有一个EditText，让我输入一个IP地址：在xml中：

        <EditText
            android:layout_width="wrap_content"
            android:layout_height="wrap_content"
            android:ems="10"
            android:id="@+id/enterIP"
            android:layout_weight="1"
            android:onClick="enterIP"
            android:inputType="textPhonetic"
            android:imeOptions="actionDone"
            />

我使用了android：imeOptions =＆＃34; actionDone＆＃34;按下完成后，输入框将消失。在Java中：

public void enterIP(View view) {
        EditText theIP = (EditText) findViewById(R.id.enterIP);
        try {
            myIP = theIP.getText().toString(); 
            validIP = ipvalidator.validate(myIP);
        } catch (NullPointerException e)
        {
            Log.d("Error", "Input address is NULL.");
        }
        Toast.makeText(getApplicationContext(), "New IP is " + myIP, Toast.LENGTH_LONG).show();
    }

然而，问题是当我按下Done时，myIP仍保留旧值。只有当我触摸EditText再次调出输入时，才会更新该值。

那么我怎样才能确保在按下Done时更新myIP。？

由于

Answer 1

试试这个

 yourEditText.setOnEditorActionListener(new TextView.OnEditorActionListener() {
            @Override
            public boolean onEditorAction(TextView v, int actionId, KeyEvent event) {
                if (actionId == EditorInfo.IME_ACTION_DONE) {
                    //do something
                    myIP = theIP.getText().toString(); 
            validIP = ipvalidator.validate(myIP);
                }
                return false;
            }
        });

Answer 2

使用此方法：

theIP.setOnEditorActionListener(new OnEditorActionListener() {
@Override
public boolean onEditorAction(TextView v, int actionId, KeyEvent event) {
    if (actionId == EditorInfo.IME_ACTION_DONE) {
        // Do whatever you want here
        return true;
    }
    return false;
}

Answer 3

你应该为EditText设置onEditorActionListener来实现你想要执行的动作。

java.lang.IllegalStateException: No WebApplicationContext found: not in a DispatcherServlet request and no ContextLoaderListener registered?

按下完成后，Android，EditText不会更新变量

3 个答案: