Question

我尝试获得软键盘状态，（显示或隐藏）使用onConfigurationChanged（此处：http://developer.android.com/guide/topics/resources/runtime-changes.html#HandlingTheChange）

但它对我不起作用。我不知道。（我也已经尝试过hardKeyboardHidden和keyboard-config的价值）

请检查我的代码。

的AndroidManifest.xml

<application
        android:allowBackup="true"
        android:icon="@mipmap/ic_launcher"
        android:label="@string/app_name"
        android:theme="@style/AppTheme">

        <activity
            android:configChanges="keyboardHidden"
            android:windowSoftInputMode="adjustResize"
            android:name=".MainActivity"
            android:label="@string/app_name" >
            <intent-filter>
                <action android:name="android.intent.action.MAIN" />

                <category android:name="android.intent.category.LAUNCHER" />
            </intent-filter>
        </activity>
    </application>

MainActivity.java

@Override
public void onConfigurationChanged(Configuration newConfig) {
    super.onConfigurationChanged(newConfig);
    // Checks whether a hardware keyboard is available
    if (newConfig.keyboardHidden== Configuration.HARDKEYBOARDHIDDEN_NO) {
        Toast.makeText(this, "keyboard visible", Toast.LENGTH_SHORT).show();
    } else if (newConfig.keyboardHidden== Configuration.HARDKEYBOARDHIDDEN_YES) {
        Toast.makeText(this, "keyboard hidden", Toast.LENGTH_SHORT).show();
    }
}

只是简单地在activity_main.xml中添加了EditText

<RelativeLayout 
xmlns:android="http://schemas.android.com/apk/res/android"
    xmlns:tools="http://schemas.android.com/tools" android:layout_width="match_parent"
    android:layout_height="match_parent" 
tools:context=".MainActivity">

    <EditText
        android:layout_width="match_parent"
        android:layout_height="match_parent" />

</RelativeLayout>

那么如何解决这个问题？

Answer 1

此方法使用// An function that may error and throw exception. function funcThatThrows(test) { throw(`oops - error in test ${test}`); } function errHandler(exception) { console.log('We got an exception: ', exception); } // Expect: We got an exception: oops - error in test 1 Promise.resolve(1).then(funcThatThrows).catch(errHandler); // Is the same as below, the errHandler tries to catch any unhandled error // from previous result. // Expect: We got an exception: oops - error in test 2 Promise.resolve(2).then(funcThatThrows, null).then(null, errHandler); // If put the function and handler in the same then, the exception won't be caught. // Expect: Uncaught (in promise) oops - error in test 3 Promise.resolve(3).then(funcThatThrows, errHandler);。它会检查活动屏幕是否较小。

How to check visibility of software keyboard in Android?

Answer 2

以下解决方案如何？

if exists (select * from INFORMATION_SCHEMA.ROUTINES where ROUTINE_NAME = 'fn_Split' )
  drop function [dbo].[fn_Split]
go
CREATE  FUNCTION fn_Split(@Text varchar(MAX), @Delimiter varchar(20) = ',')
RETURNS @Strings TABLE
(   
  Position int IDENTITY PRIMARY KEY,
  Value varchar(MAX)  
)
AS
BEGIN

DECLARE @Index int = -1

WHILE (LEN(@Text) > 0)
  BEGIN 
    SET @Index = CHARINDEX(@Delimiter , @Text) 

    IF (@Index = 0) AND (LEN(@Text) > 0) 
      BEGIN  
        INSERT INTO @Strings VALUES (@Text)
          BREAK 
      END 

    IF (@Index > 1) 
      BEGIN  
        INSERT INTO @Strings VALUES (LEFT(@Text, @Index - 1))  
        SET @Text = RIGHT(@Text, (LEN(@Text) - @Index)) 
      END 
    ELSE
      SET @Text = RIGHT(@Text, (LEN(@Text) - @Index))      
    END
  RETURN
END
go

if exists (select * from INFORMATION_SCHEMA.TABLES where TABLE_NAME = 'ContentKeyword')
  drop table ContentKeyword;
go
Create Table ContentKeyword  
(
    [KeywordId] [int] IDENTITY(1,1) NOT NULL PRIMARY KEY,
    [Name] [nvarchar](500) NOT NULL
)
go

if exists (select * from INFORMATION_SCHEMA.TABLES where TABLE_NAME = 'ContentKeyword')
  drop table JobKeyword;
go
Create Table JobKeyword 
(
    [KeywordId] [int] IDENTITY(1,1) NOT NULL PRIMARY KEY,
    [Name] [nvarchar](500) NOT NULL
)
go
if exists (select * from INFORMATION_SCHEMA.TABLES where TABLE_NAME = 'ContentKeyword')
  drop table CategoryGroup;
go
Create Table CategoryGroup 
(
    [KeywordId] [int] IDENTITY(1,1) NOT NULL PRIMARY KEY,
    [Name] [nvarchar](500) NOT NULL
)
go
  --Insert dummy data

  insert into CategoryGroup(Name)  values('First value')
go

select S.Value from dbo.fn_Split('1,2,3,4,5', ',') S
  left outer join ContentKeyword CK on CK.KeywordId = s.Value
  left outer join JobKeyword     JK on JK.KeywordId = s.Value
  left outer join CategoryGroup  JG on JG.KeywordId = s.Value
where 
  (1 = 1)
  and CK.KeywordId is null 
  and JK.KeywordId Is null 
  and JG.KeywordId is null

我想在EditTextField可见性模式下获得softkeyboard（IME）状态？

2 个答案: