我无法找到虚拟键盘的尺寸。这是一个例子：
Rectangle {
Component.onCompleted: {
Qt.inputMethod.visibleChanged.connect(resizeForKeyboard)
}
Component.onDestruction: {
Qt.inputMethod.visibleChanged.disconnect(resizeForKeyboard)
}
function resizeForKeyboard(){
console.log('Visibility changed!!!');
var keys = Object.keys(Qt.inputMethod.keyboardRectangle);
var rect = Qt.inputMethod.keyboardRectangle;
//A simple script I have for debugging, this loops
//through the keys and prints all properties
DataMethods.printObject(keys, '[INPUT]');
DataMethods.printObject(rect , '[RECTANGLE]');
}
//using the controls to save time
TextField {
focus: true //force keyboard to show up.
}
}
datamethods.js（相关方法）
/**
* This method prints an object to the console for debug purposes
* obj -> the objec to print
* prefix -> the prefix to append "[prefix] -> "...
* props -> a list of properties to use, otherwiese all will be printed
*/
function printObject(obj, prefix, props) {
if(!prefix)
prefix = "obj";
if(obj){
console.log(prefix + obj + "->" + typeof obj);
if(props){
for(var p in obj)
console.log('\t' + prefix + "["+ p + "] -> '" + obj[p] + "'");
} else {
for(var p in obj)
console.log('\t' + prefix + "["+ p + "] -> '" + obj[p] + "'");
}
} else {
console.log(prefix + "is null");
}
}
以下是输出：
[INPUT]objectName,cursorRectangle,keyboardRectangle,visible,animating,locale,inputDirection,destroyed,destroyed,objectNameChanged,deleteLater,_q_reregisterTimers,cursorRectangleChanged,keyboardRectangleChanged,visibleChanged,animatingChanged,localeChanged,inputDirectionChanged,show,hide,update,reset,commit,invokeAction->object
D/Qt (30122): qrc:js/datamethods.js:153 (printObject): [INPUT][0] -> 'objectName'
D/Qt (30122): qrc:js/datamethods.js:153 (printObject): [INPUT][1] -> 'cursorRectangle'
D/Qt (30122): qrc:js/datamethods.js:153 (printObject): [INPUT][2] -> 'keyboardRectangle'
D/Qt (30122): qrc:js/datamethods.js:153 (printObject): [INPUT][3] -> 'visible'
D/Qt (30122): qrc:js/datamethods.js:153 (printObject): [INPUT][4] -> 'animating'
D/Qt (30122): qrc:js/datamethods.js:153 (printObject): [INPUT][5] -> 'locale'
D/Qt (30122): qrc:js/datamethods.js:153 (printObject): [INPUT][6] -> 'inputDirection'
D/Qt (30122): qrc:js/datamethods.js:153 (printObject): [INPUT][7] -> 'destroyed'
D/Qt (30122): qrc:js/datamethods.js:153 (printObject): [INPUT][8] -> 'destroyed'
D/Qt (30122): qrc:js/datamethods.js:153 (printObject): [INPUT][9] -> 'objectNameChanged'
D/Qt (30122): qrc:js/datamethods.js:153 (printObject): [INPUT][10] -> 'deleteLater'
D/Qt (30122): qrc:js/datamethods.js:153 (printObject): [INPUT][11] -> '_q_reregisterTimers'
D/Qt (30122): qrc:js/datamethods.js:153 (printObject): [INPUT][12] -> 'cursorRectangleChanged'
D/Qt (30122): qrc:js/datamethods.js:153 (printObject): [INPUT][13] -> 'keyboardRectangleChanged'
D/Qt (30122): qrc:js/datamethods.js:153 (printObject): [INPUT][14] -> 'visibleChanged'
D/Qt (30122): qrc:js/datamethods.js:153 (printObject): [INPUT][15] -> 'animatingChanged'
D/Qt (30122): qrc:js/datamethods.js:153 (printObject): [INPUT][16] -> 'localeChanged'
D/Qt (30122): qrc:js/datamethods.js:153 (printObject): [INPUT][17] -> 'inputDirectionChanged'
D/Qt (30122): qrc:js/datamethods.js:153 (printObject): [INPUT][18] -> 'show'
D/Qt (30122): qrc:js/datamethods.js:153 (printObject): [INPUT][19] -> 'hide'
D/Qt (30122): qrc:js/datamethods.js:153 (printObject): [INPUT][20] -> 'update'
D/Qt (30122): qrc:js/datamethods.js:153 (printObject): [INPUT][21] -> 'reset'
D/Qt (30122): qrc:js/datamethods.js:153 (printObject): [INPUT][22] -> 'commit'
D/Qt (30122): qrc:js/datamethods.js:153 (printObject): [INPUT][23] -> 'invokeAction'
D/Qt (30122): qrc:js/datamethods.js:147 (printObject): [RECTANGLE]QRectF(0, 0, 0, 0)->object
我可能会以完全错误的方式解决这个问题，我希望有办法处理它。我需要在设备上使用键盘大小的原因是我可以在我的UI中做出响应，而不是在它下面隐藏控件（这样用户可以继续滚动表单）。
正如您所看到的，Qt.inputMethod的keyboardRectangle属性似乎是默认值。
我应该在哪里检索此信息，信号似乎正确，因为它在打开/关闭虚拟键盘时会触发？
答案 0 :(得分：2)
我能够通过使用QDesktopWidget::availableGeometry();获取可用的屏幕尺寸来获取虚拟键盘尺寸，QDesktopWidget::screenGeometry();以获取全屏尺寸和一些Java代码以获得菜单尺寸。< / p>
以下是代码中最相关的部分：
C ++方面：
QRect KeyboardInterface::rect()
{
int menuheight = (int)QAndroidJniObject::callStaticMethod<jint>("org.qtproject.example.Demo2.JavaInterface", "getHeight");
QDesktopWidget widget;
QRect rect = widget.availableGeometry();
QRect geom = widget.screenGeometry();
rect.moveTop(rect.top() + menuheight);
geom.setTop(geom.top() + menuheight);
QRect final;
if (rect != geom)
{
int ftop, fleft, fwidth, fheight;
geom.getRect(&fleft, &ftop, &fwidth, &fheight);
if (rect.top() != ftop)
fheight = rect.top();
else if (rect.left() != fleft)
fwidth = rect.left();
else if (rect.height() != fheight)
ftop = rect.height();
else if (rect.width() != fwidth)
fleft = rect.width();
final = QRect(fleft, ftop, fwidth - fleft, fheight - ftop);
}
return final;
}
Java方面：
package org.qtproject.example.Demo2;
import android.app.Activity;
import android.graphics.Rect;
import android.view.View;
import android.view.Window;
public class JavaInterface extends org.qtproject.qt5.android.bindings.QtActivity
{
private static JavaInterface instance;
public JavaInterface()
{
instance = this;
}
public static int getHeight()
{
Rect r = new Rect();
Window window = instance.getWindow();
View rootview = window.getDecorView();
rootview.getWindowVisibleDisplayFrame(r);
return r.top;
}
}
可以找到此项目的完整源代码here
希望有所帮助。
答案 1 :(得分：0)
根据此错误报告，这是Qt在Android上的预期行为，因为Android上的软键盘不限于显示在屏幕底部，因为iOS就是这种情况：https://bugreports.qt.io/browse/QTBUG-40731
答案 2 :(得分：0)
OP看到的错误“ Qt.inputMethod.keyboardRectangle始终为0 x 0”行为是a bug in Qt，已在2015-12-07修复，已修复在Qt 5.7中发布。
从那时起，Qt.inputMethod.keyboardRectangle将包含正确的大小，除非在极少数情况下使用浮动Android键盘。在iOS上，键盘只限于屏幕底部，这种情况不会发生。现在，Qt官方文档中也documented有此限制：
QInputMethod :: keyboardRectangle：在窗口坐标中虚拟键盘的几何形状。
如果无法知道键盘的几何形状，则该矩形可能为空。安卓上的浮动键盘就是这种情况。