Qt 5，获取鼠标在屏幕上的位置 [英] Qt 5, get the mouse position in a screen

问题描述

首先，我想提到的是，我发现了相关的帖子如何在Qt中获得鼠标在屏幕上的位置？，但对我来说这根本不起作用。我做了一些测试，结果却没有按预期工作，所以我决定写一篇新文章谈论我所做的测试并找到替代解决方案。

这就是我用来进行测试的代码：

  QScreen * screen0 = QApplication :: screens（）。at（ 0）; 
 QScreen * screen1 = QApplication :: screens（）。at（1）; 
 
 printf（ screen0％s \n，screen0-> name（）。toStdString（）。c_str（））; 
 printf（ screen1％s \n，screen1-> name（）。toStdString（）。c_str（））; 
 
 //在第一个屏幕上的位置。 
 QPoint pos0 = QCursor :: pos（screen0）; 
 
 //在第二个屏幕上的位置。 
 QPoint pos1 = QCursor :: pos（screen1）; 
 
 printf（ pos 0：％d，％d \n，pos0.x（），pos0.y（））; 
 printf（ pos 1：％d，％d \n，pos1.x（），pos1.y（））; 
 
 //获取不带屏幕的位置。 
 QPoint pos = QCursor :: pos（）; 
 printf（ pos：％d，％d \n，pos.x（），pos.y（））; 
  

我期望的是，只有一个屏幕会返回有效位置，因为光标仅在一个屏幕上，而不是两个屏幕上。但是事实并非如此，两个位置（ pos0 和 pos1 ）的值完全相同，如我们所见在输出上：

  screen0 DVI-D-0 
 screen1 HDMI-0 
 pos 0：1904 ，1178 
 pos 1：1904，1178 
 pos：1904，1178 
  

由于两个位置的值相同，所以我不知道光标在哪个屏幕上。我不知道这是正常现象还是错误，因为文档没有说明屏幕参数不是鼠标所在的屏幕时会发生什么。

<我的想法是将应用程序（由必须检测选定屏幕的Qt守护程序执行）打开/启动到鼠标所在的屏幕。我知道使用 libX11 是可能的，因为我过去曾经这样做过，但是我需要使用Qt 5，而且我不知道如何使用Qt检测选定的屏幕。 / p>

我还使用 QApplication 和 QDesktopWidget 进行了其他测试没有运气的类。

解决方案

那真的很奇怪。解决方法是，您可以尝试以下操作：

  QPoint globalCursorPos = QCursor :: pos（）; 
 int mouseScreen = qApp-> desktop（）-> screenNumber（globalCursorPos）; 
  

现在您知道了光标所在的屏幕。然后您可以找到该屏幕中光标的位置

  QRect mouseScreenGeometry = qApp-> desktop（）-> screen（mouseScreen）-> geometry（）; 
 QPoint localCursorPos = globalCursorPos-mouseScreenGeometry.topLeft（）; 
  

First of all, I'd like to mention that I found that related post How to get the mouse position on the screen in Qt? but it "just didn't work" for me. I made some tests, and the results didn't work as I expected, so I decided to make a new post to talk about the test I made and to find an alternative solution.

That's the code I used to make the test:

QScreen *screen0 = QApplication::screens().at(0);
QScreen *screen1 = QApplication::screens().at(1);

printf("screen0 %s \n", screen0->name().toStdString().c_str());
printf("screen1 %s \n", screen1->name().toStdString().c_str());

// Position on first screen.
QPoint pos0 = QCursor::pos(screen0);

// Position on second screen.
QPoint pos1 = QCursor::pos(screen1);

printf("pos 0: %d, %d \n", pos0.x(), pos0.y());
printf("pos 1: %d, %d \n", pos1.x(), pos1.y());

// Get position without screen.
QPoint pos = QCursor::pos();
printf("pos: %d, %d \n", pos.x(), pos.y());

What I was expecting, is that only one screen would return a valid position, since the cursor is only at one screen, not on both. But it's not the case, the both positions (pos0 and pos1) has the exactly same value, as we can see on the output:

screen0 DVI-D-0 
screen1 HDMI-0 
pos 0: 1904, 1178 
pos 1: 1904, 1178 
pos: 1904, 1178 

Since the both positions has the same values, I can't know at which screen is the cursor. I don't know if that's a normal behavior or a bug, since the documentation doesn't say what happens when the screen argument isn't the screen where the mouse is.

My idea, is to open/launch an application (executed by a Qt daemon that must detect the selected screen) to the screen where the mouse is. I know that with libX11 it's possible, because I did it in the past, but I need to work with Qt 5, and I can't figure out how to do detect the selected screen with Qt.

I also made other tests, using QApplication and QDesktopWidget classes with no luck.

解决方案

That's really weird. As a workaround, you could try this:

QPoint globalCursorPos = QCursor::pos();
int mouseScreen = qApp->desktop()->screenNumber(globalCursorPos);

Now you know which screen the cursor is in. Then you could find the cursor position within that screen doing this:

QRect mouseScreenGeometry = qApp->desktop()->screen(mouseScreen)->geometry();
QPoint localCursorPos = globalCursorPos - mouseScreenGeometry.topLeft();

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