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MFC：使对话框始终位于其他对话框的小集之上

[英]MFC: Make dialog always above A SMALL SET of other dialogs

原文 2011-10-04 05:16:15 4 1 c++/ mfc/ dialog/ z-order

Google-ing yields results on how to make a dialog "always on top" of everything. Google-ing产生了有关如何使对话框“始终位于一切之上”的结果。 This is not what I want, though. 不过，这不是我想要的。

Say I have a huge amount of dialogs. 说我有很多对话。 Within this pile of dialogs, consider 3 of them -- A, B and C. 在这堆对话框中，考虑其中的三个-A，B和C。

I want C to always be on top of A and B, but not the rest of the pile. 我希望C始终位于A和B的顶部，而不是其余部分。 In other words, if C is active, then it should be on top, otherwise it will be below another active dialog, except for when it involves A and B. 换句话说，如果C处于活动状态，则它应位于顶部，否则它将位于另一个活动对话框的下方，除非涉及A和B。

And by the way, A and B can never appear at the same time. 顺便说一下，A和B永远不会同时出现。

With that, the only way I've managed to do it is by destroying C every time, and then re-create it with a new parent (A or B, depending who is visible). 这样，我设法做到的唯一方法是每次都销毁C，然后用新的父代（A或B，取决于谁可见）重新创建它。 I've learned that once child dialogs are created, the ownership cannot be transferred, hence the need to destroy. 我了解到，一旦创建了子对话框，就无法转让所有权，因此需要销毁。

I cannot afford to keep destroying it (long story to it). 我无法继续销毁它（长话不说）。

Any other way to achieve this? 还有其他方法可以做到这一点吗？ Thanks! 谢谢！

1 个解决方案

With that, the only way I've managed to do it is by destroying C every time, and then re-create it with a new parent (A or B, depending who is visible). 这样，我设法做到的唯一方法是每次都销毁C，然后用新的父代（A或B，取决于谁可见）重新创建它。

If you have the handle to C's window in A or B, how about sending a user-defined message to C, when ever you want to set C as TopMost ? 如果您拥有A或B中C的窗口的句柄，那么当您要将C设置为TopMost时，如何向C发送用户定义的消息呢？