问题描述

我设置了一个QT菜单，它自动连接到动作函数on_actionOpen_triggered().后来我想将文件名字符串传递给这个函数，以便在特殊情况下手动调用这个函数.所以我将函数签名更改为 on_actionOpen_triggered( const char *filename_in ).修改后程序运行良好，但终端报错，

I set a QT menu, which is automatically connected with action function on_actionOpen_triggered(). Later I want to pass a filename string to this function in order to call this function manually in a special condition. So I changed the function signature to on_actionOpen_triggered( const char *filename_in ). After this change the program is running well, but there is a complain in terminal,

QMetaObject::connectSlotsByName: on_actionOpen_triggered(const char*) 没有匹配的信号

QMetaObject::connectSlotsByName: No matching signal for on_actionOpen_triggered(const char*)

我想知道发生了什么，以及如何为此菜单操作功能添加参数.

I am wondering what happened, and how I can add arguments for this menu action functions.

谢谢.

推荐答案

我遇到了同样的警告/错误 QMetaObject::connectSlotsByName: No matching signal for

并得到了简单的解决方案.例如:

And got simple solution. For Example:

问题:
QMetaObject::connectSlotsByName: on_actionOpen_triggered(const char*) 没有匹配的信号 警告您只需要更改 Slot

Problem :
QMetaObject::connectSlotsByName: No matching signal for on_actionOpen_triggered(const char*) Warning You just need to change the name of the Slot

解决方案
更改插槽名称，如 on_actionOpenTriggered，此警告消失.

提示
Qt 尝试理解其默认插槽，如 on__，因此如果您指定任何具有上述签名的插槽，Qt 将抛出警告.

Hint
Qt try to understand its default slot like on_<name_of_object>_<action>, So if you specify any slot with above signature, Qt will throw warning.

希望对大家有所帮助.

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