按下时切换 UIButton 状态，如开关-移动开发问题

toggle UIButton-state when pressing, like a switch(按下时切换 UIButton 状态，如开关)

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问题描述

-(void)setState:(id)sender
{
    UIButton* button = (UIButton*)sender;
    BOOL buttonBool = ([button state]==selected : YES ? NO);
    [sender setSelected:buttonBool++];

}

这是我的想法，但我无法弄清楚调用功能的按钮的实际状态.任何调用此功能的按钮都应在默认状态和选定状态之间切换，以便它像开关一样工作.我只需要拥有自己的背景图形，否则我使用了开关.

this is my idea, but i cant figure out the actual state of the button calling the funktion. any button, that calls this funktion, should be toggled between default and selected-state, so that it works like a switch. i just need to have own background-graphics, else i had used a switch.

那么，谁能纠正我的想法?

so, can anyone please correct my idea ?

推荐答案

试试这个方法:

-(void)setState:(id)sender
{
    UIButton* button = (UIButton*)sender;
    button.selected = !button.selected;
}

这篇关于按下时切换 UIButton 状态，如开关的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持编程学习网！