Swift中的块显示错误＆＃34;在调用＆＃34;中缺少参数＃2的参数。

Question

我现在正在项目中使用Jonas Gessner的JGActionSheet和Swift，示例是由Objective-C编写的，当我尝试将块转换为Swift时，Xcode显示错误“参数缺少参数＃2在电话中“，这是我写的代码和屏幕截图：

Objective-C Sample

JGActionSheet *sheet = [JGActionSheet actionSheetWithSections:sections];
[sheet setButtonPressedBlock:^(JGActionSheet *sheet, NSIndexPath *indexPath) 
{
    [sheet dismissAnimated:YES];
}];

我用Swift编写的代码

let actionSheet = JGActionSheet(sections: sections)
actionSheet.buttonPressedBlock {
    (sheet: JGActionSheet!, indexPath: NSIndexPath!) in
    actionSheet.dismissAnimated(true)
}

错误截图

Missing argument for parameter #2 in call

所以请帮我解决这个问题并非常感谢！

Answer 1

actionSheet.buttonPressedBlock是属性。您正在尝试设置它。那你的等号在哪里？这就是你在Swift中设置的方法：

 myThing.myProperty = myValue

您尝试将此属性设置为块（函数）的事实不会改变任何内容。所以：

let actionSheet = JGActionSheet(sections: sections)
actionSheet.buttonPressedBlock = {
    (sheet: JGActionSheet!, indexPath: NSIndexPath!) in
    actionSheet.dismissAnimated(true)
}