Question

抱歉，措辞很糟糕，但是在onAppear中，我正在更改一个State变量，但是除非在视图内部的某个地方使用该State变量，否则这不会重新渲染视图。

如果用户未登录，我试图弹出一个模式，但是除非我将变量放在如下所示的视图中，否则它将不会弹出

struct SearchView: View {
// Environment Variables
@EnvironmentObject var session: SessionStore

@State var sheetIsPresented: Bool = false

var body: some View {
    NavigationView {
        // Just having a line like this will cause it to work
        if self.sheetIsPresented || !self.sheetIsPresented {}
    }
    .onAppear {
        if self.session.userSession != nil {
            self.sheetIsPresented = true
        }
    }
    .sheet(isPresented: $sheetIsPresented) {
        WelcomeSignInModal(sheetIsPresented: self.$sheetIsPresented)
    }
}
}

Answer 1

您可以尝试以下演示：

class SessionStore: ObservableObject {
    var userSession: String? = "session"
}

struct ContentView: View {
    @EnvironmentObject var session: SessionStore

    @State var sheetIsPresented: Bool = false

    var body: some View {
        NavigationView {
            Text("SearchView")
        }
        .onAppear {
            self.sheetIsPresented = self.session.userSession != nil
        }
        .sheet(isPresented: $sheetIsPresented) {
            Text("WelcomeSignInModal")
        }
    }
}

在Xcode 11.6，iOS 13.6上进行了测试

除非在内部视图中使用，否则SwiftUI状态变量不呈现视图（修改器中的经文）

1 个答案: