SwiftUI防止onReceive在加载时开火
当我试图接收的变量是@EnvironmentObject的@Published属性时,当视图最初加载时,是否有任何方法防止onReceive触发 以下是观点:SwiftUI防止onReceive在加载时开火,swift,environmentobject,Swift,Environmentobject,当我试图接收的变量是@EnvironmentObject的@Published属性时,当视图最初加载时,是否有任何方法防止onReceive触发 以下是观点: struct ContentView: View { @EnvironmentObject var appState: AppState var body: some View { VStack { Text(appState.test) .
struct ContentView: View {
@EnvironmentObject var appState: AppState
var body: some View {
VStack {
Text(appState.test)
.padding()
}
.onAppear() {
appState.test = "World"
}
.onReceive(appState.$test) { test in
print("Hello from onReceive: \(test)")
}
}
}
以下是环境对象:
public class AppState: ObservableObject {
@Published public var test = "hello"
}
和输出:
Hello from onReceive: hello
Hello from onReceive: World
我猜第一个是在注入环境对象时触发的。有什么方法可以防止这种情况(除了我在onAppear中设置的一些粗俗的bool)?理想情况下,我只想看到“HellofromOnReceive:World”,这只是一个简单的例子。在我的实际应用程序中,我从onAppear中的服务获取数据,在我的环境对象中,我有一个错误状态,清除后,我希望它触发onReceive,执行一些其他逻辑,然后重新蚀刻。它是发布者,因此您可以使用任何组合运算符。在这种情况下,
.dropFirst
解决您的任务:
.onReceive(appState.$test.dropFirst()) { test in
print("Hello from onReceive: \(test)")
}