SwiftUI：点击NavigationView.searchable的Cancel按钮如何返回上视图？

你可以通过结合@Environment(\.isSearching)的状态监听，加上SwiftUI的导航 dismiss 机制来实现需求，以下是针对不同iOS版本的具体实现方案：

方案1：iOS 16+ 推荐使用NavigationStack #

利用@Environment(\.dismiss)直接触发当前视图的返回操作：

import SwiftUI

struct SearchContentView: View {
    @Environment(\.isSearching) private var isSearching
    @Environment(\.dismiss) private var dismiss
    @State private var searchText = ""
    
    var body: some View {
        NavigationStack {
            List(["苹果", "香蕉", "橙子"], id: \.self) { fruit in
                Text(fruit)
            }
            .searchable(text: $searchText)
            .onChange(of: isSearching) { oldState, newState in
                // 仅当搜索从激活状态变为取消状态时触发返回
                if oldState && !newState {
                    dismiss()
                }
            }
        }
    }
}

方案2：iOS 15及以下使用NavigationView #

通过@Environment(\.presentationMode)来执行返回：

import SwiftUI

struct SearchContentView: View {
    @Environment(\.isSearching) private var isSearching
    @Environment(\.presentationMode) private var presentationMode
    @State private var searchText = ""
    
    var body: some View {
        NavigationView {
            List(["苹果", "香蕉", "橙子"], id: \.self) { fruit in
                Text(fruit)
            }
            .searchable(text: $searchText)
            .onChange(of: isSearching) { oldState, newState in
                if oldState && !newState {
                    presentationMode.wrappedValue.dismiss()
                }
            }
        }
    }
}

关键注意点 #

• 必须判断oldState为true且newState为false，避免视图初始化时的状态变化误触发返回操作
• 如果使用NavigationStack的路径绑定（比如@State var path: [String] = []），也可以通过path.removeLast()来实现返回，效果和dismiss()一致

内容的提问来源于stack exchange，提问作者Engin Çelik