iOS：如何通过默认浏览器与搜索引擎发起网页搜索（URL Scheme需求）

• 放弃使用x-web-search://这个Scheme，它仅会唤起Safari，无法适配用户设置的默认浏览器。

• 正确的实现思路是：构造用户系统默认搜索引擎的搜索URL，再通过系统API打开该URL，系统会自动调用用户设置的默认浏览器加载搜索结果。

• 具体步骤：

  1. 获取系统默认搜索引擎的标识：通过NSUserDefaults读取DefaultSearchProviderIdentifier字段，该字段对应的值是类似com.apple.Safari.Google、com.apple.Safari.DuckDuckGo这样的标识符。
  2. 根据标识符匹配对应的搜索URL模板：
     • com.apple.Safari.Google → https://www.google.com/search?q=
     • com.apple.Safari.Bing → https://www.bing.com/search?q=
     • com.apple.Safari.Yahoo → https://search.yahoo.com/search?p=
     • com.apple.Safari.DuckDuckGo → https://duckduckgo.com/?q=
  3. 对搜索关键词做URL编码：使用addingPercentEncoding(withAllowedCharacters: .urlQueryAllowed)处理关键词，避免特殊字符破坏URL结构。
  4. 拼接编码后的关键词到搜索模板，生成完整URL，再调用UIApplication.shared.open(_:options:completionHandler:)打开该URL。

• 示例代码片段：

func openSystemSearch(with query: String) {
    guard let defaults = UserDefaults(suiteName: "com.apple.mobile.internetsearch"),
          let providerId = defaults.string(forKey: "DefaultSearchProviderIdentifier"),
          let encodedQuery = query.addingPercentEncoding(withAllowedCharacters: .urlQueryAllowed) else {
        // 若获取失败则使用默认搜索引擎兜底
        if let url = URL(string: "https://www.google.com/search?q=\(encodedQuery ?? "")") {
            UIApplication.shared.open(url)
        }
        return
    }
    
    var searchTemplate = ""
    switch providerId {
    case "com.apple.Safari.Google":
        searchTemplate = "https://www.google.com/search?q="
    case "com.apple.Safari.Bing":
        searchTemplate = "https://www.bing.com/search?q="
    case "com.apple.Safari.Yahoo":
        searchTemplate = "https://search.yahoo.com/search?p="
    case "com.apple.Safari.DuckDuckGo":
        searchTemplate = "https://duckduckgo.com/?q="
    default:
        searchTemplate = "https://www.google.com/search?q="
    }
    
    if let searchUrl = URL(string: searchTemplate + encodedQuery) {
        UIApplication.shared.open(searchUrl)
    }
}

内容的提问来源于stack exchange，提问作者performat