Question

我在我的应用程序中已经按照以下示例代码搜索功能 https://www.raywenderlich.com/157864/uisearchcontroller-tutorial-getting-started

一切都很好，搜索工作也很完美。但是当我以不同的顺序/顺序输入单词时，它不会给我结果。下面是我想要的例子

var String = "High, Fever"
var String = "fever"

现在，当我搜索＆＃34;发烧＆＃34;它给了我两个回应，但当我搜索类似＆＃34;发烧h＆＃34;它没有给我字符串..在我的情况下，他们只是输入不管他们的顺序的单词..

以下是我的代码

func filterContentForSearchText(_ searchText: String, scope: String = "Present") {
        filteredsymptoms = symptoms.filter({( symptoms : Symptoms) -> Bool in
            let doesCategoryMatch = (scope == "Present") || (scope == "Absent") || (scope == "Focus")

            if searchBarIsEmpty() {
                return doesCategoryMatch
            } else {
                return doesCategoryMatch && symptoms.name.lowercased().contains(searchText.lowercased())
            }
        })
        print(filteredsymptoms)
        self.tblListOfSymptoms.reloadData()
    }

Answer 1

symptoms.name.lowercased().contains(searchText.lowercased()

这部分代码将检查“symptoms.name”是否包含整个搜索文本，在您的情况下是“fever h”。如果要搜索以空格分隔的文本的所有部分，则应使用以下内容：

func filterContentForSearchText(_ searchText: String, scope: String = "Present") {
    filteredsymptoms = symptoms.filter({( symptoms : Symptoms) -> Bool in
        let doesCategoryMatch = (scope == "Present") || (scope == "Absent") || (scope == "Focus")

        if searchBarIsEmpty() {
            return doesCategoryMatch
        } else {
            var result:Bool = false
            let searchTerms = searchText.lowercased().components(separatedBy: CharacterSet.whitespacesAndNewlines)
            for searchTerm in searchTerms
            {
                result = doesCategoryMatch && symptoms.name.lowercased().contains(searchTerm)
                if result == true
                {
                    break
                }
            }
            return result
        }
    })
    print(filteredsymptoms)
    self.tblListOfSymptoms.reloadData()
}

Answer 2

我想这取决于你想要达到的结果。如果您希望用户键入多个单词并搜索其中任何一个单词，请使用以下代码：

searchText.lowercased().components(separatedBy: " ").map({ symptoms.name.lowercased().contains($0) }).reduce(false, { $0 || $1 })

搜索控制器的单词

2 个答案: