代码之家 › 专栏 › 技术社区 › Govaadiyo

从字典数组中搜索结果的最佳方法

filter cocoa-touch swift arrays ios

Govaadiyo · 技术社区 · 7 年前

正如我的标题所说,从字典数组中搜索结果的最佳方法是什么?

前任。

let array = [
    ["name": "Amit", "age": "27", "rollno": 12],
    ["name": "Ananad", "age": "26", "rollno": 10],
    ["name": "Kashyap", "age": "25", "rollno": 11],
    ["name": "Raj", "age": "25", "rollno": 07],
    ["name": "Akshya", "age": "28", "rollno": 13]
]

我想搜索 “名称” 基于用户输入。看看下面我做的事情。

let strSeach = "k"
let searchResult = myArray.filter{String(($0["name"] as! String).characters.prefix(strSeach.characters.count)).lowercased() == strSeach.lowercased()}

print("index => \(searchResult)")

输出为:

搜索结果=>[“rollno”:11,“name”:“Kashyap”,“age”:“25”]]

但困惑的是,这是正确的搜索方式吗?我用过 characters.prefix , lowercased()

如果我想应用多个搜索,比如按“姓名”、“年龄”等进行搜索,那么我们如何在 swift过滤器?

NSPredicate 使用Swift Array ? 否则,我们可以通过以下方式实现

let searchPredicate = NSPredicate(format: "name CONTAINS [c] %@ OR age CONTAINS [c]", strSeach)

像那样!

4 回复 | 直到 7 年前

Sweeper 7 年前

NSPredicate 如果需要,请注意它比 filter .

您只需将swift数组转换为 NSArray filtered .

此外,这里还有另一种使用swift过滤器的方法:

let searchResult = array.filter { 
    ($0["name"] as! String).lowercased().hasPrefix(strSeach.lowercased()) }

Prashant Tukadiya 7 年前

带过滤器

你可以用这个

let array = [
["name": "Amit", "age": "27", "rollno": 12],
["name": "Ananad", "age": "26", "rollno": 10],
["name": "Kashyap", "age": "25", "rollno": 11],
["name": "Raj", "age": "25", "rollno": 07],
["name": "Akshya", "age": "28", "rollno": 13]
]

let searchSting = "am"

var search = array.filter { ($0["name"] as!    String).lowercased().contains(searchSting.lowercased()) }

输出

[["rollno": 12, "name": "Amit", "age": "27"]]

带谓词

let searchPredicate = NSPredicate(format: "name CONTAINS [c] %@ OR age == %@", searchSting)
search = array.filter { searchPredicate.evaluate(with: $0)}

[“rollno”:12,“name”:“Amit”,“age”:“27”]]

Luca Angeletti 7 年前

您应该使用模型值

struct Person {

    let name: String
    let age: Int
    let rollno: Int

    init?(dict:[String:Any]) {
        guard
            let name = dict["name"] as? String,
            let ageText = dict["age"] as? String,
            let age = Int(ageText),
            let rollno = dict["rollno"] as? Int
            else { return nil}

        self.name = name
        self.age = age
        self.rollno = rollno
    }
}

现在,给定您的初始阵列

let data = [
    ["name": "Amit", "age": "27", "rollno": 12],
    ["name": "Ananad", "age": "26", "rollno": 10],
    ["name": "Kashyap", "age": "25", "rollno": 11],
    ["name": "Raj", "age": "25", "rollno": 07],
    ["name": "Akshya", "age": "28", "rollno": 13]
]

Person(s)

let persons = data.flatMap(Person.init)

姓名包含字符串的人员

现在过滤部分很简单,看

let personsWhereNameContainsK = persons.filter { $0.name.lowercased().contains("k") }

后果

[
    Person(name: "Kashyap", age: 25, rollno: 11),
    Person(name: "Akshya", age: 28, rollno: 13)
]

Venk 7 年前

您可以使用 contains 过滤器中的关键字如下,

let searchResult = myArray.filter{String(($0["name"] as! String).lowercased().contains(strSeach.lowercased())}

对于多个过滤器(我根据您的谓词添加条件),

let searchResult = myArray.filter{(String(($0["name"] as! String).lowercased().contains(strSeach.lowercased())) || (String(($0["age"] as! String).lowercased().contains(strSeach.lowercased()))}