Question

我正在尝试从字典数组中访问以下项目，我有两个问题（两种方法都不同）。字典数组初始化如下：

var testingArray = [[String: String]()]

testingArray.append(["name": "Ethiopia", "url": "localhost:8088"])
testingArray.append(["name": "Bugatti", "url": "localhost:8088"])
testingArray.append(["name": "Brazil", "url": "localhost:8088"])
testingArray.append(["name": "Jasmine", "url": "localhost:8088"])
testingArray.append(["name": "Hello", "url": "localhost:8088"])

第一种方法：

for (k,v) in testingArray {
    // code here
}

因为（在for循环初始化的行上显示）而无法运行：

"Expression type '[[String : String]]' is ambiguous without more context

第二种方法：

for indices in testingArray {
    for(k, v) in indices {
        print(indices.keys)
    }
}

返回以下内容：

LazyMapCollection<Dictionary<String, String>, String>(_base: ["url": "localhost:8088", "name": "Ethiopia"], _transform: (Function))

LazyMapCollection<Dictionary<String, String>, String>(_base: ["url": "localhost:8088", "name": "Ethiopia"], _transform: (Function))

LazyMapCollection<Dictionary<String, String>, String>(_base: ["url": "localhost:8088", "name": "Bugatti"], _transform: (Function))

LazyMapCollection<Dictionary<String, String>, String>(_base: ["url": "localhost:8088", "name": "Bugatti"], _transform: (Function))

LazyMapCollection<Dictionary<String, String>, String>(_base: ["url": "localhost:8088", "name": "Brazil"], _transform: (Function))

LazyMapCollection<Dictionary<String, String>, String>(_base: ["url": "localhost:8088", "name": "Brazil"], _transform: (Function))

LazyMapCollection<Dictionary<String, String>, String>(_base: ["url": "localhost:8088", "name": "Jasmine"], _transform: (Function))

LazyMapCollection<Dictionary<String, String>, String>(_base: ["url": "localhost:8088", "name": "Jasmine"], _transform: (Function))

LazyMapCollection<Dictionary<String, String>, String>(_base: ["url": "localhost:8088", "name": "Hello"], _transform: (Function))

LazyMapCollection<Dictionary<String, String>, String>(_base: ["url": "localhost:8088", "name": "Hello"], _transform: (Function))

这是我想要实现的伪代码：

for(int i = 0; i < sizeOfArray; i++ {
    print testingArray[i]."name"
    print testingArray[i]."url"
}

我已经对此感到头疼了好几天，但我不太清楚它的成语足以单独解决这个问题，任何帮助都会非常感激（特别是如果我们能弄明白如何获得＃1）工作）。

Answer 1

我同意错误信息令人困惑/误导。但是for (k,v) in testingArray没有意义，因为testingArray是一个数组，而不是字典。它的元素是字典。

我认为你正在寻找这样的东西：

for obj in testingArray {
    print(obj["name"])
    print(obj["url"])
}

Answer 2

这有效

for (k,v) in testingArray.enumerated() {
    for (_, element) in k.enumerated(){
       guard let elem = element.value as? [String: Any] else {
            continue
       }
       print elem["name"]
    }
}

Swift for-loop＆＃34;表达式类型＆＃39; [[String：String]]＆＃39;没有更多背景

2 个答案: