0

我有两种情况需要将字符串转换为不同的格式。

例如：

case 1:
  string inputs: abc, xyz, mno, & llr   // All Strings from a dictionary
  output: ["abc","xyz", "mno", "llr"]  //I need to get the String array like this.

但是当我使用这段代码时：

 var stringBuilder:[String] = [];
 for i in 0..<4 {
   stringBuilder.append("abc"); //Appends all four Strings from a Dictionary
 }

print(stringBuilder); //Output is 0: abc, 1:xyz like that, how to get desired format of that array like ["abc", "xyz"];

实际使用：

let arr = Array(stringReturn.values);
//print(arr)  // Great, it prints ["abc","xyz"];
let context = JSContext()
context?.evaluateScript(stringBuilder)   
let testFunction = context?.objectForKeyedSubscript("KK")
let result = testFunction?.call(withArguments:arr); // Here when I debugger enabled array is passed to call() like 0:"abc" 1:"xyz". where as it should be passed as above print.

其次，如何快速替换转义字符：我使用了“\”，replaceOccurances(of:"\\'" with:"'");但它没有改变。为什么以及如何逃避该序列。

case 2:
  string input: \'abc\'
  output: 'abc'
4

3 回答 3

1

要将字典的所有值作为数组获取，您可以使用values字典的属性：

let dictionary: Dictionary<String, Any> = [
    "key_a": "value_a",
    "key_b": "value_b",
    "key_c": "value_c",
    "key_d": "value_d",
    "key_e": 3
]

let values = Array(dictionary.values)
// values: ["value_a", "value_b", "value_c", "value_d", 3]

您filter可以忽略字典中所有非类型的值String：

let stringValues = values.filter({ $0 is String }) as! [String]
// stringValues: ["value_a", "value_b", "value_c", "value_d"]

使用 map 您可以转换stringValues并应用您的replacingOccurrences函数的值：

let adjustedValues = stringValues.map({ $0.replacingOccurrences(of: "value_", with: "") })
// adjustedValues: ["a", "b", "c", "d"]
于 2018-03-05T10:19:10.930 回答
0

案例1： 我已经实施了这个解决方案，希望这能解决你的问题

   let dict: [String: String] = ["0": "Abc", "1": "CDF", "2": "GHJ"]                
   var array: [String] = []

for (k, v) in dict.enumerated() {
    print(k)
    print(v.value)
    array.append(v.value)  
}
print(array)

案例2：

    var str = "\'abc\'"
    print(str.replacingOccurrences(of: "\'", with: ""))
于 2018-03-05T10:14:52.223 回答
0

为什么不尝试这样的事情呢？对于问题的第 1 部分：

var stringReturn: Dictionary = Dictionary<String,Any>()
stringReturn = ["0": "abc","1": "def","2": "ghi"]
print(stringReturn)

var stringBuilder = [String]()
for i in stringReturn {
  stringBuilder.append(String(describing: i.value))
}
print(stringBuilder)

此外，除非我没记错，否则第 2 部分似乎微不足道

var escaped: String = "\'abc\'"
print(escaped)
于 2018-03-05T12:14:51.407 回答