使用Swift来取消unicode字符,即\ u1234

Question

在使用swift在xcode 6中使用JSON时,我遇到了特殊字符的问题

我在Cocoa/objective C中找到了这些代码来解决转换重音的一些问题但却无法在Swift中运行.有关如何使用它的任何建议？......最好的替代建议也很酷......

谢谢

NSString *input = @"\\u5404\\u500b\\u90fd";
NSString *convertedString = [input mutableCopy];

CFStringRef transform = CFSTR("Any-Hex/Java");
CFStringTransform((__bridge CFMutableStringRef)convertedString, NULL, transform, YES);

NSLog(@"convertedString: %@", convertedString);
// prints: ???, tada!

Answer 1

它在Swift中非常相似,但你仍然需要使用Foundation字符串类:

let transform = "Any-Hex/Java"
let input = "\\u5404\\u500b\\u90fd" as NSString
var convertedString = input.mutableCopy() as NSMutableString

CFStringTransform(convertedString, nil, transform as NSString, 1)

println("convertedString: \(convertedString)")
// convertedString: ???

(最后一个参数引发了我的循环,直到我意识到Boolean在Swift中是UInt的类型别名 - 对于这些类型的方法,Objective-C中的YES在Swift中变为1.)

Answer 2

快速String扩展：

extension String {
    var unescapingUnicodeCharacters: String {       
        let mutableString = NSMutableString(string: self)
        CFStringTransform(mutableString, nil, "Any-Hex/Java" as NSString, true)
        
        return mutableString as String
    }
}

Answer 3

雨燕3

\n\n

let transform = "Any-Hex/Java"\nlet input = "\\\\u5404\\\\u500b\\\\u90fd" as NSString\nvar convertedString = input.mutableCopy() as! NSMutableString\n\nCFStringTransform(convertedString, nil, transform as NSString, true)\n\nprint("convertedString: \\(convertedString)")\n// convertedString: \xe5\x90\x84\xe5\x80\x8b\xe9\x83\xbd\n

\n