如果我像这样编码一个字符串:
var escapedString = originalString.stringByAddingPercentEscapesUsingEncoding(NSUTF8StringEncoding)
它没有逃脱斜杠/。
我搜索并找到了这段Objective C代码:
NSString *encodedString = (NSString *)CFURLCreateStringByAddingPercentEscapes(
NULL,
(CFStringRef)unencodedString,
NULL,
(CFStringRef)@"!*'();:@&=+$,/?%#[]",
kCFStringEncodingUTF8 );
是否有一个更简单的方法来编码一个URL，如果没有，我怎么写在Swift?
当前回答
Swift 4和5(谢谢@sumizome的建议。感谢@FD_和@derickito的测试)
var allowedQueryParamAndKey = NSCharacterSet.urlQueryAllowed
allowedQueryParamAndKey.remove(charactersIn: ";/?:@&=+$, ")
paramOrKey.addingPercentEncoding(withAllowedCharacters: allowedQueryParamAndKey)
斯威夫特3
let allowedQueryParamAndKey = NSCharacterSet.urlQueryAllowed.remove(charactersIn: ";/?:@&=+$, ")
paramOrKey.addingPercentEncoding(withAllowedCharacters: allowedQueryParamAndKey)
Swift 2.2(借用Zaph's，修正url查询键和参数值)
var allowedQueryParamAndKey = NSCharacterSet(charactersInString: ";/?:@&=+$, ").invertedSet
paramOrKey.stringByAddingPercentEncodingWithAllowedCharacters(allowedQueryParamAndKey)
例子:
let paramOrKey = "https://some.website.com/path/to/page.srf?a=1&b=2#top"
paramOrKey.addingPercentEncoding(withAllowedCharacters: allowedQueryParamAndKey)
// produces:
"https%3A%2F%2Fsome.website.com%2Fpath%2Fto%2Fpage.srf%3Fa%3D1%26b%3D2%23top"
以下是陈布莱恩回答的一个简短版本。我猜urlQueryAllowed允许控制字符通过，这很好，除非它们在查询字符串中形成键或值的一部分，在这一点上它们需要转义。
其他回答
这些答案对我都没用。当url包含非英语字符时，我们的应用程序崩溃了。
let unreserved = "-._~/?%$!:"
let allowed = NSMutableCharacterSet.alphanumeric()
allowed.addCharacters(in: unreserved)
let escapedString = urlString.addingPercentEncoding(withAllowedCharacters: allowed as CharacterSet)
根据您要做的事情的参数，您可能只想创建自己的字符集。上面允许使用英文字符，并且-._~/?%$!：
Swift 4 & 5
要在URL中编码参数，我发现使用. alphanumics字符集是最简单的选择:
let urlEncoded = value.addingPercentEncoding(withAllowedCharacters: .alphanumerics)
let url = "http://www.example.com/?name=\(urlEncoded!)"
使用URL编码的任何标准字符集(如. urlqueryallowed或. urlhostallowed)都不起作用，因为它们不排除=或&字符。
注意，通过使用. alphannumeric，它将编码一些不需要编码的字符(如-，.，_或~ -参见2.3。RFC 3986中的非保留字符)。我发现使用. alphanumics比构造一个自定义字符集更简单，并且不介意编码一些额外的字符。如果这困扰你，构造一个自定义字符集，如描述如何百分比编码URL字符串，例如:
// Store allowed character set for reuse (computed lazily).
private let urlAllowed: CharacterSet =
.alphanumerics.union(.init(charactersIn: "-._~")) // as per RFC 3986
extension String {
var urlEncoded: String? {
return addingPercentEncoding(withAllowedCharacters: urlAllowed)
}
}
let url = "http://www.example.com/?name=\(value.urlEncoded!)"
警告:urlEncoded参数是强制打开的。对于无效的unicode字符串，它可能会崩溃。参见为什么String.addingPercentEncoding()的返回值是可选的?而不是强制展开urlEncoded!你可以使用urlEncoded ??或者if let urlEncoded = urlEncoded{…}。
一切都是一样的
var str = CFURLCreateStringByAddingPercentEscapes(
nil,
"test/test",
nil,
"!*'();:@&=+$,/?%#[]",
CFStringBuiltInEncodings.UTF8.rawValue
)
// test%2Ftest
版本:斯威夫特5
// space convert to +
let mstring = string.replacingOccurrences(of: " ", with: "+")
// remove special character
var allowedQueryParamAndKey = NSCharacterSet.urlQueryAllowed
allowedQueryParamAndKey.remove(charactersIn: "!*'\"();:@&=+$,/?%#[]%")
return mstring.addingPercentEncoding(withAllowedCharacters: allowedQueryParamAndKey) ?? mstring
斯威夫特4.2
有时发生这种情况只是因为在段码中有空间或没有URL编码的参数通过API URL。
let myString = self.slugValue
let csCopy = CharacterSet(bitmapRepresentation: CharacterSet.urlPathAllowed.bitmapRepresentation)
let escapedString = myString!.addingPercentEncoding(withAllowedCharacters: csCopy)!
//always "info:hello%20world"
print(escapedString)
注意:不要忘记探索位图表示。
