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)@"!*'();:@&=+$,/?%#[]",
kCFStringEncodingUTF8 );
是否有一个更简单的方法来编码一个URL，如果没有，我怎么写在Swift?
当前回答
让Url = Url。加德林编。“”)
其他回答
斯威夫特4.2
一个快速的解决方案。将originalString替换为要编码的String。
var encodedString = originalString.addingPercentEncoding(withAllowedCharacters: CharacterSet(charactersIn: "!*'();:@&=+$,/?%#[]{} ").inverted)
网上游乐场演示
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: ";/?:@&=+$, ").invertedSet
paramOrKey.stringByAddingPercentEncodingWithAllowedCharacters(allowedQueryParamAndKey)
例子:
let paramOrKey = "https://some.website.com/path/to/page.srf?a=1&b=2#top"
paramOrKey.addingPercentEncoding(withAllowedCharacters: allowedQueryParamAndKey)
// produces:
"https%3A%2F%2Fsome.website.com%2Fpath%2Fto%2Fpage.srf%3Fa%3D1%26b%3D2%23top"
以下是陈布莱恩回答的一个简短版本。我猜urlQueryAllowed允许控制字符通过，这很好，除非它们在查询字符串中形成键或值的一部分，在这一点上它们需要转义。
版本:斯威夫特5
// space convert to +
let mstring = string.replacingOccurrences(of: " ", with: "+")
// remove special character
var allowedQueryParamAndKey = NSCharacterSet.urlQueryAllowed
allowedQueryParamAndKey.remove(charactersIn: "!*'\"();:@&=+$,/?%#[]%")
return mstring.addingPercentEncoding(withAllowedCharacters: allowedQueryParamAndKey) ?? mstring
斯威夫特5 你可以尝试.afURLQueryAllowed选项，如果你想编码字符串如下
let testString = "6hAD9/RjY+SnGm&B"
let escodedString = testString.addingPercentEncoding(withAllowedCharacters: .afURLQueryAllowed)
print(escodedString!)
//编码后的字符串将像en6hAD9%2FRjY%2BSnGm%26B
这个对我有用。
func stringByAddingPercentEncodingForFormData(plusForSpace: Bool=false) -> String? {
let unreserved = "*-._"
let allowed = NSMutableCharacterSet.alphanumericCharacterSet()
allowed.addCharactersInString(unreserved)
if plusForSpace {
allowed.addCharactersInString(" ")
}
var encoded = stringByAddingPercentEncodingWithAllowedCharacters(allowed)
if plusForSpace {
encoded = encoded?.stringByReplacingOccurrencesOfString(" ", withString: "+")
}
return encoded
}
我发现上述功能从这个链接:http://useyourloaf.com/blog/how-to-percent-encode-a-url-string/。
