문자열을 URL 인코딩하는 방법
NSString공백과 &문자 가있는 URL 문자열 ( )이 있습니다 . &앰퍼샌드 및 공백을 포함하여 전체 문자열을 어떻게 URL 인코딩 합니까?
불행히도 stringByAddingPercentEscapesUsingEncoding항상 100 % 작동하지는 않습니다. URL이 아닌 문자를 인코딩하지만 예약 문자 (슬래시 /및 앰퍼샌드 등 &)는 그대로 둡니다. 분명히 이것은 Apple이 알고 있는 버그 이지만 아직 수정하지 않았기 때문에이 범주를 사용하여 문자열을 URL 인코딩했습니다.
@implementation NSString (NSString_Extended)
- (NSString *)urlencode {
NSMutableString *output = [NSMutableString string];
const unsigned char *source = (const unsigned char *)[self UTF8String];
int sourceLen = strlen((const char *)source);
for (int i = 0; i < sourceLen; ++i) {
const unsigned char thisChar = source[i];
if (thisChar == ' '){
[output appendString:@"+"];
} else if (thisChar == '.' || thisChar == '-' || thisChar == '_' || thisChar == '~' ||
(thisChar >= 'a' && thisChar <= 'z') ||
(thisChar >= 'A' && thisChar <= 'Z') ||
(thisChar >= '0' && thisChar <= '9')) {
[output appendFormat:@"%c", thisChar];
} else {
[output appendFormat:@"%%%02X", thisChar];
}
}
return output;
}
이런 식으로 사용 :
NSString *urlEncodedString = [@"SOME_URL_GOES_HERE" urlencode];
// Or, with an already existing string:
NSString *someUrlString = @"someURL";
NSString *encodedUrlStr = [someUrlString urlencode];
이것은 또한 작동합니다 :
NSString *encodedString = (NSString *)CFURLCreateStringByAddingPercentEscapes(
NULL,
(CFStringRef)unencodedString,
NULL,
(CFStringRef)@"!*'();:@&=+$,/?%#[]",
kCFStringEncodingUTF8 );
주제에 대한 좋은 독서 :
Objective-c iPhone 퍼센트가 문자열을 인코딩합니까?
Objective-C 및 Swift URL 인코딩
http://cybersam.com/programming/proper-url-percent-encoding-in-ios
https://devforums.apple.com/message/15674#15674 http://simonwoodside.com/weblog/2009/4/ 22 / how_to_really_url_encode /
도움이 될 수 있습니다.
NSString *sampleUrl = @"http://www.google.com/search.jsp?params=Java Developer";
NSString* encodedUrl = [sampleUrl stringByAddingPercentEscapesUsingEncoding:
NSUTF8StringEncoding];
iOS 7 이상에서 권장되는 방법은 다음과 같습니다.
NSString* encodedUrl = [sampleUrl stringByAddingPercentEncodingWithAllowedCharacters:[NSCharacterSet URLQueryAllowedCharacterSet]];
URL 구성 요소의 요구 사항에 따라 허용되는 문자 집합을 선택할 수 있습니다.
답변이 선택된 이후에 새로운 API가 추가되었습니다. 이제 NSURLUtilities를 사용할 수 있습니다. URL의 다른 부분에서는 다른 문자를 사용할 수 있으므로 해당 문자 세트를 사용하십시오. 다음 예제는 쿼리 문자열에 포함되도록 인코딩합니다.
encodedString = [myString stringByAddingPercentEncodingWithAllowedCharacters:NSCharacterSet.URLQueryAllowedCharacterSet];
'&'를 구체적으로 변환하려면 URL 쿼리에서 '&'가 허용되므로 URL 쿼리 세트에서 제거하고 다른 세트를 사용해야합니다.
NSMutableCharacterSet *chars = NSCharacterSet.URLQueryAllowedCharacterSet.mutableCopy;
[chars removeCharactersInRange:NSMakeRange('&', 1)]; // %26
encodedString = [myString stringByAddingPercentEncodingWithAllowedCharacters:chars];
스위프트 2.0 예제 (iOS 9 호환 가능)
extension String {
func stringByURLEncoding() -> String? {
let characters = NSCharacterSet.URLQueryAllowedCharacterSet().mutableCopy() as! NSMutableCharacterSet
characters.removeCharactersInString("&")
guard let encodedString = self.stringByAddingPercentEncodingWithAllowedCharacters(characters) else {
return nil
}
return encodedString
}
}
iOS 7 업데이트
NSString *encode = [string stringByAddingPercentEncodingWithAllowedCharacters:[NSCharacterSet URLQueryAllowedCharacterSet]];
NSString *decode = [encode stringByReplacingPercentEscapesUsingEncoding:NSUTF8StringEncoding];
CFURLCreateStringByAddingPercentEscapes허용 된 답변으로 주어진 호출 을 사용하기로 선택 했지만 최신 버전의 XCode (및 IOS)에서는 오류가 발생하여 다음을 대신 사용했습니다.
NSString *apiKeyRaw = @"79b|7Qd.jW=])(fv|M&W0O|3CENnrbNh4}2E|-)J*BCjCMrWy%dSfGs#A6N38Fo~";
NSString *apiKey = (NSString *)CFBridgingRelease(CFURLCreateStringByAddingPercentEscapes(NULL, (CFStringRef)apiKeyRaw, NULL, (CFStringRef)@"!*'();:@&=+$,/?%#[]", kCFStringEncodingUTF8));
모든 경우를 다루는 stringByAddingPercentEncodingWithAllowedCharacters방법 을 사용하십시오.[NSCharacterSet URLUserAllowedCharacterSet]
목표 C
NSString *value = @"Test / Test";
value = [value stringByAddingPercentEncodingWithAllowedCharacters:[NSCharacterSet URLUserAllowedCharacterSet]];
빠른
var value = "Test / Test"
value.stringByAddingPercentEncodingWithAllowedCharacters(NSCharacterSet.URLUserAllowedCharacterSet())
산출
Test%20%2F%20Test
이 주제에 대한 모든 답변과 ( 잘못된 ) 수락 된 답변을 읽은 후 내 기여를 추가하고 싶습니다.
경우 대상은 iOS7에 +이며, 2017 년은 엑스 코드는 빠르고 iOS8의 가장 좋은 방법, 스레드 안전에서 호환성을 제공하기 위해 정말 열심히하게해야하기 때문에, AMD는 전체 UTF-8 지원이가 할 것입니다 :
(목표 C 코드)
@implementation NSString (NSString_urlencoding)
- (NSString *)urlencode {
static NSMutableCharacterSet *chars = nil;
static dispatch_once_t pred;
if (chars)
return [self stringByAddingPercentEncodingWithAllowedCharacters:chars];
// to be thread safe
dispatch_once(&pred, ^{
chars = NSCharacterSet.URLQueryAllowedCharacterSet.mutableCopy;
[chars removeCharactersInString:@"!*'();:@&=+$,/?%#[]"];
});
return [self stringByAddingPercentEncodingWithAllowedCharacters:chars];
}
@end
NSString을 확장하고 RFC 금지 문자를 제외하고 UTF-8 문자를 지원하며 다음과 같은 것을 사용할 수 있습니다.
NSString *myusername = "I'm[evil]&want(to)break!!!$->àéìòù";
NSLog(@"Source: %@ -> Dest: %@", myusername, [myusername urlencode]);
디버그 콘솔에 인쇄됩니다.
출처 : I [m] & want (to) break !!! $-> àéìòù-> Dest : I % 27m % 5Bevil % 5D % 26want % 28to % 29break % 21 % 21 % 21 % 24- % 3E % C3 % A0 % C3 % A9 % C3 % AC % C3 % B2 % C3 % B9
... 또한 멀티 스레드 환경에서 다중 초기화를 피하기 위해 dispatch_once를 사용하십시오.
NSURLComponents를 사용하여 HTTP GET 매개 변수를 인코딩하십시오.
var urlComponents = NSURLComponents(string: "https://www.google.de/maps/")!
urlComponents.queryItems = [
NSURLQueryItem(name: "q", value: String(51.500833)+","+String(-0.141944)),
NSURLQueryItem(name: "z", value: String(6))
]
urlComponents.URL // returns https://www.google.de/maps/?q=51.500833,-0.141944&z=6
http://www.ralfebert.de/snippets/ios/encoding-nsurl-get-parameters/
이 코드는 특수 문자를 인코딩하는 데 도움이되었습니다.
NSString* encPassword = [password stringByAddingPercentEncodingWithAllowedCharacters:[NSCharacterSet alphanumericCharacterSet]];
다음은 Swift 5.x 의 프로덕션 지원 유연한 접근 방식입니다 .
public extension CharacterSet {
static let urlQueryParameterAllowed = CharacterSet.urlQueryAllowed.subtracting(CharacterSet(charactersIn: "&?"))
static let urlQueryDenied = CharacterSet.urlQueryAllowed.inverted()
static let urlQueryKeyValueDenied = CharacterSet.urlQueryParameterAllowed.inverted()
static let urlPathDenied = CharacterSet.urlPathAllowed.inverted()
static let urlFragmentDenied = CharacterSet.urlFragmentAllowed.inverted()
static let urlHostDenied = CharacterSet.urlHostAllowed.inverted()
static let urlDenied = CharacterSet.urlQueryDenied
.union(.urlQueryKeyValueDenied)
.union(.urlPathDenied)
.union(.urlFragmentDenied)
.union(.urlHostDenied)
func inverted() -> CharacterSet {
var copy = self
copy.invert()
return copy
}
}
public extension String {
func urlEncoded(denying deniedCharacters: CharacterSet = .urlDenied) -> String? {
return addingPercentEncoding(withAllowedCharacters: deniedCharacters.inverted())
}
}
사용법 예 :
print("Hello, World!".urlEncoded()!)
print("You&Me?".urlEncoded()!)
print("#Blessed 100%".urlEncoded()!)
print("Pride and Prejudice".urlEncoded(denying: .uppercaseLetters)!)
산출:
Hello,%20World!
You%26Me%3F
%23Blessed%20100%25
%50ride and %50rejudice
이 글타래에서 chown 의 objc 답변을 기반으로 한 신속한 코드 .
extension String {
func urlEncode() -> String {
return CFURLCreateStringByAddingPercentEscapes(
nil,
self,
nil,
"!*'();:@&=+$,/?%#[]",
CFStringBuiltInEncodings.UTF8.rawValue
)
}
}
에서 스위프트 3 , 아래 시도하십시오
let stringURL = "YOUR URL TO BE ENCODE";
let encodedURLString = stringURL.addingPercentEncoding(withAllowedCharacters: .urlHostAllowed)
print(encodedURLString)
, 이후 stringByAddingPercentEscapesUsingEncoding(같은 비 URL 문자하지만 잎 예약 문자 인코딩 !*'();:@&=+$,/?%#[]), 다음 코드와 같이 URL을 인코딩 할 수 있습니다 :
let stringURL = "YOUR URL TO BE ENCODE";
let characterSetTobeAllowed = (CharacterSet(charactersIn: "!*'();:@&=+$,/?%#[] ").inverted)
if let encodedURLString = stringURL.addingPercentEncoding(withAllowedCharacters: characterSetTobeAllowed) {
print(encodedURLString)
}
신속한 3 :
// exclude alpha and numeric == "full" encoding
stringUrl = stringUrl.addingPercentEncoding(withAllowedCharacters: .alphanumerics)!;
// exclude hostname and symbols &,/ and etc
stringUrl = stringUrl.addingPercentEncoding(withAllowedCharacters: .urlHostAllowed)!;
Apple's advice, in the 10.11 release notes, is:
If you need to percent-encode an entire URL string, you can use this code to encode a NSString intended to be a URL (in urlStringToEncode):
NSString *percentEncodedURLString =
[[NSURL URLWithDataRepresentation:[urlStringToEncode dataUsingEncoding:NSUTF8StringEncoding] relativeToURL:nil] relativeString];
In my case where the last component was Arabic letters I did the following in Swift 2.2:
extension String {
func encodeUTF8() -> String? {
//If I can create an NSURL out of the string nothing is wrong with it
if let _ = NSURL(string: self) {
return self
}
//Get the last component from the string this will return subSequence
let optionalLastComponent = self.characters.split { $0 == "/" }.last
if let lastComponent = optionalLastComponent {
//Get the string from the sub sequence by mapping the characters to [String] then reduce the array to String
let lastComponentAsString = lastComponent.map { String($0) }.reduce("", combine: +)
//Get the range of the last component
if let rangeOfLastComponent = self.rangeOfString(lastComponentAsString) {
//Get the string without its last component
let stringWithoutLastComponent = self.substringToIndex(rangeOfLastComponent.startIndex)
//Encode the last component
if let lastComponentEncoded = lastComponentAsString.stringByAddingPercentEncodingWithAllowedCharacters(NSCharacterSet.alphanumericCharacterSet()) {
//Finally append the original string (without its last component) to the encoded part (encoded last component)
let encodedString = stringWithoutLastComponent + lastComponentEncoded
//Return the string (original string/encoded string)
return encodedString
}
}
}
return nil;
}
}
usage:
let stringURL = "http://xxx.dev.com/endpoint/nonLatinCharacters"
if let encodedStringURL = stringURL.encodeUTF8() {
if let url = NSURL(string: encodedStringURL) {
...
}
}
-(NSString *)encodeUrlString:(NSString *)string {
return CFBridgingRelease(
CFURLCreateStringByAddingPercentEscapes(
kCFAllocatorDefault,
(__bridge CFStringRef)string,
NULL,
CFSTR("!*'();:@&=+$,/?%#[]"),
kCFStringEncodingUTF8)
);
}
according to the following blog
For individual www form-encoded query parameters, I made a category on NSString:
- (NSString*)WWWFormEncoded{
NSMutableCharacterSet *chars = NSCharacterSet.alphanumericCharacterSet.mutableCopy;
[chars addCharactersInString:@" "];
NSString* encodedString = [self stringByAddingPercentEncodingWithAllowedCharacters:chars];
encodedString = [encodedString stringByReplacingOccurrencesOfString:@" " withString:@"+"];
return encodedString;
}
//This is without test
NSMutableCharacterSet* set = [[NSCharacterSet alphanumericCharacterSet] mutableCopy];
[set addCharactersInString:@"-_.~"];
NSString *encode = [test stringByAddingPercentEncodingWithAllowedCharacters:set];
I faced a similar problem passing complex strings as a POST parameter. My strings can contain Asian characters, spaces, quotes and all sorts of special characters. The solution I eventually found was to convert my string into the matching series of unicodes, e.g. "Hu0040Hu0020Hu03f5...." using [NSString stringWithFormat:@"Hu%04x",[string characterAtIndex:i]] to get the Unicode from each character in the original string. The same can be done in Java.
This string can be safely passed as a POST parameter.
On the server side (PHP), I change all the "H" to "\" and I pass the resulting string to json_decode. Final step is to escape single quotes before storing the string into MySQL.
This way I can store any UTF8 string on my server.
This one is working for me.
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
}
I found the above function from this link: http://useyourloaf.com/blog/how-to-percent-encode-a-url-string/
You can also use this function with swift extension. Please let me know if there is any issue.
참고URL : https://stackoverflow.com/questions/8088473/how-do-i-url-encode-a-string
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