如何对字符串进行URL编码
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我有一个带有空格和NSString字符的URL字符串（&）。我如何url编码整个字符串（包括&＆符号和空格）？
答案
不幸的是，stringByAddingPercentEscapesUsingEncoding并不总是100％工作。它对非URL字符进行编码，但仅保留保留字符（如斜线/和＆符号&）。显然这是Apple意识到的错误，但由于他们还没有修复它，我一直在使用这个类别来对字符串进行url编码：
@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;
}
像这样使用：
NSString *urlEncodedString = [@"SOME_URL_GOES_HERE" urlencode];
// Or, with an already existing string:
NSString *someUrlString = @"someURL";
NSString *encodedUrlStr = [someUrlString urlencode];
这也有效：
NSString *encodedString = (NSString *)CFURLCreateStringByAddingPercentEscapes(
NULL,
(CFStringRef)unencodedString,
NULL,
(CFStringRef)@"!*'();:@&=+$,/?%#[]",
kCFStringEncodingUTF8 );
关于这个主题的一些好的阅读：
Objective-c iPhone percent encode a string? Objective-C and Swift URL encoding
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/
另一答案
使用NSURLComponents对HTTP GET参数进行编码：
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/
另一答案
这是Swift 4中的一种生产就绪的灵活方法：
public extension CharacterSet {
public static let urlQueryParameterAllowed = CharacterSet.urlQueryAllowed.subtracting(CharacterSet(charactersIn: "&?"))
public static let urlQueryDenied = CharacterSet.urlQueryAllowed.inverted()
public static let urlQueryKeyValueDenied = CharacterSet.urlQueryParameterAllowed.inverted()
public static let urlPathDenied = CharacterSet.urlPathAllowed.inverted()
public static let urlFragmentDenied = CharacterSet.urlFragmentAllowed.inverted()
public static let urlHostDenied = CharacterSet.urlHostAllowed.inverted()
public static let urlDenied = CharacterSet.urlQueryDenied
.union(.urlQueryKeyValueDenied)
.union(.urlPathDenied)
.union(.urlFragmentDenied)
.union(.urlHostDenied)
public 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())
}
}
Example usage:
print("Hello, World!".urlEncoded()!)
print("You&Me?".urlEncoded()!)
print("#Blessed 100%".urlEncoded()!)
print("Pride and Prejudice".urlEncoded(denying: .uppercaseLetters)!)
Output:
Hello,%20World!
You%26Me%3F
%23Blessed%20100%25
%50ride and %50rejudice
另一答案
这段代码帮助我编码特殊字符
NSString* encPassword = [password stringByAddingPercentEncodingWithAllowedCharacters:[NSCharacterSet alphanumericCharacterSet]];
另一答案
在Swift 3中，请尝试以下内容：
let stringURL = "YOUR URL TO BE ENCODE";
let encodedURLString = stringURL.addingPercentEncoding(withAllowedCharacters: .urlHostAllowed)
print(encodedURLString)
因为，stringByAddingPercentEscapesUsingEncoding编码非URL字符但保留保留字符（如!*'();:@&=+$,/?%#[]），您可以像下面的代码一样对url进行编码：
let stringURL = "YOUR URL TO BE ENCODE";
let characterSetTobeAllowed = (CharacterSet(charactersIn: "!*'();:@&=+$,/?%#[] ").inverted)
if let encodedURLString = stringURL.addingPercentEncoding(withAllowedCharacters: characterSetTobeAllowed) {
print(encodedURLString)
}
另一答案
在快速3：
// exclude alpha and numeric == "full" encoding
stringUrl = stringUrl.addingPercentEncoding(withAllowedCharacters: .alphanumerics)!;
// exclude hostname and symbols &,/ and etc
stringUrl = stringUrl.addingPercentEncoding(withAllowedCharacters: .urlHostAllowed)!;
另一答案
在我的情况下，最后一个组件是阿拉伯语字母，我在Swift 2.2中执行了以下操作：
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;
}
}
用法：
let stringURL = "http://xxx.dev.com/endpoint/nonLatinCharacters"
if let encodedStringURL = stringURL.encodeUTF8() {
if let url = NSURL(string: encodedStringURL) {
...
}
}
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