Ios 如何在Swift中对URL进行编码

这是我的

URL

问题是，

地址

字段没有附加到

urlpath

有人知道为什么吗

var address:string
address = "American Tourister, Abids Road, Bogulkunta, Hyderabad, Andhra Pradesh, India"
let urlpath = NSString(format: "http://maps.googleapis.com/maps/api/geocode/json?address="+"\(address)")

Swift 4.2 Swift 3.0 使用：

---

在iOS 9和OS X v10.11中使用不推荐的

---

如果您要添加到URL的值可能包含保留字符（如的第2节所定义），则可能需要优化转义百分比。值得注意的是，虽然

&

和

+

是URL中的有效字符，但它们在URL查询参数值中无效（因为

&

用作查询参数之间的分隔符，这将提前终止您的值，

+

转换为空格字符）。不幸的是，标准的百分比转义使这些分隔符无法转义

因此，您可能希望对不在RFC 3986的无保留字符列表中的所有字符进行百分比转义：

URI中允许但没有保留名称的字符 目的被称为无保留。这些包括大写和小写 字母、十进制数字、连字符、句点、下划线和波浪号

unreserved = ALPHA / DIGIT / "-" / "." / "_" / "~"

  unreserved  = ALPHA / DIGIT / "-" / "." / "_" / "~"

---

顺便说一句，虽然上述RFC中没有考虑到这一点，但W3C HTML规范规定，

应用程序/x-www-form-urlencoded

请求还应将空格字符替换为

+

字符（并在不应转义的字符中包含星号）。不幸的是，

URLComponents

无法正确地进行百分比转义，因此苹果建议您在检索

URLComponents

对象的

url

属性之前手动进行百分比转义：

// configure `components` as shown above, and then:

components.percentEncodedQuery = components.percentEncodedQuery?.replacingOccurrences(of: "+", with: "%2B")
let url = components.url!

---

对于Swift 2格式副本，我自己手动执行所有百分比转义，请参见。

Swift 2.0

let needsLove = "string needin some URL love"
let safeURL = needsLove.stringByAddingPercentEncodingWithAllowedCharacters(NSCharacterSet.URLQueryAllowedCharacterSet())!

---

URLQueryAllowedCharacterSet

不应用于查询参数的URL编码，因为此字符集包括

&

、

？

、

/

等，它们在URL查询中用作分隔符，例如

/?paramname=paramvalue&paramname=paramvalue

这些字符可以作为一个整体出现在URL查询中，但不能出现在参数值中

具体谈到未保留的字符，这与允许的字符不同：

2.3。无保留字符

URI中允许但没有保留的字符
目的被称为无保留。这些包括大写和小写 字母、十进制数字、连字符、句点、下划线和波浪号

unreserved = ALPHA / DIGIT / "-" / "." / "_" / "~"

  unreserved  = ALPHA / DIGIT / "-" / "." / "_" / "~"

因此：

extension String {
    var URLEncoded:String {

        let unreservedChars = "ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789-._~"
        let unreservedCharsSet: CharacterSet = CharacterSet(charactersIn: unreservedChars)
        let encodedString = self.addingPercentEncoding(withAllowedCharacters: unreservedCharsSet)!
        return encodedString
    }
}

上面的代码没有调用

alphanumericCharacterSet

，因为它返回的字符集非常大（103806个字符）。考虑到Unicode字符

alphanumericCharacterSet

允许多少个字符，将其用于URL编码是完全错误的

用法：

let URLEncodedString = myString.URLEncoded

"aouäöü!".urlPercentEncoded(withAllowedCharacters: .urlQueryAllowed,
                                         encoding: .isoLatin1) 
// Results in -> "aou%E4%F6%FC!"

---

Mac OS 10.9中引入了Maverics和iOS 7

NSURLComponents

，它以非常方便的方式处理不同URL部分的编码

NSURLComponents类是一个用于解析URL的类 基于RFC3986并从其组成部分构造URL。 它的行为与NSURL类略有不同，后者符合 旧的RFC。但是，您可以基于以下内容轻松获得NSURL对象： URL组件对象的内容，反之亦然

---

在最后一个组件是非拉丁字符的情况下，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;
}
}

---

除了上面Bryan Chen的回答之外：

以防万一，其他人也会从阿拉莫菲尔身上得到类似的东西：

错误：Alamofire是使用优化编译的-单步执行可能会出现问题 奇怪地变量可能不可用

这不是一个非常描述性的错误。我在为谷歌地理服务构建URL时遇到了这个错误。我在URL的末尾附加了一个街道地址，而没有首先对街道地址本身进行编码。 我能够使用Bryan Chen的解决方案修复它：

var streetAdress = "123 fake street, new york, ny"
var escapedStreetAddress = streetAddress.stringByAddingPercentEncodingWithAllowedCharacters(.URLHostAllowedCharacterSet())
let url = "\(self.baseUrl)&address=\(escapedAddress!)"

那帮我修好了！它不喜欢地址有空格和逗号等

---

希望这对其他人有帮助

Swift 3:

let escapedString = originalString.addingPercentEncoding(withAllowedCharacters:NSCharacterSet.urlQueryAllowed)

代码8，SWIFT 3.0

来自grokswift

从字符串创建URL是bug的雷区。只是错过了一个或偶然的URL编码？在查询中，您的API调用将失败，并且您的应用程序将不会显示任何数据（如果您没有预料到这种可能性，甚至会崩溃）自从iOS 8以来，有一种更好的方法可以使用

NSURLComponents

和

NSURLQueryItems

构建URL

func createURLWithComponents() -> URL? {
    var urlComponents = URLComponents()
    urlComponents.scheme = "http"
    urlComponents.host = "maps.googleapis.com"
    urlComponents.path = "/maps/api/geocode/json"

    let addressQuery = URLQueryItem(name: "address", value: "American Tourister, Abids Road, Bogulkunta, Hyderabad, Andhra Pradesh, India")
    urlComponents.queryItems = [addressQuery]

    return urlComponents.url
}

http://maps.googleapis.com/maps/api/geocode/json?address=American%20Tourister,%20Abids%20Road,%20Bogulkunta,%20Hyderabad,%20Andhra%20Pradesh,%20India

下面是使用

guard

语句访问url的代码

输出：

func createURLWithComponents() -> URL? {
    var urlComponents = URLComponents()
    urlComponents.scheme = "http"
    urlComponents.host = "maps.googleapis.com"
    urlComponents.path = "/maps/api/geocode/json"

    let addressQuery = URLQueryItem(name: "address", value: "American Tourister, Abids Road, Bogulkunta, Hyderabad, Andhra Pradesh, India")
    urlComponents.queryItems = [addressQuery]

    return urlComponents.url
}

http://maps.googleapis.com/maps/api/geocode/json?address=American%20Tourister,%20Abids%20Road,%20Bogulkunta,%20Hyderabad,%20Andhra%20Pradesh,%20India

---

阅读更多信息：

刚刚完成Desmond Hume的回答，以扩展RFC 3986无保留字符有效编码函数的字符串类（如果要对查询表单参数进行编码，则需要）：

斯威夫特3 为Swift 3更新：

var escapedAddress = address.addingPercentEncoding(
    withAllowedCharacters: CharacterSet.urlQueryAllowed)

---

Swift 4.1

根据所需的选项创建“字符集”（urlQueryAllowed）。然后删除不需要的其他字符（+&）。然后将该字符集传递给“addingPercentEncoding”

---

我需要用ISO-8859-1对参数进行编码，因此addingPercentEncoding（）方法对我不起作用。 我自己在Swift 4中提出了一个解决方案：

extension String {

  // Url percent encoding according to RFC3986 specifications
  // https://tools.ietf.org/html/rfc3986#section-2.1
  func urlPercentEncoded(withAllowedCharacters allowedCharacters: 
    CharacterSet, encoding: String.Encoding) -> String {
    var returnStr = ""

    // Compute each char seperatly
    for char in self {
      let charStr = String(char)
      let charScalar = charStr.unicodeScalars[charStr.unicodeScalars.startIndex]
      if allowedCharacters.contains(charScalar) == false,
        let bytesOfChar = charStr.data(using: encoding) {
        // Get the hexStr of every notAllowed-char-byte and put a % infront of it, append the result to the returnString
        for byte in bytesOfChar {
          returnStr += "%" + String(format: "%02hhX", byte as CVarArg)
        }
      } else {
        returnStr += charStr
      }
    }

    return returnStr
  }

}

用法：

let URLEncodedString = myString.URLEncoded

"aouäöü!".urlPercentEncoded(withAllowedCharacters: .urlQueryAllowed,
                                         encoding: .isoLatin1) 
// Results in -> "aou%E4%F6%FC!"

---

让urlpath=”http://maps.googleapis.com/maps/api/geocode/json?address=\（地址）“）

就足够了，我想“某个字符串”+“\（aString）”太过分了。。。只需使用“Some string\（aString）”或“Some string”+aString我们可以通过添加PercentEncoding with AllowedCharacters来使用

string方法。您可以在此答案中找到更多详细信息：您可能希望修改该字符集，以便如果
+
或
&
出现在
地址
值中，它们将得到转义百分比。不幸的是，
URLQueryAllowedCharacterSetvar address = "American Tourister, Abids Road, Bogulkunta, Hyderabad, Andhra Pradesh, India"
var queryCharSet = NSCharacterSet.urlQueryAllowed
queryCharSet.remove(charactersIn: "+&")
let escapedAddress = address.addingPercentEncoding(withAllowedCharacters: queryCharSet)!
let urlpath = String(format: "http://maps.googleapis.com/maps/api/geocode/json?address=\(escapedAddress)")

extension String {

  // Url percent encoding according to RFC3986 specifications
  // https://tools.ietf.org/html/rfc3986#section-2.1
  func urlPercentEncoded(withAllowedCharacters allowedCharacters: 
    CharacterSet, encoding: String.Encoding) -> String {
    var returnStr = ""

    // Compute each char seperatly
    for char in self {
      let charStr = String(char)
      let charScalar = charStr.unicodeScalars[charStr.unicodeScalars.startIndex]
      if allowedCharacters.contains(charScalar) == false,
        let bytesOfChar = charStr.data(using: encoding) {
        // Get the hexStr of every notAllowed-char-byte and put a % infront of it, append the result to the returnString
        for byte in bytesOfChar {
          returnStr += "%" + String(format: "%02hhX", byte as CVarArg)
        }
      } else {
        returnStr += charStr
      }
    }

    return returnStr
  }

}

"aouäöü!".urlPercentEncoded(withAllowedCharacters: .urlQueryAllowed,
                                         encoding: .isoLatin1) 
// Results in -> "aou%E4%F6%FC!"