I have the following URL query:
encodedMessage=PD94bWwgdmVyNlPg%3D%3D&signature=kcig33sdAOAr%2FYYGf5r4HGN
How can I split the query to get the of encodedMessage and signature values?
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2URLComponents: developer.apple.com/reference/foundation/urlcomponentsEric Aya– Eric Aya2017-03-30 13:10:37 +00:00Commented Mar 30, 2017 at 13:10
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3 Answers
The right way to achieve this is to work with URLComponents:
A structure designed to parse URLs based on RFC 3986 and to construct URLs from their constituent parts.
By getting the url components host string and queryItems array, as follows:
if let urlComponents = URLComponents(string: "http://mydummysite.com?encodedMessage=PD94bWwgdmVyNlPg%3D%3D&signature=kcig33sdAOAr%2FYYGf5r4HGN"), let host = urlComponents.host, let queryItems = urlComponents.queryItems {
print(host) // mydummysite.com
print(queryItems) // [encodedMessage=PD94bWwgdmVyNlPg==, signature=kcig33sdAOAr/YYGf5r4HGN]
}
queryItems array contains URLQueryItem objects, which have name and value properties:
if let urlComponents = URLComponents(string: "http://mydummysite.com?encodedMessage=PD94bWwgdmVyNlPg%3D%3D&signature=kcig33sdAOAr%2FYYGf5r4HGN"),let queryItems = urlComponents.queryItems {
// for example, we will get the first item name and value:
let name = queryItems[0].name // encodedMessage
let value = queryItems[0].value // PD94bWwgdmVyNlPg==
}
Also:
In case of you are getting the query without the full url, I'd suggest to do a pretty simple trick, by adding a dummy host as a prefix to your query string, as follows:
let myQuery = "encodedMessage=PD94bWwgdmVyNlPg%3D%3D&signature=kcig33sdAOAr%2FYYGf5r4HGN"
let myDummyUrlString = "http://stackoverflow.com?" + myQuery
if let urlComponents = URLComponents(string: myDummyUrlString),let queryItems = urlComponents.queryItems {
// for example, we will get the first item name and value:
let name = queryItems[0].name // encodedMessage
let value = queryItems[0].value // PD94bWwgdmVyNlPg==
} else {
print("invalid url")
}
5 Comments
pavlos
Nothing happens.. It doesnt return anything when i execute it
Ahmad F
make sure that the used url is a valid one, what are showing in the question is just the query, it is not a full url.
Ahmad F
It is kind of weird to me that you are getting a query without the full url. However, please check the edit.
RickiG
This should be the accepted answer. The above will follow RFC 3986 and will correctly handle all the edge cases - like the initial '?' and so on in a complete URL.
Naishta
Brilliant ! thanks for a detailed explanation with a perfect example.
You can get the key value pairs this way:
let str = "encodedMessage=PD94bWwgdmVyNlPg%3D%3D&signature=kcig33sdAOAr%2FYYGf5r4HGN"
let arr = str.components(separatedBy:"&")
var data = [String:Any]()
for row in arr {
let pairs = row.components(separatedBy:"=")
data[pairs[0]] = pairs[1]
}
let message = data["encodedMessage"]
let sig = data["signature"]
I am not sure if that's what you were looking for or not. If it is not, could you please clarify a bit further as to what you are looking for?
2 Comments
This is the method I am using to break URLs with swift5
func breakURL(_ urlString: String) -> URLComponents? {
// Create a URL from the given urlString
guard let url = URL(string: urlString) else {
print("Invalid URL")
return nil
}
// Get the components of the URL
if let urlComponents = URLComponents(url: url, resolvingAgainstBaseURL: false) {
return urlComponents
} else {
print("Failed to parse URL components")
return nil
}
}
Following is the way to use it.
if let urlComponents = breakURL("https://www.example.com/path/to/resource?param1=value1¶m2=value2") {
print("Scheme: \(urlComponents.scheme ?? "N/A")") // https
print("Host: \(urlComponents.host ?? "N/A")") //www.example.com
print("Path: \(urlComponents.path)") //path/to/resource
print("Query: \(urlComponents.query ?? "N/A")") //param1=value1¶m2=value2
}
if you want to break path or query you can get specific value with index number or just put loop on it ...
for queryItem in urlComponents.queryItems ?? [] {
print("\(queryItem.name): \(queryItem.value ?? "N/A")")
}