I have this kind of string: "POINT(101.650577657408 3.1653186153213)".
Anyone know how can I get the first and second value of POINT from this String?
You can easily spit the string using componentsSeparatedByString function.
let myStr = "POINT(101.650577657408 3.1653186153213)"
let characterSet = NSCharacterSet(charactersInString: "( )")
var splitString = myStr.componentsSeparatedByCharactersInSet(characterSet)
print(splitString[0])
print(splitString[1])
print(splitString[2])
The above only works if you have the complete one String.
Although the NSCharacterSet solution is correct but here is another solution using the most powerful regex.
var error: NSError?
// Initialise the regex for a float value
let regex: NSRegularExpression = NSRegularExpression(pattern: "(\\d*\\.\\d*)", options: NSRegularExpressionOptions.CaseInsensitive, error: &error)!
// Matches array contains all the match found for float in given string
let matches: NSArray = regex.matchesInString(str as String, options: NSMatchingOptions.ReportProgress, range: NSMakeRange(0, str.length))
// You can easily get all values by enumeration
for match in matches {
println(str.substringWithRange(match.range))
}
The benefit of this solution is it will scan all the float values and will also work in case of pattern got changed.
Try this, this will work for your string
let myStr = "POINT(101.650577657408 3.1653186153213)"
let strWithout_POINT_openingBrace = myStr.stringByReplacingOccurrencesOfString("POINT(", withString: "")//"101.650577657408 3.1653186153213)"
let strWithout_closingBrace = strWithout_POINT_openingBrace.stringByReplacingOccurrencesOfString(")", withString: "")//"101.650577657408 3.1653186153213"
//now you have only space between two values
//so split string by space
let arrStringValues = strWithout_closingBrace.componentsSeparatedByString(" ");//["101.650577657408","3.1653186153213"]
print(arrStringValues[0]);//first value "101.650577657408"
print(arrStringValues[1]);//second value "3.1653186153213"
Related
I have Strings with the form string \ string example
"some sting with random length\233"
I want to deletes the last \ and get the value after it, so the result will be
"some sting with random length"
"233"
I tried this code but it's not working
let regex = try! NSRegularExpression(pattern: "\\\s*(\\S[^,]*)$")
if let match = regex.firstMatch(in: string, range: string.nsRange) {
let result = string.substring(with: match.rangeAt(1))
}
You did not correctly adapt the pattern from How to get substring after last occurrence of character in string: Swift IOS to your case. Both instances of the comma must be replaced by a backslash,
and that must be "double-escaped":
let regex = try! NSRegularExpression(pattern: "\\\\\\s*(\\S[^\\\\]*)$")
(once be interpreted as a literal backslash in the regex pattern, and
once more in the definition of a Swift string literal).
However, a simpler solution is to find the last occurrence of the
backslash and extract the suffix from that position:
let string = "some sting with random length\\233"
let separator = "\\" // A single(!) backslash
if let r = string.range(of: separator, options: .backwards) {
let prefix = string.substring(to: r.lowerBound)
let suffix = string.substring(from: r.upperBound)
print(prefix) // some sting with random length
print(suffix) // 233
}
Update for Swift 4:
if let r = string.range(of: separator, options: .backwards) {
let prefix = string[..<r.lowerBound]
let suffix = string[r.upperBound...]
print(prefix) // some sting with random length
print(suffix) // 233
}
prefix and suffix are a String.SubSequence, which can be used
in many places instead of a String. If necessary, create a real
string:
let prefix = String(string[..<r.lowerBound])
let suffix = String(string[r.upperBound...])
You could do this with regex, but I think this solution is better:
yourString.components(separatedBy: "\\").last!
It splits the string with \ as the separator and gets the last split.
I am trying to read the string from a Label and remove the last character form it.
This is how I am trying:
#IBAction func del(sender: UIButton) {
let str = telephone.text!;
let newstr = str.remove(at: str.index(before: str.endIndex))
telephone.text = newstr;
}
When I run, I get an error:
"String" does not have a member named "remove"
Can someone help me figure out the problem?
Just started learning swift :(
remove(at:) mutates the receiver which must therefore be a variable
string:
var str = telephone.text!
str.remove(at: str.index(before: str.endIndex))
telephone.text = str
Alternatively use substring(to:), which returns the new string
instead of modifying the receiver:
let str = telephone.text!
let newstr = str.substring(to: str.index(before: str.endIndex))
telephone.text = newstr
remove is defined as follows:
public mutating func remove(at i: String.Index) -> Character
See the mutating modifier? That means it mutates the instance on which the method is called. In your case, the instance is str, a constant. Since constants cannot be mutated, the code does not compile.
And since remove returns the removed character,
let newstr = str.remove(at: str.index(before: str.endIndex))
here newstr will not be storing the string with the last character removed.
You should rewrite the method like this:
telephone.text!.remove(at: telephone.text!.index(before: telephone.text!.endIndex))
You can use:
let idx = str.index(before: str.endIndex) // compute the index
let s = str.substring(to: idx) // get the substring
I am very much new to swift language. I am performing some business logic which needs to take NSRange from given String.
Here is my requirement,
Given Amount = "144.44"
Need NSRange of only cent part i.e. after "."
Is there any API available for doing this?
You can do a regex-based search to find the range:
let str : NSString = "123.45"
let rng : NSRange = str.range("(?<=[.])\\d*$", options: .RegularExpressionSearch)
Regular expression "(?<=[.])\\d*$" means "zero or more digits following a dot character '.' via look-behind, all the way to the end of the string $."
If you want a substring from a given string you can use componentsSeparatedByString
Example :
var number: String = "144.44";
var numberresult= number.componentsSeparatedByString(".")
then you can get components as :
var num1: String = numberresult [0]
var num2: String = numberresult [1]
hope it help !!
Use rangeOfString and substringFromIndex:
let string = "123.45"
if let index = string.rangeOfString(".") {
let cents = string.substringFromIndex(index.endIndex)
print("\(cents)")
}
Another version that uses Swift Ranges, rather than NSRange
Define the function that returns an optional Range:
func centsRangeFromString(str: String) -> Range<String.Index>? {
let characters = str.characters
guard let dotIndex = characters.indexOf(".") else { return nil }
return Range(dotIndex.successor() ..< characters.endIndex)
}
Which you can test with:
let r = centsRangeFromString(str)
// I don't recommend force unwrapping here, but this is just an example.
let cents = str.substringWithRange(r!)
I have a string: "Hey #username that's funny". For a given string, how can I search the string to return all ranges of string with first character # and last character to get the username?
I suppose I can get all indexes of # and for each, get the substringToIndex of the next space character, but wondering if there's an easier way.
If your username can contain only letters and numbers, you can use regular expression for that:
let s = "Hey #username123 that's funny"
if let r = s.rangeOfString("#\\w+", options: NSStringCompareOptions.RegularExpressionSearch) {
let name = s.substringWithRange(r) // #username123"
}
#Vladimir's answer is correct, but if you're trying to find multiple occurrences of "username", this should also work:
let s = "Hey #username123 that's funny"
let ranges: [NSRange]
do {
// Create the regular expression.
let regex = try NSRegularExpression(pattern: "#\\w+", options: [])
// Use the regular expression to get an array of NSTextCheckingResult.
// Use map to extract the range from each result.
ranges = regex.matchesInString(s, options: [], range: NSMakeRange(0, s.characters.count)).map {$0.range}
}
catch {
// There was a problem creating the regular expression
ranges = []
}
for range in ranges {
print((s as NSString).substringWithRange(range))
}
I have an array in Swift:
["\"100003866283798-2\"", "\"100001986741004-2\"",
"\"100003455181526-2\"", "\"100002261472542-2\"",
"\"100003866283798-3\"", "\"100003866283798-0\"",
"\"100001986741004-3\"", "\"100001986741004-0\"",
"\"100003455181526-3\"", "\"100003455181526-0\"",
"\"100002261472542-3\"", "\"100002261472542-0\""]
and I only want the numbers, not the quotes and the -2.
I can't figure out to do this, because when I type
let cleanStr = numberArray.stringByReplacingOccurrencesOfString("\"", withString: "", options: nil, range: nil)
it gives the error:
Value of type '[String]' has no member
'stringByReplacingOccurrencesOfString'
I know it's very confusing with all the quotes etc but I want these numbers without their quotes etc, anyone a solution?
You can do it like this:
let newArray = array.map { (string: String) -> String in
let woFirst = string.characters.dropFirst()
let tilDash = woFirst.prefixUpTo(
woFirst.indexOf("-") ?? woFirst.endIndex)
return String(tilDash)
}
(Weirdly it doesn't compile without the type annotation). Or if you feel fancy:
let newArray = array
.map{ {$0.prefixUpTo($0.indexOf("-") ?? $0.endIndex)}($0.characters.dropFirst()) }
.map(String.init)
(XCode 7 beta 6)
stringByReplacingOccurrencesOfString
works only with NSString not Array. Then you must iterate your array remove excess char.