Howto split a string into character groups in actionscript? - string

I need a string splitted into groups of characters in ActionScript like this:
var txt:String = "Hello World";
var arr:Array = txt.split(3);
// Now arr should contain a value like: ["Hel", "lo ", "Wor", "ld"]
This is possible in PHP like this:
$arr = str_split(txt, 3);
But I need a ActionScript equivalent of this PHP code.

You can use a regex:
var txt:String = "Hello World";
var arr:Array = txt.match(/.{3}|.+/g);
trace(arr);
// Hel,lo ,Wor,ld

Related

how separate string into key/value pair in dart?

how can a string be separated into key/value pair in dart? The string is separated by a "=". And how can the pair value be extracted?
main(){
var stringTobeSeparated = ['ab = cd','ef = gh','ld = kg'];
Map<String ,dynamic> map = {};
for (String s in stringTobeSeparated) {
var keyValue = s.split("=");
//failed to add to a map , to many positiona arguments error
map.addAll(keyValue[0],keyValue[1]);
}
}
The split() function gives you a List of Strings, so you just need to check if the length of this List is equal to 2 and then you can add those values in a Map like this:
Map<String, String> map = {};
for (String s in stringTobeSeparated) {
var list = s.split("=");
if(list.length == 2) {
// list[0] is your key and list[1] is your value
map[list[0]] = list[1];
}
}
You can use map for this, the accepted answer is correct, but since your string looks like this
var stringTobeSeparated = ['ab = cd','ef = gh','ld = kg'];
I would rather use regex to remove spaces from final result (replace the line with split with this):
var list = s.split(RegExp(r"\s+=\s+"));

Remove some text from a string after some constant value(string)

Input: String str="Fund testing testing";
Output: str="Fund";
After fund whatever the text is there need to remove that text.
Please suggest some solution.
The easiest way to solve this is a .Substring() method, as you can provide it the start index of your original string and length of the string you need:
var length = "Fund".Length;
var str = "Fund testing testing";
Console.WriteLine(str.Substring(0, length)); //returns "Fund"
var str1 = "testFund testing testing";
Console.WriteLine(str1.Substring(4, length)); //returns "Fund"
var str2 = "testFund testing testing";
Console.WriteLine(str2.Substring(str2.IndexOf("Fund"), length)); //returns "Fund"
You can also use regular expression like this:
string strRegex = #".*?(Fund).*";
Regex myRegex = new Regex(strRegex, RegexOptions.Singleline);
string strTargetString = #"Fund testing testing";
string strReplace = #"$1";
return myRegex.Replace(strTargetString, strReplace);
As mentioned in comments below, replace can lack performance and is kind of overkill, so regex Match can be better. Here is how it looks like:
string strRegex = #".*?(Fund).*";
Regex myRegex = new Regex(strRegex, RegexOptions.None);
string strTargetString = "\n\n" + #" Fund testing testing";
foreach (Match myMatch in myRegex.Matches(strTargetString))
{
if (myMatch.Success)
{
var fund = myMatch.Groups[1].Value;
Console.WriteLine(fund);
}
}
Note that Groups first element is your entire match

Selecting a tuple index using a variable in Swift

That is what i am trying to do:
var i = 0
var string = "abcdef"
for value in string
{
value.[Put value of variable i here] = "a"
i++
}
How can i insert the value of i in the code?
Easiest is probably just convert it to an NSMutableString:
let string = "abcdef".mutableCopy() as NSMutableString
println( "\(string)")
for var i = 0; i < string.length; ++i {
string.replaceCharactersInRange(NSMakeRange(i, 1), withString: "a")
}
println( "\(string)")
Yes, it's a bit ugly but it works.
A much cleaner way is to use Swifts map function:
var string = "abcdef"
let result = map(string) { (c) -> Character in
"a"
}
println("\(result)") // aaaaaa
You should just be able to use the following but this doesn't compile:
map(string) { "a" }
In you comments you mention you want to split up the string on a space, you can just use this for that:
let stringWithSpace = "abcdef 012345"
let splitString = stringWithSpace.componentsSeparatedByString(" ")
println("\(splitString[0])") // abcdef
println("\(splitString[1])") // 012345

Remove last character from string. Swift language

How can I remove last character from String variable using Swift? Can't find it in documentation.
Here is full example:
var expression = "45+22"
expression = expression.substringToIndex(countElements(expression) - 1)
Swift 4.0 (also Swift 5.0)
var str = "Hello, World" // "Hello, World"
str.dropLast() // "Hello, Worl" (non-modifying)
str // "Hello, World"
String(str.dropLast()) // "Hello, Worl"
str.remove(at: str.index(before: str.endIndex)) // "d"
str // "Hello, Worl" (modifying)
Swift 3.0
The APIs have gotten a bit more swifty, and as a result the Foundation extension has changed a bit:
var name: String = "Dolphin"
var truncated = name.substring(to: name.index(before: name.endIndex))
print(name) // "Dolphin"
print(truncated) // "Dolphi"
Or the in-place version:
var name: String = "Dolphin"
name.remove(at: name.index(before: name.endIndex))
print(name) // "Dolphi"
Thanks Zmey, Rob Allen!
Swift 2.0+ Way
There are a few ways to accomplish this:
Via the Foundation extension, despite not being part of the Swift library:
var name: String = "Dolphin"
var truncated = name.substringToIndex(name.endIndex.predecessor())
print(name) // "Dolphin"
print(truncated) // "Dolphi"
Using the removeRange() method (which alters the name):
var name: String = "Dolphin"
name.removeAtIndex(name.endIndex.predecessor())
print(name) // "Dolphi"
Using the dropLast() function:
var name: String = "Dolphin"
var truncated = String(name.characters.dropLast())
print(name) // "Dolphin"
print(truncated) // "Dolphi"
Old String.Index (Xcode 6 Beta 4 +) Way
Since String types in Swift aim to provide excellent UTF-8 support, you can no longer access character indexes/ranges/substrings using Int types. Instead, you use String.Index:
let name: String = "Dolphin"
let stringLength = count(name) // Since swift1.2 `countElements` became `count`
let substringIndex = stringLength - 1
name.substringToIndex(advance(name.startIndex, substringIndex)) // "Dolphi"
Alternatively (for a more practical, but less educational example) you can use endIndex:
let name: String = "Dolphin"
name.substringToIndex(name.endIndex.predecessor()) // "Dolphi"
Note: I found this to be a great starting point for understanding String.Index
Old (pre-Beta 4) Way
You can simply use the substringToIndex() function, providing it one less than the length of the String:
let name: String = "Dolphin"
name.substringToIndex(countElements(name) - 1) // "Dolphi"
The global dropLast() function works on sequences and therefore on Strings:
var expression = "45+22"
expression = dropLast(expression) // "45+2"
// in Swift 2.0 (according to cromanelli's comment below)
expression = String(expression.characters.dropLast())
Swift 4:
let choppedString = String(theString.dropLast())
In Swift 2, do this:
let choppedString = String(theString.characters.dropLast())
I recommend this link to get an understanding of Swift strings.
Swift 4/5
var str = "bla"
str.removeLast() // returns "a"; str is now "bl"
This is a String Extension Form:
extension String {
func removeCharsFromEnd(count_:Int) -> String {
let stringLength = count(self)
let substringIndex = (stringLength < count_) ? 0 : stringLength - count_
return self.substringToIndex(advance(self.startIndex, substringIndex))
}
}
for versions of Swift earlier than 1.2:
...
let stringLength = countElements(self)
...
Usage:
var str_1 = "Maxim"
println("output: \(str_1.removeCharsFromEnd(1))") // "Maxi"
println("output: \(str_1.removeCharsFromEnd(3))") // "Ma"
println("output: \(str_1.removeCharsFromEnd(8))") // ""
Reference:
Extensions add new functionality to an existing class, structure, or enumeration type. This includes the ability to extend types for which you do not have access to the original source code (known as retroactive modeling). Extensions are similar to categories in Objective-C. (Unlike Objective-C categories, Swift extensions do not have names.)
See DOCS
Use the function removeAtIndex(i: String.Index) -> Character:
var s = "abc"
s.removeAtIndex(s.endIndex.predecessor()) // "ab"
Swift 4
var welcome = "Hello World!"
welcome = String(welcome[..<welcome.index(before:welcome.endIndex)])
or
welcome.remove(at: welcome.index(before: welcome.endIndex))
or
welcome = String(welcome.dropLast())
The easiest way to trim the last character of the string is:
title = title[title.startIndex ..< title.endIndex.advancedBy(-1)]
import UIKit
var str1 = "Hello, playground"
str1.removeLast()
print(str1)
var str2 = "Hello, playground"
str2.removeLast(3)
print(str2)
var str3 = "Hello, playground"
str3.removeFirst(2)
print(str3)
Output:-
Hello, playgroun
Hello, playgro
llo, playground
let str = "abc"
let substr = str.substringToIndex(str.endIndex.predecessor()) // "ab"
var str = "Hello, playground"
extension String {
var stringByDeletingLastCharacter: String {
return dropLast(self)
}
}
println(str.stringByDeletingLastCharacter) // "Hello, playgroun"
Short answer (valid as of 2015-04-16): removeAtIndex(myString.endIndex.predecessor())
Example:
var howToBeHappy = "Practice compassion, attention and gratitude. And smile!!"
howToBeHappy.removeAtIndex(howToBeHappy.endIndex.predecessor())
println(howToBeHappy)
// "Practice compassion, attention and gratitude. And smile!"
Meta:
The language continues its rapid evolution, making the half-life for many formerly-good S.O. answers dangerously brief. It's always best to learn the language and refer to real documentation.
With the new Substring type usage:
Swift 4:
var before: String = "Hello world!"
var lastCharIndex: Int = before.endIndex
var after:String = String(before[..<lastCharIndex])
print(after) // Hello world
Shorter way:
var before: String = "Hello world!"
after = String(before[..<before.endIndex])
print(after) // Hello world
Use the function advance(startIndex, endIndex):
var str = "45+22"
str = str.substringToIndex(advance(str.startIndex, countElements(str) - 1))
A swift category that's mutating:
extension String {
mutating func removeCharsFromEnd(removeCount:Int)
{
let stringLength = count(self)
let substringIndex = max(0, stringLength - removeCount)
self = self.substringToIndex(advance(self.startIndex, substringIndex))
}
}
Use:
var myString = "abcd"
myString.removeCharsFromEnd(2)
println(myString) // "ab"
Another way If you want to remove one or more than one character from the end.
var myStr = "Hello World!"
myStr = (myStr as NSString).substringToIndex((myStr as NSString).length-XX)
Where XX is the number of characters you want to remove.
Swift 3 (according to the docs) 20th Nov 2016
let range = expression.index(expression.endIndex, offsetBy: -numberOfCharactersToRemove)..<expression.endIndex
expression.removeSubrange(range)
The dropLast() function removes the last element of the string.
var expression = "45+22"
expression = expression.dropLast()
Swift 4.2
I also delete my last character from String (i.e. UILabel text) in IOS app
#IBOutlet weak var labelText: UILabel! // Do Connection with UILabel
#IBAction func whenXButtonPress(_ sender: UIButton) { // Do Connection With X Button
labelText.text = String((labelText.text?.dropLast())!) // Delete the last caracter and assign it
}
I'd recommend using NSString for strings that you want to manipulate. Actually come to think of it as a developer I've never run into a problem with NSString that Swift String would solve... I understand the subtleties. But I've yet to have an actual need for them.
var foo = someSwiftString as NSString
or
var foo = "Foo" as NSString
or
var foo: NSString = "blah"
And then the whole world of simple NSString string operations is open to you.
As answer to the question
// check bounds before you do this, e.g. foo.length > 0
// Note shortFoo is of type NSString
var shortFoo = foo.substringToIndex(foo.length-1)
Swift 3: When you want to remove trailing string:
func replaceSuffix(_ suffix: String, replacement: String) -> String {
if hasSuffix(suffix) {
let sufsize = suffix.count < count ? -suffix.count : 0
let toIndex = index(endIndex, offsetBy: sufsize)
return substring(to: toIndex) + replacement
}
else
{
return self
}
}
complimentary to the above code I wanted to remove the beginning of the string and could not find a reference anywhere. Here is how I did it:
var mac = peripheral.identifier.description
let range = mac.startIndex..<mac.endIndex.advancedBy(-50)
mac.removeRange(range) // trim 17 characters from the beginning
let txPower = peripheral.advertisements.txPower?.description
This trims 17 characters from the beginning of the string (he total string length is 67 we advance -50 from the end and there you have it.
I prefer the below implementation because I don't have to worry even if the string is empty
let str = "abc"
str.popLast()
// Prints ab
str = ""
str.popLast() // It returns the Character? which is an optional
// Print <emptystring>

Best way to replace all spaces, symbols, numbers, uppercase letters from a string in actionscript?

What would be the best way to simply take a string like
var myString:String = "Thi$ i$ a T#%%Ible Exam73#";
and make myString = "thiiatibleeam";
or another example
var myString:String = "Totally Awesome String";
and make myString = "totallyawesomestring";
In actionscript 3 Thanks!
Extending #Sam OverMars' answer, you can use a combination of String's replace method with a Regex and String's toLowerCase method to get what you're looking for.
var str:String = "Thi$ i$ a T#%%Ible Exam73#";
str = str.toLowerCase(); //thi$ i$ a t#%%ible exam73#
str = str.replace(/[^a-z]/g,""); //thiiatibleexam
The regular expression means:
[^a-z] -- any character *not* in the range a-z
/g -- global tag means find all, not just find one
I think this is the regex you're looking for:
[Bindable]
var myString:String = "Thi$ i$ a T#%%Ible Exam73#";
[Bindable]
var anotherString:String = "";
protected function someFunction():void
{
anotherString = myString.replace(/[^a-zA-Z]/g, "");
anotherString = anotherString.toLowerCase();
}
I belive what your looking for is:
var myString = str.replace("find", "replace");
or in your case:
str.replace("$", "");
also, it might be:
str.replace('$', ' ');
//EDIT
How about:
var mySearch:RegExp = /(\t|\n|\s{1,})/g;
var myString = str.replace(mySearch, "");

Resources