newbie programmer here, can't figure out why it does it...
let a1 = "This is some text"
let x : Int = 1
var stringValue = "a\(x)"
print(stringValue)
I want it to print "This is some text", but it only ever prints a1.
Rather than storing 'This is some text' in a1 string, save it in a dictionary so you can create the key (x1,x2 etc) to access it:
let stringDictionary = ["a1": "This is some text", "a2": "This is more text"]
You can then pull it out using:
let x : Int = 1
let stringKey = "a\(x)"
let stringValue = stringDictionary[stringKey]
print(stringValue)
What you want is an eval or evaluate type function, but that doesn't exist. What you're actually getting is a description of the x variable, which is the number 1 converted into text, hence a1.
If you want to get a value for a key you should use a dictionary.
Related
My Swift app involves searching through text in a UITextView. The user can search for a certain substring within that text view, then jump to any instance of that string in the text view (say, the third instance). I need to find out the integer value of which character they are on.
For example:
Example 1: The user searches for "hello" and the text view reads "hey hi hello, hey hi hello", then the user presses down arrow to view second instance. I need to know the integer value of the first h in the second hello (i.e. which # character that h in hello is within the text view). The integer value should be 22.
Example 2: The user searches for "abc" while the text view reads "abcd" and they are looking for the first instance of abc, so the integer value should be 1 (which is the integer value of that a since it's the first character of the instance they're searching for).
How can I get the index of the character the user is searching for?
Xcode 11 • Swift 5 or later
let sentence = "hey hi hello, hey hi hello"
let query = "hello"
var searchRange = sentence.startIndex..<sentence.endIndex
var indices: [String.Index] = []
while let range = sentence.range(of: query, options: .caseInsensitive, range: searchRange) {
searchRange = range.upperBound..<searchRange.upperBound
indices.append(range.lowerBound)
}
print(indices) // "[7, 21]\n"
Another approach is NSRegularExpression which is designed to easily iterate through matches in an string. And if you use the .ignoreMetacharacters option, it will not apply any sophisticated wildcard/regex logic, but will just look for the string in question. So consider:
let string = "hey hi hello, hey hi hello" // string to search within
let searchString = "hello" // string to search for
let matchToFind = 2 // grab the second occurrence
let regex = try! NSRegularExpression(pattern: searchString, options: [.caseInsensitive, .ignoreMetacharacters])
You could use enumerateMatches:
var count = 0
let range = NSRange(string.startIndex ..< string.endIndex, in: string)
regex.enumerateMatches(in: string, range: range) { result, _, stop in
count += 1
if count == matchToFind {
print(result!.range.location)
stop.pointee = true
}
}
Or you can just find all of them with matches(in:range:) and then grab the n'th one:
let matches = regex.matches(in: string, range: range)
if matches.count >= matchToFind {
print(matches[matchToFind - 1].range.location)
}
Obviously, if you were so inclined, you could omit the .ignoreMetacharacters option and allow the user to perform regex searches, too (e.g. wildcards, whole word searches, start of word, etc.).
For Swift 2, see previous revision of this answer.
Is it possible to take a string and do a split based on an character in a character set?
let sourceString = "This is the original string/sentence."
let nonAlphabet = NSCharacterSet(charactersInString: "ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz").invertedSet
// let words = sourceString.split(charSet:nonAlphabet)
In this case words = ["This","is","the","original","string","sentence"]
Is this possible?
(NS)String has a method componentsSeparatedByCharactersInSet
let words = sourceString.componentsSeparatedByCharactersInSet(nonAlphabet)
The result will have an empty string at the end of the array because of the period.
let s = "ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz"
let ss = "This is the original string/sentence."
let arr = ss.characters.split{!s.characters.contains($0)}.map{String($0)}
// ["This", "is", "the", "original", "string", "sentence"]
I have a code :
var i : AnyObject!
i = 10
println(i as String)
println(i.stringValue)
it get crashed on as String line but runs in second i.stringValue.
What is the difference between as String and stringValue in the above lines?
.stringValue is a way to extract Integer into string value but as String will not work for that And if you use as String then Xcode will force you to add ! with as which is not good and it will never succeed and it would crash your app. you can't cast Int to String. It will always fail. Thats why when you do as! String it crashes the app.
So casting is not a good idea here.
And here is some more ways to extract Integer into string value:
let i : Int = 5 // 5
let firstWay = i.description // "5"
let anotherWay = "\(i)" // "5"
let thirdWay = String(i) // "5"
Here you can not use let forthway = i.stringValue because Int Doesn't have member named stringValue
But you can do same thing with anyObject as shown below:
let i : AnyObject = 5 // 5
let firstWay = i.description // "5"
let anotherWay = "\(i)" // "5"
let thirdWay = String(i) // "5"
let forthway = i.stringValue // "5" // now this will work.
Both are casting an Int to String but this will not work anymore.
In Swift 2 its not possible to do it like that.
U should use:
let i = 5
print(String(format: "%i", i))
This will specifically write the int value as a String
with as String, you can not cast the value but you define that the variable contains String but in you case it is Int.so it crashes.
while the other way i.e. of i.stringValue cast your value into String.So it doesn't gives you any crash and successfully cast into String value.
Note: As you are using AnyObject, variable have member stringvalue...but Int doesn't have...To cast Int value check out #Dharmesh Kheni answer
Basic question
I have 2 strings. I want to add one string to another? Here's an example:
var secondString= "is your name."
var firstString = "Mike, "
Here I have 2 strings. I want to add firstString to secondString, NOT vice versa. (Which would be: firstString += secondString.)
More detail
I have 5 string
let first = "7898"
let second = "00"
let third = "5481"
let fourth = "4782"
var fullString = "\(third):\(fourth)"
I know for sure that third and fourth will be in fullString, but I don't know about first and second.
So I will make an if statement checking if second has 00. If it does, first and second won't go in fullString. If it doesn't, second will go intofullString`.
Then I will check if first has 00. If it does, then first won't go inside of fullString, and if not, it will go.
The thing is, I need them in the same order: first, second, third fourth. So in the if statement, I need a way to potentially add first and second at the beginning of fullString.
Re. your basic question:
secondString = "\(firstString)\(secondString)"
or
secondString = firstString + secondString
Here is a way to insert string at the beginning "without resetting" per your comment (first at front of second):
let range = second.startIndex..<second.startIndex
second.replaceRange(range, with: first)
Re. your "more detail" question:
var fullString: String
if second == "00" {
fullString = third + fourth
} else if first == "00" {
fullString = second + third + fourth
} else {
fullString = first + second + third + fourth
}
From the Apple documentation:
String values can be added together (or concatenated) with the addition operator (+) to create a new String value:
let string1 = "hello"
let string2 = " there"
var welcome = string1 + string2
// welcome now equals "hello there"
You can also append a String value to an existing String variable with the addition assignment operator (+=):
var instruction = "look over"
instruction += string2
// instruction now equals "look over there"
You can append a Character value to a String variable with the String type’s append() method:
let exclamationMark: Character = "!"
welcome.append(exclamationMark)
// welcome now equals "hello there!"
So you are pretty much free to add these in any way shape or form.
Which includes
secondstring += firststring
Edit to accommodate the new information:
Strings in Swift are mutable which means you can always add to a string in-place without recreating any objects.
Something like (pseudo-code)
if(second != "00")
{
fullstring = second + fullstring
//only do something with first if second != 00
if(first != "00")
{
fullstring = first + fullstring
}
}
I have a question. I know that this can be done in Ruby, not sure about Swift, but never did anything similar.
What I'm working on is a program that works similar to this: if the user writes in a TextView "a + b", my code should elaborate the value of the sum between the variable "a" and "b".
What I mean is, how can I get variable names from a string? Is it possible at all?
If it's not, I'd use a switch (variable names are limited), but I'd need to "connect" a variable to the one written by the user. In the example, if the user writes "a + b", my sum function is something like "sum(par1,par2)", so, how can I make variable "par1" point to "a" and "par2" point to "b"? I can do that with pointers right? Is there any other way?
I'm not sure I've been clear, but it's quite hard to explain
Check this out. You can do with NSExpression and KVC.
class myclass:NSObject {
var a = 10;
var b = 20;
override init() {
super.init();
}
}
var obj = myclass();
// Expression from user
var expr:NSString = "a+b" ;
//Operators..
let opSet = NSMutableCharacterSet()
opSet.addCharactersInString("+-*/");
let components:NSArray = expr.componentsSeparatedByCharactersInSet(opSet)
let uniqueChars:NSSet = NSSet(array: components)
for variable in uniqueChars
{
expr = expr.stringByReplacingOccurrencesOfString ( variable as String,
withString : NSString(format: "%d", obj.valueForKey(variable as String) as Int),
options : NSStringCompareOptions.LiteralSearch,
range : NSMakeRange(0,expr.length) );
}
var result: AnyObject = NSExpression(format: expr).expressionValueWithObject(nil, context: nil)
print (result);