I want to search an element of an array of strings in a string. Like this:
let array:[String] = ["dee", "kamal"]
let str:String = "Hello all how are you, I m here for deepak."
so, I want
str.contain("dee") == true
any possible search in string?
You can do it in one line by composing a regular expression pattern "(item1|item2|item3)"
let array = ["dee", "kamal"]
let str = "Hello all how are you, I m here for deepak."
let success = str.range(of: "(" + array.joined(separator: "|") + ")", options: .regularExpression) != nil
You should iterate over the array and for each element, call str.contains.
for word in array {
if str.contains(word) {
print("\(word) is part of the string")
} else {
print("Word not found")
}
}
You can do like this:
array.forEach { (item) in
var isContains:Bool = str.contains(item)
print(isContains)
}
Related
Swift 3 introduced String.range(of:options). Then, with this function, is possible match a part of string without creating a NSRegularExpression object, for example:
let text = "it is need #match my both #hashtag!"
let match = text.range(of: "(?:^#|\\s#)[\\p{L}0-9_]*", options: .regularExpression)!
print(text[match]) // #math
But, is possible match both occurrences of the regexp (that is, #match and #hashtag), instead of only the first?
let text = "it is need #match my both #hashtag!"
// create an object to store the ranges found
var ranges: [Range<String.Index>] = []
// create an object to store your search position
var start = text.startIndex
// create a while loop to find your regex ranges
while let range = text.range(of: "(?:^#|\\s#)[\\p{L}0-9_]*", options: .regularExpression, range: start..<text.endIndex) {
// append your range found
ranges.append(range)
// and change the startIndex of your string search
start = range.lowerBound < range.upperBound ? range.upperBound : text.index(range.lowerBound, offsetBy: 1, limitedBy: text.endIndex) ?? text.endIndex
}
ranges.forEach({print(text[$0])})
This will print
#match
#hashtag
If you need to use it more than once in your code you should add this extension to your project:
extension StringProtocol {
func ranges<S: StringProtocol>(of string: S, options: String.CompareOptions = []) -> [Range<Index>] {
var result: [Range<Index>] = []
var start = startIndex
while start < endIndex,
let range = self[start...].range(of: string, options: options) {
result.append(range)
start = range.lowerBound < range.upperBound ?
range.upperBound : index(after: range.lowerBound)
}
return result
}
}
usage:
let text = "it is need #match my both #hashtag!"
let pattern = "(?<!\\S)#[\\p{L}0-9_]*"
let ranges = text.ranges(of: pattern, options: .regularExpression)
let matches = ranges.map{text[$0]}
print(matches) // ["#match", "#hashtag"]
My main string is "hello Swift Swift and Swift" and substring is Swift.
I need to get the number of times the substring "Swift" occurs in the mentioned string.
This code can determine whether the pattern exists.
var string = "hello Swift Swift and Swift"
if string.rangeOfString("Swift") != nil {
println("exists")
}
Now I need to know the number of occurrence.
A simple approach would be to split on "Swift", and subtract 1 from the number of parts:
let s = "hello Swift Swift and Swift"
let tok = s.components(separatedBy:"Swift")
print(tok.count-1)
This code prints 3.
Edit: Before Swift 3 syntax the code looked like this:
let tok = s.componentsSeparatedByString("Swift")
Should you want to count characters rather than substrings:
extension String {
func count(of needle: Character) -> Int {
return reduce(0) {
$1 == needle ? $0 + 1 : $0
}
}
}
Optimising dwsolbergs solution to count faster. Also faster than componentsSeparatedByString.
extension String {
/// stringToFind must be at least 1 character.
func countInstances(of stringToFind: String) -> Int {
assert(!stringToFind.isEmpty)
var count = 0
var searchRange: Range<String.Index>?
while let foundRange = range(of: stringToFind, options: [], range: searchRange) {
count += 1
searchRange = Range(uncheckedBounds: (lower: foundRange.upperBound, upper: endIndex))
}
return count
}
}
Usage:
// return 2
"aaaa".countInstances(of: "aa")
If you want to ignore accents, you may replace options: [] with options: .diacriticInsensitive like dwsolbergs did.
If you want to ignore case, you may replace options: [] with options: .caseInsensitive like ConfusionTowers suggested.
If you want to ignore both accents and case, you may replace options: [] with options: [.caseInsensitive, .diacriticInsensitive] like ConfusionTowers suggested.
If, on the other hand, you want the fastest comparison possible and you can guarantee some canonical form for composed character sequences, then you may consider option .literal and it will only perform exact matchs.
Swift 5 Extension
extension String {
func numberOfOccurrencesOf(string: String) -> Int {
return self.components(separatedBy:string).count - 1
}
}
Example use
let string = "hello Swift Swift and Swift"
let numberOfOccurrences = string.numberOfOccurrencesOf(string: "Swift")
// numberOfOccurrences = 3
I'd recommend an extension to string in Swift 3 such as:
extension String {
func countInstances(of stringToFind: String) -> Int {
var stringToSearch = self
var count = 0
while let foundRange = stringToSearch.range(of: stringToFind, options: .diacriticInsensitive) {
stringToSearch = stringToSearch.replacingCharacters(in: foundRange, with: "")
count += 1
}
return count
}
}
It's a loop that finds and removes each instance of the stringToFind, incrementing the count on each go-round. Once the searchString no longer contains any stringToFind, the loop breaks and the count returns.
Note that I'm using .diacriticInsensitive so it ignore accents (for example résume and resume would both be found). You might want to add or change the options depending on the types of strings you want to find.
I needed a way to count substrings that may contain the start of the next matched substring. Leveraging dwsolbergs extension and Strings range(of:options:range:locale:) method I came up with this String extension
extension String
{
/**
Counts the occurrences of a given substring by calling Strings `range(of:options:range:locale:)` method multiple times.
- Parameter substring : The string to search for, optional for convenience
- Parameter allowOverlap : Bool flag indicating whether the matched substrings may overlap. Count of "🐼🐼" in "🐼🐼🐼🐼" is 2 if allowOverlap is **false**, and 3 if it is **true**
- Parameter options : String compare-options to use while counting
- Parameter range : An optional range to limit the search, default is **nil**, meaning search whole string
- Parameter locale : Locale to use while counting
- Returns : The number of occurrences of the substring in this String
*/
public func count(
occurrencesOf substring: String?,
allowOverlap: Bool = false,
options: String.CompareOptions = [],
range searchRange: Range<String.Index>? = nil,
locale: Locale? = nil) -> Int
{
guard let substring = substring, !substring.isEmpty else { return 0 }
var count = 0
let searchRange = searchRange ?? startIndex..<endIndex
var searchStartIndex = searchRange.lowerBound
let searchEndIndex = searchRange.upperBound
while let rangeFound = range(of: substring, options: options, range: searchStartIndex..<searchEndIndex, locale: locale)
{
count += 1
if allowOverlap
{
searchStartIndex = index(rangeFound.lowerBound, offsetBy: 1)
}
else
{
searchStartIndex = rangeFound.upperBound
}
}
return count
}
}
why not just use some length maths??
extension String {
func occurences(of search:String) -> Int {
guard search.count > 0 else {
preconditionFailure()
}
let shrunk = self.replacingOccurrences(of: search, with: "")
return (self.count - shrunk.count)/search.count
}
}
Try this
var mainString = "hello Swift Swift and Swift"
var count = 0
mainString.enumerateSubstrings(in: mainString.startIndex..<mainString.endIndex, options: .byWords) { (subString, subStringRange, enclosingRange, stop) in
if case let s? = subString{
if s.caseInsensitiveCompare("swift") == .orderedSame{
count += 1
}
}
}
print(count)
For the sake of completeness – and because there is a regex tag – this is a solution with Regular Expression
let string = "hello Swift Swift and Swift"
let regex = try! NSRegularExpression(pattern: "swift", options: .caseInsensitive)
let numberOfOccurrences = regex.numberOfMatches(in: string, range: NSRange(string.startIndex..., in: string))
The option .caseInsensitive is optional.
My solution, maybe it will be better to use String.Index instead of Int range but I think in such way it is a bit easier to read.
extension String {
func count(of char: Character, range: (Int, Int)? = nil) -> Int {
let range = range ?? (0, self.count)
return self.enumerated().reduce(0) {
guard ($1.0 >= range.0) && ($1.0 < range.1) else { return $0 }
return ($1.1 == char) ? $0 + 1 : $0
}
}
}
Solution which uses a higher order functions
func subStringCount(str: String, substr: String) -> Int {
{ $0.isEmpty ? 0 : $0.count - 1 } ( str.components(separatedBy: substr))
}
Unit Tests
import XCTest
class HigherOrderFunctions: XCTestCase {
func testSubstringWhichIsPresentInString() {
XCTAssertEqual(subStringCount(str: "hello Swift Swift and Swift", substr: "Swift"), 3)
}
func testSubstringWhichIsNotPresentInString() {
XCTAssertEqual(subStringCount(str: "hello", substr: "Swift"), 0)
}
}
Another way using RegexBuilder in iOS 16+ & swift 5.7+.
import RegexBuilder
let text = "hello Swift Swift and Swift"
let match = text.matches(of: Regex{"Swift"})
print(match.count) // prints 3
Using this as a function
func countSubstrings(string : String, subString : String)-> Int{
return string.matches(of: Regex{subString}).count
}
print(countSubstrings(string: text, subString: "Swift")) //prints 3
Using this as an Extension
extension String {
func countSubstrings(subString : String)-> Int{
return self.matches(of: Regex{subString}).count
}
}
print(text.countSubstrings(subString: "Swift")) // prints 3
Using JavaScript & regex I want to split a string on every %20 that is not within quotes, example:
Here%20is%20"a%20statement%20"%20for%20Testing%20"%20The%20Values%20"
//easy to read version: Here is "a statement " for Testing " The Values "
______________ ______________
would return
{"Here","is","a statement ","for","Testing"," The Values "}
but it seems my regex are no longer strong enough to build the expression. Thanks for any help!
A way using the replace method, but without using the replacement result. The idea is to use a closure to fill the result variable at each occurence:
var txt = 'Here%20is%20"a%20statement%20"%20for%20Testing%20"%20The%20Values%20"';
var result = Array();
txt.replace(/%20/g, ' ').replace(/"([^"]+)"|\S+/g, function (m,g1) {
result.push( (g1==undefined)? m : g1); });
console.log(result);
Just try with:
var input = 'Here%20is%20"a%20statement%20"%20for%20Testing%20"%20The%20Values%20"',
tmp = input.replace(/%20/g, ' ').split('"'),
output = []
;
for (var i = 0; i < tmp.length; i++) {
var part = tmp[i].trim();
if (!part) continue;
if (i % 2 == 0) {
output = output.concat(part.split(' '));
} else {
output.push(part);
}
}
Output:
["Here", "is", "a statement", "for", "Testing", "The Values"]
my regex query below does an exact match of a word say Bob or Bill for example
var regExp = new RegExp("^" + inputVal + "$", 'i');
what i want it to do is match anything exactly (Bob or Bill Etc) but not match Fred
so match anything exactly except for Fred, does that make sense?
anyone help me out as to how i do that?
Thanks
EDIT 2:
i thought id show my actual script instead, what im doing is searching a table, and im page load i want to hide rows that contain a string. so if exlucde lenght is greater than 0 hide that row...
function searchPagingTable(inputVal, tablename, fixedsearch, exclude) {
var table = $(tablename);
table.find('tr:not(.header)').each(function (index, row) {
var allCells = $(row).find('td');
if (allCells.length > 0) {
var found = false;
allCells.each(function (index, td) {
if (fixedsearch == 1) {
var regExp = new RegExp("^" + inputVal + "$", 'i');
}
else if (exclude.length > 0)
{
var regExp = new RegExp("^(?!" + exclude + ")", "i");
}
else {
var regExp = new RegExp(inputVal, 'i');
}
if (regExp.test($(td).text())) {
found = true;
return false;
}
});
if (found == true) $(row).show().removeClass('exclude'); else $(row).hide().addClass('exclude');
}
});
pa
ginate();
}
That would be
var exclude = "Fred"
var regExp = new RegExp("^(?!.*" + exclude + ")", "i");
This regex matches any string except those that contain Fred. It doesn't actually match any characters in the string, but that's sufficient if you're just looking for a true/false result.
This will also find strings that contain Alfred or Fredo, so if you don't want that, you need to tell the regex only to look for entire words using word boundaries:
var regExp = new RegExp("^(?!.*\\b" + exclude + "\\b)", "i");
You need to make sure that your exclude string only contains ASCII letters/digits (or underscores) for this to work correctly.
You could populate a list of names you wish to match against:
var validNames = ['bob', 'bill'];
Then lowercase each input and match against the list:
if (validNames.indexOf(inputVal.toLowerCase()) != -1) {
// it's a good name
}
For older browsers you have to shim Array.indexOf()
var re = new RegExp('^\\s*Fred\\s*$','i');
if (inputVal.match(re)) {
// Fred has been found
} else {
// Anything has been found
}
I have the following string:
str1 = "cat-one,cat2,cat-3";
OR
str1 = "catone,cat-2,cat3";
OR
str1 = "catone";
OR
str1 = "cat-one";
The point here is words may/may not have "-"s in it
Using regex:
How could I extract the 1st word?
Appreciate any help on this.
Thanks,
L
It's pretty easy, just include allowed characters in brackets:
^([\w\-]+)
An approach not using a regex: assuming the first word is delimited always by a comma "," you can do this:
var str1 = "cat-one";
var i = str1.indexOf(",");
var firstTerm = i == -1 ? str1 : str1.substring(0, i);
Edit: Assumed this was a javascript question, for some reason.
If someone, one day would like to do it in Swift here you go with an extension :
extension String {
func firstWord() -> String? {
var error : NSError?
let internalExpression = NSRegularExpression(pattern: "^[a-zA-Z0-9]*", options: .CaseInsensitive, error: &error)!
let matches = internalExpression.matchesInString(self, options: nil, range:NSMakeRange(0, countElements(self)))
if (matches.count > 0) {
let range = (matches[0] as NSTextCheckingResult).range
return (self as NSString).substringWithRange(range)
}
return nil
}
}
To use it just write:
myString.firstWord()