NSString hasPrefix: function with case sensitivity - ios

Question

Simple and may asked many time but little trick with this. We know, NSString doesn't work with case sensitivity for hasPrefix: method.
NSString *string = #"Xyzabcdedfghij";
NSString *substring = #"xyz";
if ([string hasPrefix:substring])
NSLog(#"string has prefix "); // won't get here.
Question is: Is there any built-in method for resolve this issue? I mean, hasPrefix: with case sensitive?
I could use below answer at least case. But want to know if there is any method which better than this..?
Known answer:(lease case)
if ([[test substringWithRange:NSMakeRange(0,3)] caseInsensitiveCompare:#"xyz"] == NSOrderedSame) {
// ....
}

Answer 1

From Apple themselves:
NSString *searchString = #"age";
NSString *beginsTest = #"Agencies";
NSRange prefixRange = [beginsTest rangeOfString:searchString
options:(NSAnchoredSearch | NSCaseInsensitiveSearch)];
// prefixRange = {0, 3}
NSString *endsTest = #"BRICOLAGE";
NSRange suffixRange = [endsTest rangeOfString:searchString
options:(NSAnchoredSearch | NSCaseInsensitiveSearch | NSBackwardsSearch)];
// suffixRange = {6, 3}
This could be wrapped into an easy-to-use method:
- (BOOL) string:(NSString *)string
hasPrefix:(NSString *)prefix
caseInsensitive:(BOOL)caseInsensitive {
if (!caseInsensitive)
return [string hasPrefix:prefix];
const NSStringCompareOptions options = NSAnchoredSearch|NSCaseInsensitiveSearch;
NSRange prefixRange = [string rangeOfString:prefix
options:options];
return prefixRange.location == 0 && prefixRange.length > 0;
}

Answer 2

You can always use lowercaseString on both strings and thus forcing the same case. So for example
[[string lowercaseString] hasPrefix:[substring lowercaseString]];

Answer 3

Nasty way to do is to lower case the both string and than use hasPrefix
e.g.
[[mainString lowercaseString] hasPrefix:[stringToFind lowercaseString]];

Answer 4

Swift 5:
extension String {
public func hasPrefixIgnoringCase(_ prefix: String) -> Bool {
range(of: prefix, options: [.anchored, .caseInsensitive]) != nil
}
}
Usage:
"Hello".hasPrefixIgnoringCase("hEl") // return true

Answer 5

you can do this by
if([[string lowercaseString] hasPrefix:[substring lowercaseString]])
{
NSLog(#"found");
}

Answer 6

A Swift 4.2 version of accepted answer:
extension String {
public func hasPrefix<Prefix>(_ prefix: Prefix, caseSensitive: Bool) -> Bool where Prefix : StringProtocol {
if caseSensitive { return self.hasPrefix(prefix) }
let prefixRange = self.range(of: prefix, options: [.anchored, .caseInsensitive])
return prefixRange != nil
}
}
Alternatively, using .lowercased()
extension String {
public func hasPrefix<Prefix>(_ prefix: Prefix, caseSensitive: Bool) -> Bool where Prefix : StringProtocol {
if caseSensitive { return self.hasPrefix(prefix) }
return self.lowercased().hasPrefix(prefix.lowercased())
}
}
The extension is to use as following
let string = "Hello World"
let caseSensitiveSearch = string.hasPrefix("hello", caseSensitive: true) // return false
let caseNotSensitiveSearch = string.hasPrefix("hello", caseSensitive: false) // return true

Answer 7

For those of you who like functional style one-liners (like I do):
extension String {
func hasPrefix<Prefix>(_ prefix: Prefix, caseSensitive: Bool) -> Bool where Prefix: StringProtocol {
return caseSensitive ? hasPrefix(prefix) :
self.range(of: prefix, options: [.anchored, .caseInsensitive]) != nil
}
}

Answer 8

Swift 5
extension StringProtocol {
public func hasPrefix<T: StringProtocol>(caseInsensitive prefix: T) -> Bool {
lowercased().starts(with: prefix.lowercased())
}
}
Use:
"Hello World".hasPrefix(caseInsensitive: "hello") //true