我如何检查一个字符串(NSString)包含另一个较小的字符串?

我希望是这样的:

NSString *string = @"hello bla bla";
NSLog(@"%d",[string containsSubstring:@"hello"]);

但我能找到的最接近的是:

if ([string rangeOfString:@"hello"] == 0) {
    NSLog(@"sub string doesnt exist");
} 
else {
    NSLog(@"exists");
}

不管怎样,这是找出字符串是否包含另一个字符串的最好方法吗?


当前回答

在Swift 4中:

let a = "Hello, how are you?"
a.contains("Hello")   //will return true

其他回答

如果你需要这样做,可以这样写:

NSString *stringToSearchThrough = @"-rangeOfString method finds and returns the range of the first occurrence of a given string within the receiver.";
BOOL contains = [stringToSearchThrough rangeOfString:@"occurence of a given string"].location != NSNotFound;

联机(更少的代码。DRY,因为你只有一个NSLog):

NSString *string = @"hello bla bla";
NSLog(@"String %@", ([string rangeOfString:@"bla"].location == NSNotFound) ? @"not found" : @"cotains bla"); 

试试这个:

Swift 4.1、4.2:

let stringData = "Black board"

//swift quick way and case sensitive
if stringData.contains("bla") {
    print("data contains string");
}

//case sensitive
if stringData.range(of: "bla",options: .caseInsensitive) != nil {
    print("data contains string");
}else {
    print("data does not contains string");
}

objective - c:

NSString *stringData = @"Black board";

//Quick way and case sensitive
if ([stringData containsString:@"bla"]) {
    NSLog(@"data contains string");
}

//Case Insensitive
if ([stringData rangeOfString:@"bla" options:NSCaseInsensitiveSearch].location != NSNotFound) {
   NSLog(@"data contains string");
}else {
   NSLog(@"data does not contain string");
}

因为这似乎是谷歌中的一个高级结果,我想添加这个:

iOS 8和OS X 10.10在NSString中添加了containsString:方法。Dave DeLong关于这些系统的例子的更新版本:

NSString *string = @"hello bla bla";
if ([string containsString:@"bla"]) {
    NSLog(@"string contains bla!");
} else {
    NSLog(@"string does not contain bla");
}

在iOS 8和Swift中,我们可以使用localizedCaseInsensitiveContainsString方法

 let string: NSString = "Café"
 let substring: NSString = "É"

 string.localizedCaseInsensitiveContainsString(substring) // true