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的情况下,可以使用这个

let string = "Package #23"
if string.containsString("Package #") {
    //String contains substring
}
else {
    //String does not contain substring
}

其他回答

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

请使用此代码

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

试试这个,

NSString *string = @"test Data";
if ([[string lowercaseString] rangeOfString:@"data"].location == NSNotFound) 
{
    NSLog(@"string does not contain Data");
}   
else 
{
    NSLog(@"string contains data!");
}

Swift 4及以上

let str = "Hello iam midhun"

if str.contains("iam") {
  //contains substring
}
else {
  //doesn't contain substring
}

objective - c

NSString *stringData = @"Hello iam midhun";

if ([stringData containsString:@"iam"]) {
    //contains substring
}
else {
    //doesn't contain substring
}
NSString *string = @"hello bla bla";
if ([string rangeOfString:@"bla"].location == NSNotFound) {
  NSLog(@"string does not contain bla");
} else {
  NSLog(@"string contains bla!");
}

关键是注意到rangeOfString:返回一个NSRange结构体,并且文档说如果“haystack”不包含“needle”,它返回结构体{NSNotFound, 0}。


如果你在iOS 8或OS X Yosemite上,你现在可以这样做:*(注意:如果在iOS7设备上调用此代码,这将使你的应用程序崩溃)。

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

(这也是它在Swift中的工作方式)