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
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;
其他回答
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");
}
NSString *categoryString = @"Holiday Event";
if([categoryString rangeOfString:@"Holiday"].location == NSNotFound)
{
//categoryString does not contains Holiday
}
else
{
//categoryString contains Holiday
}
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");
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!");
}
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");
}
