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
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");
}
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;
如果不介意区分大小写的字符串。 试试这个。
NSString *string = @"Hello World!";
if([string rangeOfString:@"hello" options:NSCaseInsensitiveSearch].location !=NSNotFound)
{
NSLog(@"found");
}
else
{
NSLog(@"not found");
}
使用选项nsscaseinsensitivesearch与rangeOfString:options:
NSString *me = @"toBe" ;
NSString *target = @"abcdetobe" ;
NSRange range = [target rangeOfString: me options: NSCaseInsensitiveSearch];
NSLog(@"found: %@", (range.location != NSNotFound) ? @"Yes" : @"No");
if (range.location != NSNotFound) {
// your code
}
输出结果:Yes
选项可以用“或”加在一起，包括:
NSCaseInsensitiveSearch NSLiteralSearch NSBackwardsSearch等等
下面是一个复制粘贴函数片段:
-(BOOL)Contains:(NSString *)StrSearchTerm on:(NSString *)StrText
{
return [StrText rangeOfString:StrSearchTerm
options:NSCaseInsensitiveSearch].location != NSNotFound;
}
NSString *categoryString = @"Holiday Event";
if([categoryString rangeOfString:@"Holiday"].location == NSNotFound)
{
//categoryString does not contains Holiday
}
else
{
//categoryString contains Holiday
}
