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
NSString *categoryString = @"Holiday Event";
if([categoryString rangeOfString:@"Holiday"].location == NSNotFound)
{
//categoryString does not contains Holiday
}
else
{
//categoryString contains Holiday
}
其他回答
NSString *categoryString = @"Holiday Event";
if([categoryString rangeOfString:@"Holiday"].location == NSNotFound)
{
//categoryString does not contains Holiday
}
else
{
//categoryString contains Holiday
}
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
}
对于iOS 8.0+和macOS 10.10+，你可以使用NSString的本地containsString:。
对于旧版本的iOS和macOS，你可以为NSString创建自己的(过时的)类别:
@interface NSString ( SubstringSearch )
- (BOOL)containsString:(NSString *)substring;
@end
// - - - -
@implementation NSString ( SubstringSearch )
- (BOOL)containsString:(NSString *)substring
{
NSRange range = [self rangeOfString : substring];
BOOL found = ( range.location != NSNotFound );
return found;
}
@end
注意:观察Daniel Galasko下面关于命名的评论
最佳解决方案。就这么简单!如果你想找个词或者 字符串的一部分。您可以使用此代码。在本例中，我们将检查word的值是否包含"acter"。
NSString *word =@"find a word or character here";
if ([word containsString:@"acter"]){
NSLog(@"It contains acter");
} else {
NSLog (@"It does not contain acter");
}
