[英]How do I check if a string contains another string in Objective-C?
如何检查字符串( NSString
)是否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");
}
无论如何,这是查找字符串是否包含另一个字符串的最佳方法吗?
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,并且文档说如果“haystack”不包含“needle”,它返回结构{NSNotFound, 0}
。
如果您使用的是 iOS 8 或 OS X Yosemite,您现在可以执行以下操作: (*注意:如果在 iOS7 设备上调用此代码,这将使您的应用程序崩溃)。
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)
👍
对于 iOS 8.0+ 和 macOS 10.10+,您可以使用 NSString 的原生containsString:
。
对于旧版本的 iOS 和 macOS,您可以为 NSString 创建您自己的(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:请注意以下 Daniel Galasko 关于命名的评论
由于这似乎是 Google 中排名靠前的结果,因此我想添加以下内容:
iOS 8 和 OS X 10.10 将containsString:
方法添加到NSString
。 这些系统的 Dave DeLong 示例的更新版本:
NSString *string = @"hello bla bla";
if ([string containsString:@"bla"]) {
NSLog(@"string contains bla!");
} else {
NSLog(@"string does not contain bla");
}
NSString *myString = @"hello bla bla";
NSRange rangeValue = [myString rangeOfString:@"hello" options:NSCaseInsensitiveSearch];
if (rangeValue.length > 0)
{
NSLog(@"string contains hello");
}
else
{
NSLog(@"string does not contain hello!");
}
//您也可以使用以下方法:
if (rangeValue.location == NSNotFound)
{
NSLog(@"string does not contain hello");
}
else
{
NSLog(@"string contains hello!");
}
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
所以我个人真的很讨厌NSNotFound
但理解它的必要性。
但是有些人可能不明白对比 NSNotFound 的复杂性
例如,这段代码:
- (BOOL)doesString:(NSString*)string containString:(NSString*)otherString {
if([string rangeOfString:otherString].location != NSNotFound)
return YES;
else
return NO;
}
有它的问题:
1) 显然,如果otherString = nil
这段代码会崩溃。 一个简单的测试是:
NSLog(@"does string contain string - %@", [self doesString:@"hey" containString:nil] ? @"YES": @"NO");
结果是 !! 碰撞 !!
2)对于Objective-c的新手来说不太明显的是,当string = nil
时,相同的代码不会崩溃。 例如,这段代码:
NSLog(@"does string contain string - %@", [self doesString:nil containString:@"hey"] ? @"YES": @"NO");
和这个代码:
NSLog(@"does string contain string - %@", [self doesString:nil containString:nil] ? @"YES": @"NO");
两者都会导致
does string contains string - YES
这显然不是你想要的。
所以我认为更好的解决方案是使用 rangeOfString 返回长度为 0 的事实,所以更好更可靠的代码是这样的:
- (BOOL)doesString:(NSString*)string containString:(NSString*)otherString {
if(otherString && [string rangeOfString:otherString].length)
return YES;
else
return NO;
}
或者简单地:
- (BOOL)doesString:(NSString*)string containString:(NSString*)otherString {
return (otherString && [string rangeOfString:otherString].length);
}
对于情况 1 和 2 将返回
does string contains string - NO
那是我的 2 美分 ;-)
请查看我的Gist以获得更多有用的代码。
P i的解决方案的改进版本,NSString 上的一个类别,它不仅会告诉,是否在另一个字符串中找到一个字符串,而且还会通过引用获取一个范围,是:
@interface NSString (Contains)
-(BOOL)containsString: (NSString*)substring
atRange:(NSRange*)range;
-(BOOL)containsString:(NSString *)substring;
@end
@implementation NSString (Contains)
-(BOOL)containsString:(NSString *)substring
atRange:(NSRange *)range{
NSRange r = [self rangeOfString : substring];
BOOL found = ( r.location != NSNotFound );
if (range != NULL) *range = r;
return found;
}
-(BOOL)containsString:(NSString *)substring
{
return [self containsString:substring
atRange:NULL];
}
@end
像这样使用它:
NSString *string = @"Hello, World!";
//If you only want to ensure a string contains a certain substring
if ([string containsString:@"ello" atRange:NULL]) {
NSLog(@"YES");
}
// Or simply
if ([string containsString:@"ello"]) {
NSLog(@"YES");
}
//If you also want to know substring's range
NSRange range;
if ([string containsString:@"ello" atRange:&range]) {
NSLog(@"%@", NSStringFromRange(range));
}
这是一个复制和粘贴功能片段:
-(BOOL)Contains:(NSString *)StrSearchTerm on:(NSString *)StrText
{
return [StrText rangeOfString:StrSearchTerm
options:NSCaseInsensitiveSearch].location != NSNotFound;
}
Oneliner(代码量较少。DRY,因为您只有一个NSLog
):
NSString *string = @"hello bla bla";
NSLog(@"String %@", ([string rangeOfString:@"bla"].location == NSNotFound) ? @"not found" : @"cotains bla");
NSString *categoryString = @"Holiday Event";
if([categoryString rangeOfString:@"Holiday"].location == NSNotFound)
{
//categoryString does not contains Holiday
}
else
{
//categoryString contains Holiday
}
尝试这个,
NSString *string = @"test Data";
if ([[string lowercaseString] rangeOfString:@"data"].location == NSNotFound)
{
NSLog(@"string does not contain Data");
}
else
{
NSLog(@"string contains data!");
}
最佳解决方案。 就这么简单! 如果要查找单词或字符串的一部分。 您可以使用此代码。 在这个例子中,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");
}
在 Swift 4 中:
let a = "Hello, how are you?"
a.contains("Hello") //will return true
如果你需要一次写:
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;
在快速的情况下,这可以使用
let string = "Package #23"
if string.containsString("Package #") {
//String contains substring
}
else {
//String does not contain substring
}
如果不关心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");
}
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!");
}
使用选项 NSCaseInsensitiveSearch 和 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
}
输出结果找到:是
选项可以“或”在一起,包括:
NSCaseInsensitiveSearch NSLiteralSearch NSBackwardsSearch 等
第一个字符串包含或不包含第二个字符串,
NSString *first = @"Banana";
NSString *second = @"BananaMilk";
NSRange range = [first rangeOfString:second options:NSCaseInsensitiveSearch];
if (range.length > 0) {
NSLog(@"Detected");
}
else {
NSLog(@"Not detected");
}
Swift 4 及以上
let str = "Hello iam midhun"
if str.contains("iam") {
//contains substring
}
else {
//doesn't contain substring
}
目标-C
NSString *stringData = @"Hello iam midhun";
if ([stringData containsString:@"iam"]) {
//contains substring
}
else {
//doesn't contain substring
}
尝试这个:
斯威夫特 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");
}
对于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");
}
-(Bool)checkIf:(String)parentString containsSubstring:(String)checkString {
NSRange textRange =[parentString rangeOfString:checkString];
return textRange.location != NSNotFound // returns true if parent string contains substring else returns false
}
如果需要字符串的某个位置,这段代码会出现在Swift 3.0 中:
let string = "This is my string"
let substring = "my"
let position = string.range(of: substring)?.lowerBound
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