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 *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!");
}
//你也可以使用下面的语句:
if (rangeValue.location == NSNotFound)
{
NSLog(@"string does not contain hello");
}
else
{
NSLog(@"string contains hello!");
}
其他回答
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!");
}
在Swift 4中:
let a = "Hello, how are you?"
a.contains("Hello") //will return true
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");
}
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)
NSString *categoryString = @"Holiday Event";
if([categoryString rangeOfString:@"Holiday"].location == NSNotFound)
{
//categoryString does not contains Holiday
}
else
{
//categoryString contains Holiday
}
