ios - 如何检测 NSString 是否包含特定字符？

Question

我有一个 NSString 对象，例如：- ($45,0000)

现在我想知道这个字符串是否包含 ()

我怎样才能做到这一点？

Answer 1

您是否要查找它是否包含至少一个(or )？您可以使用-rangeOfCharacterFromSet:：

NSCharacterSet *cset = [NSCharacterSet characterSetWithCharactersInString:@"()"];
NSRange range = [mystr rangeOfCharacterFromSet:cset];
if (range.location == NSNotFound) {
    // no ( or ) in the string
} else {
    // ( or ) are present
}

Answer 2

如果给定的字符串包含给定的字符，下面的方法将返回 Yes

-(BOOL)doesString:(NSString *)string containCharacter:(char)character
{
    return [string rangeOfString:[NSString stringWithFormat:@"%c",character]].location != NSNotFound;
}

您可以按如下方式使用它：

 NSString *s = @"abcdefg";

    if ([self doesString:s containCharacter:'a'])
        NSLog(@"'a' found");
    else
        NSLog(@"No 'a' found");


    if ([self doesString:s containCharacter:'h'])
        NSLog(@"'h' found");
    else
        NSLog(@"No 'h' found");

输出：

2013-01-11 11:15:03.830 CharFinder[17539:c07] 'a' 发现

2013-01-11 11:15:03.831 CharFinder[17539:c07] 找不到“h”

Answer 3

- (bool) contains: (NSString*) substring {
    NSRange range = [self rangeOfString:substring];
    return range.location != NSNotFound;
}

Answer 4

我在代码中使用了您的问题的通用答案。此代码包括以下规则： 1. 没有特殊字符 2. 至少一个大写字母和一个小英文字母 3. 至少一个数字数字

 BOOL lowerCaseLetter,upperCaseLetter,digit,specialCharacter;
    int asciiValue;
    if([txtPassword.text length] >= 5)
    {
        for (int i = 0; i < [txtPassword.text length]; i++)
        {
            unichar c = [txtPassword.text characterAtIndex:i];
            if(!lowerCaseLetter)
            {
                lowerCaseLetter = [[NSCharacterSet lowercaseLetterCharacterSet] characterIsMember:c];
            }
            if(!upperCaseLetter)
            {
                upperCaseLetter = [[NSCharacterSet uppercaseLetterCharacterSet] characterIsMember:c];
            }
            if(!digit)
            {
                digit = [[NSCharacterSet decimalDigitCharacterSet] characterIsMember:c];
            }

            asciiValue = [txtPassword.text characterAtIndex:i];
            NSLog(@"ascii value---%d",asciiValue);
            if((asciiValue >=33&&asciiValue < 47)||(asciiValue>=58 && asciiValue<=64)||(asciiValue>=91 && asciiValue<=96)||(asciiValue>=91 && asciiValue<=96))
            {
                specialCharacter=1;
            }
            else
            {
                specialCharacter=0;
            }

        }

        if(specialCharacter==0 && digit && lowerCaseLetter && upperCaseLetter)
        {
            //do what u want
            NSLog(@"Valid Password %d",specialCharacter);
        }
        else
        {
            NSLog(@"Invalid Password %d",specialCharacter);
            UIAlertView *alert = [[UIAlertView alloc] initWithTitle:@"Error"
                                                            message:@"Please Ensure that you have at least one lower case letter, one upper case letter, one digit and No Any special character"
                                                           delegate:nil cancelButtonTitle:@"OK" otherButtonTitles:nil];
            [alert show];
        }

Answer 5

为什么总是使用 NSCharactorSet ？这是简单而强大的解决方案

NSString *textStr = @"This is String Containing / Character";

if ([textStr containsString:@"/"])
{
    NSLog(@"Found!!");
}
else
{
   NSLog(@"Not Found!!");

Answer 6

您可以使用

NSString rangeOfString: (NSString *) string

见这里：https ://developer.apple.com/library/mac/#documentation/Cocoa/Reference/Foundation/Classes/NSString_Class/Reference/NSString.html%23//apple_ref/occ/instm/NSString/rangeOfString ：

如果 NSRange 返回的属性 'location' 等于 NSNotFound，则该字符串不包含传递的字符串（或字符）。