Question

给出以下字符串：

s1 = '000001234'
s2 = '123400'

如何剥离拖尾和前导零？

在python中，我可以这样做：

s1 = s1.lstrip('0') # 1234
s2 = s2.rstrip('0') # 1234
s1, s2 = s1.strip('0'), s2.strip('0') # 1234, 1234

如果给出上面的两个字符串，我如何在Objective-C中执行等效操作？

Answer 1

更简单，更清洁的解决方案：

    NSString *s1 = @"0001234";
    NSString *s2 = @"123400000";

    while ([[s1 substringToIndex:1] isEqualToString:@"0"]) 
        s1 = [s1 substringFromIndex:1];

    while ([[s2 substringFromIndex:[s2 length]-1] isEqualToString:@"0"]) 
       s2 = [s2 substringToIndex:[s2 length]-1];

    NSLog(@"s1: %@", s1);
    NSLog(@"s2: %@", s2);

Answer 2

有各种可能的解决方案，这里有一个使用正则表达式。（NSScanner可以替代）。

NSString *s1 = @"000120340";
NSString *t1 = [s1 stringByReplacingOccurrencesOfString:@"^0*" withString:@""
        options:NSRegularExpressionSearch range:NSMakeRange(0, [s1 length])];
NSLog(@"%@", t1); // "120340"

NSString *s2 = @"012034000";
NSString *t2 = [s2 stringByReplacingOccurrencesOfString:@"0*$" withString:@""
        options:NSRegularExpressionSearch range:NSMakeRange(0, [s2 length])];
NSLog(@"%@", t2); // "012034"

同时从字符串的两个末尾删除零被称为“修剪”，可以用

完成

NSString *s3 = @"00012034000";
NSCharacterSet *zeroCharset = [NSCharacterSet characterSetWithCharactersInString:@"0"];
NSString *t3 = [s3 stringByTrimmingCharactersInSet:zeroCharset];
NSLog(@"%@", t3); // "12034"

Answer 3

有一个简单的解决方案：数字本身没有前导零（但导致尾随），所以要获得前导零，我们只需要将它们转换为数字对象然后再转换为字符串。这将删除它们。但这并不适用于尾随零。

NSNumberFormatter *f = [[NSNumberFormatter alloc] init];
NSString *leftStripped = [f stringFromNumber:[f numberFromString:self]];

但我们可以使它工作，我们只是反转字符串，而不是创建一些字符串，从中创建一个字符串并再次反转。

首先剥离左边的零

NSString *leftStripped = [f stringFromNumber:[f numberFromString:aNumberString]];

而不是反转

NSMutableString *reversedLeftStripped = [@"" mutableCopy];

[leftStripped enumerateSubstringsInRange:NSMakeRange(0,[leftStripped length])
                                 options:(NSStringEnumerationReverse | NSStringEnumerationByComposedCharacterSequences)
                              usingBlock:^(NSString *substring, NSRange substringRange, NSRange enclosingRange, BOOL *stop) {
                                  [reversedLeftStripped appendString:substring];
                              }];

创建一个数字对象和一个来自数字

的字符串

NSString *rightLeftStrippedReversed = [f stringFromNumber:[f numberFromString:reversedLeftStripped]];

并将其反转

NSMutableString *rightLeftStripped = [@"" mutableCopy];

[rightLeftStrippedReversed enumerateSubstringsInRange:NSMakeRange(0, [rightLeftStrippedReversed length]) options:NSStringEnumerationReverse| NSStringEnumerationByComposedCharacterSequences usingBlock:^(NSString *substring, NSRange substringRange, NSRange enclosingRange, BOOL *stop) {
    [rightLeftStripped appendString:substring];
}];

我会使用一个类别：

@interface NSString (Stripping)

-(NSString *)stringByStrippingLeadingAndTrailingZeros;

@end

@implementation NSString (Stripping)

-(NSString *)stringByStrippingLeadingAndTrailingZeros
{
    static NSNumberFormatter *f;
    static dispatch_once_t onceToken;
    dispatch_once(&onceToken, ^{
        f = [[NSNumberFormatter alloc] init];
    });

    NSString *leftStripped = [f stringFromNumber:[f numberFromString:self]];
    NSMutableString *reversedLeftStripped = [@"" mutableCopy];

    [leftStripped enumerateSubstringsInRange:NSMakeRange(0,[leftStripped length])
                                     options:(NSStringEnumerationReverse | NSStringEnumerationByComposedCharacterSequences)
                                  usingBlock:^(NSString *substring, NSRange substringRange, NSRange enclosingRange, BOOL *stop) {
                                      [reversedLeftStripped appendString:substring];
                                  }];

    NSString *rightLeftStrippedReversed = [f stringFromNumber:[f numberFromString:reversedLeftStripped]];
    NSMutableString *rightLeftStripped = [@"" mutableCopy];

    [rightLeftStrippedReversed enumerateSubstringsInRange:NSMakeRange(0, [rightLeftStrippedReversed length]) options:NSStringEnumerationReverse| NSStringEnumerationByComposedCharacterSequences usingBlock:^(NSString *substring, NSRange substringRange, NSRange enclosingRange, BOOL *stop) {
        [rightLeftStripped appendString:substring];
    }];

    return rightLeftStripped;
}

@end

所以你可以轻松打电话

NSString *s1 = @"0001234";
NSString *s2 = @"123400000";

for (NSString *string in @[s1, s2]) {
    NSLog(@"%@", [string stringByStrippingLeadingAndTrailingZeros]);
}

在生产代码中，我会创建一个辅助方法来反转字符串。

在NSString中删除尾随或前导字符

3 个答案: