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我得到一个 NSString 包含连接在一起的不同电子邮件，例如

def_ghi@hotmail.com_abc_1@me.com

每封电子邮件都由下划线分隔。问题是，如果我尝试使用下划线字符分隔字符串，它也会细分单个电子邮件地址，因为下划线字符也可以出现在单个电子邮件中。我试过的给了我这个结果

定义

ghi@hotmail.com

美国广播公司

1@me.com

这是我的代码

NSString *string = //The string I am receiving.
NSArray *chunks = [string componentsSeparatedByString: @"_"];

请帮我。

编辑： 我问了一位前辈，他告诉我应该先在字符串中搜索“@”字符。当我找到这个时，我会搜索一个“_”，如果存在则替换它。作为“@”之后的第一个下划线" 是分隔符。然后我应该从这个位置开始并再次重复上一步。我这样做直到字符串结束。请有人帮我解决这个问题。

4

4 回答 4

3

使用正则表达式的解决方案，

NSString *yourString = @"def_ghi@hotmail.com_abc_1@me.com";
NSError *error = NULL;
NSRegularExpression *regex = [NSRegularExpression
                              regularExpressionWithPattern:@"[A-Z0-9a-z._%+-]+@[A-Za-z0-9.-]+\\.[A-Za-z]{2,4}"
                              options:NSRegularExpressionCaseInsensitive
                              error:&error];
[regex enumerateMatchesInString:yourString options:0 range:NSMakeRange(0, [yourString length]) usingBlock:^(NSTextCheckingResult *match, NSMatchingFlags flags, BOOL *stop){

    // detect email addresses
    NSString *email = [yourString substringWithRange:match.range];

    //this part remove the '_' between email addresses
    if(match.range.location != 0){
        if([email characterAtIndex:0]=='_'){
            email = [email substringFromIndex:1];
        }
    }

    //print the email address
    NSLog(@"%@",email);

}];

编辑：如何收集它们，

像这样声明一个变量，

@property(nonatomic, strong) NSMutableArray *emailsArray;



 _emailsArray = [[NSMutableArray alloc] init];

NSString *yourString = @"def_ghi@hotmail.com_abc_1@me.com";
NSError *error = NULL;
NSRegularExpression *regex = [NSRegularExpression
                              regularExpressionWithPattern:@"[A-Z0-9a-z._%+-]+@[A-Za-z0-9.-]+\\.[A-Za-z]{2,4}"
                              options:NSRegularExpressionCaseInsensitive
                              error:&error];
[regex enumerateMatchesInString:yourString options:0 range:NSMakeRange(0, [yourString length]) usingBlock:^(NSTextCheckingResult *match, NSMatchingFlags flags, BOOL *stop){

    // detect email addresses
    NSString *email = [yourString substringWithRange:match.range];

    //this part remove the '_' between email addresses
    if(match.range.location != 0){
        if([email characterAtIndex:0]=='_'){
            email = [email substringFromIndex:1];
        }
    }

    //print the email address
    NSLog(@"%@",email);
    [self.emailsArray addObject:email];
}];


NSLog(@"%@",self.emailsArray);
于 2013-05-01T05:10:15.393 回答
1

这里有很多很好的答案，关于如何从你发现自己的那个有点凌乱的字符串中重建邮件地址的原始列表。

我会提出一个NSScanner基于解决方案，它似乎很适合：

NSString *messyString = @"def_ghi@hotmail.com_abc_1@me.com";

NSScanner *mailScanner = [NSScanner scannerWithString:messyString];

NSMutableArray *mailAddresses = [NSMutableArray array];

while (YES) {

    NSString *recipientName;
    NSString *serverName;
    BOOL found = [mailScanner scanUpToString:@"@" intoString:&recipientName];
    found |= [mailScanner scanUpToString:@"_" intoString:&serverName];
    if ( !found ) break;

    [mailAddresses addObject:[recipientName stringByAppendingString:serverName]];

    // Consume the delimiting underscore
    found = [mailScanner scanString:@"_" intoString:nil];
    if ( !found ) break;
}
于 2013-05-01T05:31:05.777 回答
0

Given your requirements then something like this should work for you:

NSString *string = @"def_ghi@hotmail.com_abc_1@me.com";
NSMutableArray *addresses = [NSMutableArray array];
NSUInteger currentIndex = 0; // start from beginning
// Stop when we are past the end of the string
while (currentIndex < string.length) {
    // Find the next @ symbol
    NSRange atRange = [string rangeOfString:@"@" options:0 range:NSMakeRange(currentIndex, string.length - currentIndex)];
    if (atRange.location != NSNotFound) {
        // We found another @, not look for the first underscore after the @
        NSRange underRange = [string rangeOfString:@"_" options:0 range:NSMakeRange(atRange.location, string.length - atRange.location)];
        if (underRange.location != NSNotFound) {
            // We found an underscore after the @, extract the email address
            NSString *address = [string substringWithRange:NSMakeRange(currentIndex, underRange.location - currentIndex)];
            [addresses addObject:address];
            currentIndex = underRange.location + 1;
        } else {
            // No underscore so this must be the last address in the string
            NSString *address = [string substringFromIndex:currentIndex];
            [addresses addObject:address];
            currentIndex = string.length;
        }
    } else {
        // no more @ symbols
        currentIndex = string.length;
    }
}

NSLog(@"Addresses: %@", addresses);
于 2013-05-01T05:20:29.463 回答
0

尝试

NSString *string = @"def_ghi@hotmail.com_abc_1@me.com";
NSArray *stringComponents = [string componentsSeparatedByString:@"_"];

NSMutableString *mutableString = [NSMutableString string];
NSMutableArray *emailIDs = [NSMutableArray array];
for (NSString *component in stringComponents) {
    if (!mutableString) {
        mutableString = [NSMutableString string];
    }
    [mutableString appendFormat:@"_%@",component];

    if ([component rangeOfString:@"@"].location != NSNotFound) {
        [emailIDs addObject:[mutableString substringFromIndex:1]];
        mutableString = nil;
    }
}

NSLog(@"%@",emailIDs);
于 2013-05-01T04:58:29.327 回答