Question

我正在尝试使用电子邮件清理NSString以删除域名。

所以，让我们说

“testEmail@gmail.com”

需要

“testemail子@ ****。com”

OR

testEmail @ **** （我还需要一个方法来替换“@”后的所有字符

但对于任何域名变体。

如果有人有简单的解决方案，我会非常感激。

谢谢！

Answer 1

这绝不是防弹的，但它比以前的答案更有力。

- (NSString *)sanitizedStringForEmailAddress:(NSString *)address elideEntireDomain:(BOOL)elideEntireDomain
{
    NSRange asperandRange = [address rangeOfString:@"@" options:NSBackwardsSearch];
    NSRange domainRange = [address rangeOfString:@"." options:NSBackwardsSearch];

    if (asperandRange.location != NSNotFound &&
        domainRange.location != NSNotFound &&
        asperandRange.location < domainRange.location)
    {
        NSRange elisionRange = NSMakeRange(asperandRange.location + 1, domainRange.location - asperandRange.location - 1);
        if (elideEntireDomain)
        {
            elisionRange.length = address.length - asperandRange.location - 1;
        }

        NSString *elisionString = @"****";

        return [address stringByReplacingCharactersInRange:elisionRange withString:elisionString];
    }

    return nil;
}

请记住，对于＆＃34; local＆＃34;没有任何限制。电子邮件地址的一部分，包含＆＃34; @＆＃34;只要根据RFC 2822引用本地部分。

Answer 2

这应该可以做到，这将成为一个很好的类别方法。

- (NSString *)stringByReplacingEmailDomainWithAsterisks:(NSString *)emailAddress
{
    NSCharacterSet *charSet = [NSCharacterSet characterSetWithCharactersInString:@"@."];
    NSArray *allComponents = [emailAddress componentsSeparatedByCharactersInSet:charSet];

    NSMutableString *newString = [NSMutableString new];
    [newString appendFormat:@"%@@", [allComponents firstObject]];

    NSRange range = NSMakeRange(0, 1);
    for (__unused int i = range.location; range.location < [allComponents[1] length]; range.location++) {
        [newString appendString:@"*"];
    }
    [newString appendFormat:@".%@", allComponents[2]];

    return [NSString stringWithFormat:@"%@", newString];
}

示例：

NSString *someEmail = @"This_isAtest@wipeOut.com";

NSLog(@"%@", [self stringByReplacingEmailDomainWithAsterisks:someEmail]);

返回：

This_isAtest@*******.com

编辑：你需要添加一些错误检查以确保你有一个好的字符串/电子邮件等。

Answer 3

string email = "testEmail@gmail.com";
int atIndex = email.IndexOf('@');
string scrubbed = email.substring(0, atIndex + 1) + "****.com"

将所有角色抓到@，然后追加＆＃34; ****。com＆＃34;到最后。

Answer 4

用4个星号替换域：

NSString *email = @"testEmail@gmail.com";

NSRange at = [email rangeOfString:@"@"];
NSRange com = [email rangeOfString:@".com"];
email = [email stringByReplacingCharactersInRange:NSMakeRange(at.location + at.length, com.location - at.location - at.length) withString:@"****"];

并替换@之后的所有内容：

NSString *email = @"testEmail@gmail.com";

NSRange at = [email rangeOfString:@"@"];
email = [email stringByReplacingCharactersInRange:NSMakeRange(at.location + at.length, email.length - at.location - at.length) withString:@"****"];

替换2个特定字符之间的字符串

4 个答案: