NSString：寻找上标的 Unicode：1,2,3

发布于 2024-11-26 23:51:36 字数 156 浏览 1 评论 0原文

我正在寻找 Unicode 或某种方法或字节语言，可以为 1, 2, 3 提供上标。无论出于何种原因，Unicode 都有 0, 4, 5, 6, 7, 8, 9 的上标，但没有 1, 2 , 3.

你能在 NSString 中像 HTML 那样做上标吗？

原文

分享到QQ

分享到微博

如果你对这篇内容有疑问，欢迎到本站社区发帖提问参与讨论，获取更多帮助，或者扫码二维码加入 Web 技术交流群。

发布评论

需要登录才能够评论，你可以免费注册一个本站的账号。

苍风燃霜 2024-12-03 23:51:36

从字符调色板：

¹
SUPERSCRIPT ONE
Unicode: U+00B9, UTF-8: C2 B9

²
SUPERSCRIPT TWO
Unicode: U+00B2, UTF-8: C2 B2

³
SUPERSCRIPT THREE
Unicode: U+00B3, UTF-8: C2 B3

要将它们放入 NSStrings 中，您需要执行以下操作：

NSString *superscript1 = @"\u00B9";
NSString *superscript2 = @"\u00B2";
NSString *superscript3 = @"\u00B3";

From the character palette:

¹
SUPERSCRIPT ONE
Unicode: U+00B9, UTF-8: C2 B9

²
SUPERSCRIPT TWO
Unicode: U+00B2, UTF-8: C2 B2

³
SUPERSCRIPT THREE
Unicode: U+00B3, UTF-8: C2 B3

This, to make them in to NSStrings, you'd do:

NSString *superscript1 = @"\u00B9";
NSString *superscript2 = @"\u00B2";
NSString *superscript3 = @"\u00B3";

回复收藏 0 原文

合久必婚 2024-12-03 23:51:36

我知道这个问题已经得到解答，但如果有人想重用我的代码来从标准数字字符串转换为上标，就在这里。

-(NSString *)superScriptOf:(NSString *)inputNumber{

    NSString *outp=@"";
    for (int i =0; i<[inputNumber length]; i++) {
    unichar chara=[inputNumber characterAtIndex:i] ;
    switch (chara) {
        case '1':
            NSLog(@"1");
            outp=[outp stringByAppendingFormat:@"\u00B9"];
            break;
        case '2':
            NSLog(@"2");
            outp=[outp stringByAppendingFormat:@"\u00B2"];
            break;
        case '3':
            NSLog(@"3");
            outp=[outp stringByAppendingFormat:@"\u00B3"];
            break;
        case '4':
            NSLog(@"4");
            outp=[outp stringByAppendingFormat:@"\u2074"];
            break;
        case '5':
            NSLog(@"5");
                            outp=[outp stringByAppendingFormat:@"\u2075"];
            break;
        case '6':
            NSLog(@"6");
                            outp=[outp stringByAppendingFormat:@"\u2076"];
            break;
        case '7':
            NSLog(@"7");
            outp=[outp stringByAppendingFormat:@"\u2077"];
            break;
        case '8':
            NSLog(@"8");
            outp=[outp stringByAppendingFormat:@"\u2078"];
            break;
        case '9':
            NSLog(@"9");
            outp=[outp stringByAppendingFormat:@"\u2079"];
            break;
        case '0':
            NSLog(@"0");
            outp=[outp stringByAppendingFormat:@"\u2070"];
            break;
        default:
            break;
    }
}
return outp;   
}

给定一个数字输入字符串，它只返回等效的上标字符串。

编辑（感谢jrturton）：

-(NSString *)superScriptOf:(NSString *)inputNumber{

    NSString *outp=@"";
    unichar superScripts[] = {0x2070, 0x00B9, 0x00B2,0x00B3,0x2074,0x2075,0x2076,0x2077,0x2078,0x2079};

    for (int i =0; i<[inputNumber length]; i++) {

        NSInteger x =[[inputNumber substringWithRange:NSMakeRange(i, 1)]  integerValue];
        outp=[outp stringByAppendingFormat:@"%C", superScripts[x]];

    }

    return outp;   
}

I know this question has already been answered but if anyone wants to reuse my code for converting from a standard string of numbers to superscript, here it is.

-(NSString *)superScriptOf:(NSString *)inputNumber{

    NSString *outp=@"";
    for (int i =0; i<[inputNumber length]; i++) {
    unichar chara=[inputNumber characterAtIndex:i] ;
    switch (chara) {
        case '1':
            NSLog(@"1");
            outp=[outp stringByAppendingFormat:@"\u00B9"];
            break;
        case '2':
            NSLog(@"2");
            outp=[outp stringByAppendingFormat:@"\u00B2"];
            break;
        case '3':
            NSLog(@"3");
            outp=[outp stringByAppendingFormat:@"\u00B3"];
            break;
        case '4':
            NSLog(@"4");
            outp=[outp stringByAppendingFormat:@"\u2074"];
            break;
        case '5':
            NSLog(@"5");
                            outp=[outp stringByAppendingFormat:@"\u2075"];
            break;
        case '6':
            NSLog(@"6");
                            outp=[outp stringByAppendingFormat:@"\u2076"];
            break;
        case '7':
            NSLog(@"7");
            outp=[outp stringByAppendingFormat:@"\u2077"];
            break;
        case '8':
            NSLog(@"8");
            outp=[outp stringByAppendingFormat:@"\u2078"];
            break;
        case '9':
            NSLog(@"9");
            outp=[outp stringByAppendingFormat:@"\u2079"];
            break;
        case '0':
            NSLog(@"0");
            outp=[outp stringByAppendingFormat:@"\u2070"];
            break;
        default:
            break;
    }
}
return outp;   
}

Given an input string of numbers it just returns the equivalent superscript string.

Edit (thanks to jrturton):

-(NSString *)superScriptOf:(NSString *)inputNumber{

    NSString *outp=@"";
    unichar superScripts[] = {0x2070, 0x00B9, 0x00B2,0x00B3,0x2074,0x2075,0x2076,0x2077,0x2078,0x2079};

    for (int i =0; i<[inputNumber length]; i++) {

        NSInteger x =[[inputNumber substringWithRange:NSMakeRange(i, 1)]  integerValue];
        outp=[outp stringByAppendingFormat:@"%C", superScripts[x]];

    }

    return outp;   
}

回复收藏 0 原文