NSString格式问题

发布于 2024-12-05 13:55:21 字数 205 浏览 0 评论 0原文

我正在使用 Google Place API 并获得了成功的 JSON 响应。但是一个 NSString 是 L\U00c3\U00b6wenbr\U00c3\U00a4u Keller。我想将其转换为正确的 NSString，例如 Lowenbrau Keller。我怎样才能进行这种转换？

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抱猫软卧 2024-12-12 13:55:21

正确的格式是在 Coocoa 中使用小写的 u 来表示 unicode：

错误：

NSString *string1 = @"L\U00c3\U00b6wenbr\U00c3\U00a4u Keller";

正确：

NSString *string2 = @"L\u00c3\u00b6wenbr\u00c3\u00a4u Keller";

要使其正确打印，请将 \u00 替换为 \x

NSString *string3 = @"L\xc3\xb6wenbr\xc3\xa4u Keller";
NSLog(@"string3: '%@'", string4);

NSLog 输出： string3: 'Löwenbräu Keller'

The correct format is to use a lowercase u to denote unicode in Coocoa:

Wrong:

NSString *string1 = @"L\U00c3\U00b6wenbr\U00c3\U00a4u Keller";

Correct:

NSString *string2 = @"L\u00c3\u00b6wenbr\u00c3\u00a4u Keller";

To get it to print correctly replace \u00 with \x

NSString *string3 = @"L\xc3\xb6wenbr\xc3\xa4u Keller";
NSLog(@"string3: '%@'", string4);

NSLog output: string3: 'Löwenbräu Keller'

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爱要勇敢去追 2024-12-12 13:55:21

测试代码：100 % 有效

注意：

\U 和 \u 不是同一件事。 \U 转义符需要 8 个（十六进制）数字而不是 4 个。

NSString *inputString =@"L\u00c3\u00b6wenbr\u00c3\u00a4u Keller";



NSString *outputString=[[NSString stringWithFormat:@"%@",inputString] stringByFoldingWithOptions:NSDiacriticInsensitiveSearch locale:[NSLocale currentLocale]];

NSLog(@"outputString : %@ \n\n",outputString);

输出：

outputString : LA¶wenbrA¤u Keller

TESTED CODE:100 % WORKS

NOTE:

\U and \u are not the same thing. The \U escape expects 8 (hex) digits instead of 4.

NSString *inputString =@"L\u00c3\u00b6wenbr\u00c3\u00a4u Keller";



NSString *outputString=[[NSString stringWithFormat:@"%@",inputString] stringByFoldingWithOptions:NSDiacriticInsensitiveSearch locale:[NSLocale currentLocale]];

NSLog(@"outputString : %@ \n\n",outputString);

OUTPUT:

outputString : LA¶wenbrA¤u Keller

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自此以后，行同陌路 2024-12-12 13:55:21

您可以通过以下方式使用它。

NSString *localStr = @"L\U00c3\U00b6wenbr\U00c3\U00a4u 凯勒";

localStr = [localStr stringByReplacingOccurrencesOfString:@"'" withString:@"'"];
localStr = [localStr stringByReplacingOccurrencesOfString:@" " withString:@" "];
localStr = [localStr stringByReplacingOccurrencesOfString:@""" withString:@"'"];
localStr = [localStr stringByReplacingPercentEscapesUsingEncoding:NSUTF8StringEncoding];
localStr = [localStr stringByTrimmingCharactersInSet:[NSCharacterSet whitespaceCharacterSet]];

我希望它对您有所帮助，并且会显示准确的字符串。

干杯。

You can use it in following way.

NSString *localStr = @"L\U00c3\U00b6wenbr\U00c3\U00a4u Keller";

localStr = [localStr stringByReplacingOccurrencesOfString:@"'" withString:@"'"];
localStr = [localStr stringByReplacingOccurrencesOfString:@" " withString:@" "];
localStr = [localStr stringByReplacingOccurrencesOfString:@""" withString:@"'"];
localStr = [localStr stringByReplacingPercentEscapesUsingEncoding:NSUTF8StringEncoding];
localStr = [localStr stringByTrimmingCharactersInSet:[NSCharacterSet whitespaceCharacterSet]];

I hope it will be helpful to you and will display exact string.

Cheers.

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