如何将8位字符转换为7位字符? (即Ü到U) [英] How do I translate 8bit characters into 7bit characters? (i.e. Ü to U)

问题描述

我正在寻找将更高位的ascii字符(例如Ü，扩展为ascii 154)转换为U(即ascii 85)的伪代码或示例代码.

I'm looking for pseudocode, or sample code, to convert higher bit ascii characters (like, Ü which is extended ascii 154) into U (which is ascii 85).

我最初的猜测是，由于只有大约25个ascii字符与7位ascii字符相似，因此必须使用转换数组.

My initial guess is that since there are only about 25 ascii characters that are similar to 7bit ascii characters, a translation array would have to be used.

让我知道您是否还能想到其他事情.

Let me know if you can think of anything else.

推荐答案

实际上是unexist提出的: "iconv"功能可以为您处理所有怪异的转换，几乎所有编程语言都提供该功能，并且具有一个特殊选项，可尝试近似转换目标集中缺少的字符.

Indeed as proposed by unexist : "iconv" function exists to handle all weird conversion for you, is available in almost all programming language and has a special option which tries to convert characters missing in the target set with approximations.

使用iconv可以简单地将输入的UTF-8字符串转换为7位ASCII.

Use iconv to simply convert your input UTF-8 string to 7bit ASCII.

否则，您将总是遇到麻烦:使用不同代码页的8位输入使用不同的字符集(因此根本无法与转换表一起使用)，却忘记映射最后一个愚蠢的带重音符号的字符(您已映射所有重音/重音，但忘记绘制捷克卡通或北欧的°")，等等.

Otherwise, you'll always end hitting corner case : a 8bit input using a different codepage with a different set of characters (thus not working at all with your conversion table), forgot to map one last stupid accented caracter (you mapped all grave/acute accent, but forgot to map Czech caron or the nordic '°'), etc.

当然，如果您想将解决方案应用于特定的小问题(为音乐收藏创建文件系统友好的文件名)，则查找数组是必经之路(可以是上面每个代码编号的数组) 128映射了JeeBee提出的128以下的近似值，或vIceBerg提出的源/目标对，这取决于您选择的语言中已经提供了哪些替换功能)，因为它会很快被黑在一起并迅速检查丢失的元素.

Of course if you want to apply the solution to a small specific problem (making file-system friendly filenames for your music collection) the the look-up arrays are the way to go (either an array which for each code number above 128 maps an approximation under 128 as proposed by JeeBee, or the source/target pairs proposed by vIceBerg depending on which substitution functions are already available in your language of choice), because it's quickly hacked together and quickly check for missing elements.

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