计算 zsh 提示的用户可见字符串的长度 [英] Count length of user-visible string for zsh prompt

问题描述

我想把我当前的 git 分支放到我的多行 ZSH 提示符中.然而，这把两行搞乱了——我希望他们能很好地排队.

<预><代码>┌─(simon@charmander:s000)─[大师*]──────────────────────(~)──┐└─(127:15:44)── ──(Sat, May12)─┘

应该是:

<预><代码>┌─(simon@charmander:s000)─[大师*]─────────(~)─┐└─(127:15:44)── ──(Sat, May12)─┘

git 分支是从 oh-my-zsh 函数中获取的，git_prompt_info()，它给了我分支，脏状态，还有一堆提示转义，可以很好地为事物着色.

我如何计算将可见插入到 ZSH 提示中的字符 - 而不是提示转义序列?

解决方案

假设提示转义的字符串存储在变量 FOO 中，这将只计算用户可见的字符:

<预><代码>FOO=$(git_prompt_info)本地零='%([BSUbfksu]|([FK]|){*})'FOOLENGTH=${#${(S%%)FOO//$~零/}}

这来自这个.zshrc.

这是对其工作原理的粗略解释，自由地引用了 man zshexpn，PARAMETER EXPANSION 部分.我不是 100% 确定细节，因此，如果您使用它来开发自己的等价物，请阅读相关的 man zshall 部分.

从 FOOLENGTH=${#${(S%%)FOO//$~zero/}} 行开始，我们得到了许多位.由内而外:

1. $~zero:~ 确保 zero，我们将其定义为 '%([BSUbfksu]|([FB]|){*})'，被视为一个模式而不是一个普通的字符串.

2. ${(S%%)FOO//$~zero/}:这匹配${name//pattern/repl}:<块引用>

   用字符串repl替换参数名扩展中最长可能匹配的模式

   注意我们没有repl；我们将 pattern 的最长可能匹配替换为空，从而将其删除.
   (S%%)FOO 对 FOO 进行扩展，并设置了几个标志.我不太遵循它.

3. ${#${(S%%)FOO//$~zero/}}: ${#spec} 将替换长度替换结果的字符 spec，如果 spec 是替换.在我们的例子中，spec 是替换的结果 ${(S%%)FOO//$~zero/};所以这基本上返回了 FOO 上的正则表达式 s/zero// 结果中的字符长度，其中 zero 是上面的模式.

I'd like to put my current git branch into my multi-line ZSH prompt. However, this messes up the two lines - I'd like them to line up nicely.

┌─(simont@charmander:s000)─[master *]────────────────
───(~  )─┐  
└─(127:15:44)──                       ──(Sat,May12)─┘

should be:

┌─(simont@charmander:s000)─[master *]─────────(~  )─┐  
└─(127:15:44)──                       ──(Sat,May12)─┘

The git branch is grabbed from an oh-my-zsh function, git_prompt_info(), which gives me the branch, dirty status, and a bunch of prompt-escapes to color things nicely.

How do I count the characters that will be visibly inserted into the ZSH prompt - not the prompt escape sequences?

解决方案

Assuming that the prompt-escaped string is stored in a variable FOO, this will count only user-visible characters:

                                                                                                                                
FOO=$(git_prompt_info)                                                                                                                     
local zero='%([BSUbfksu]|([FK]|){*})'
FOOLENGTH=${#${(S%%)FOO//$~zero/}} 

This comes from this .zshrc.

This is a rough explanation of why it works, liberally quoting from man zshexpn, section PARAMETER EXPANSION. I'm not 100% sure of the details, so, if you're using this to develop your own equivalent, read the relevant man zshall sections.

Working from the line FOOLENGTH=${#${(S%%)FOO//$~zero/}}, we've got a number of bits. Going from the inside out:

1. $~zero: The ~ ensures that zero, which we've defined as '%([BSUbfksu]|([FB]|){*})', is treated as a pattern rather than as a plain string.

2. ${(S%%)FOO//$~zero/}: This matches ${name//pattern/repl}:

   Replace the longest possible match of pattern in the expansion of parameter name by string repl

   Note that we don't have a repl; we replace the longest possible match of pattern with nothing, thereby removing it.
   (S%%)FOO conducts an expansion on FOO with several flags set. I don't quite follow it.

3. ${#${(S%%)FOO//$~zero/}}: ${#spec} will substitue the length in characters of the result of the substitution spec, if spec is a substitution. In our case, spec is the result of the substitution ${(S%%)FOO//$~zero/}; so this basically returns the length of characters in the result of the regular expression s/zero// on FOO, where zero is the pattern above.

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