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Editorial Team
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Asked: 2026-06-03T16:19:13+00:00

I’d like to put my current git branch into my multi-line ZSH prompt. However,

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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?

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1 Answer

  1. Editorial Team

    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 substitute 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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