如何检查 homebrew-core 是否是 bash 脚本中的浅克隆？

Question

我如何在 bash 脚本中检查当前 homebrew-core 和 homebrew-cask 是否是浅层克隆，如果是则取消浅层克隆？如果他们已经不浅了，我也想打印一条消息。

运行 git -C "/usr/local/Homebrew/Library/Taps/homebrew/homebrew-cask" fetch --unshallow 在已经非浅层克隆上只给出错误消息而不是优雅处理。

我试过了

if [[ "git rev-parse --is-shallow-repository /usr/local/Homebrew/Library/Taps/homebrew/homebrew-core" == "false" ]] 
then
    echo "homebrew core is already non shallow. Moving to next step."
else
    git -C "/usr/local/Homebrew/Library/Taps/homebrew/homebrew-core" fetch --unshallow
fi

但是得到了

fatal: --unshallow on a complete repository does not make sense

Answer 1

你可以使用像

这样的东西

$ for d in homebrew-core homebrew-cask
do
    (cd /usr/local/Homebrew/Library/Taps/homebrew/$d && git rev-parse --is-shallow-repository)
done

如果 repo 很浅，它将打印 true 或 false。

Answer 2

这似乎有效

if [[ "git rev-parse --is-shallow-repository /usr/local/Homebrew/Library/Taps/homebrew/homebrew-core | grep -q 'false'" ]] 
then
    echo "homebrew core is already non shallow. Moving to next step."
else
    git -C "/usr/local/Homebrew/Library/Taps/homebrew/homebrew-core" fetch --unshallow
fi

如何检查 homebrew-core 是否是 bash 脚本中的浅克隆？

How can I check if homebrew-core is a shallow clone in a bash script?

git

bash

homebrew