"${x%% *}" 在 sh 中是什么意思？

Question

我刚刚在 makefile 中看到了“$${x%% *}”，这意味着 sh 中的“${x%% *}”。为什么这样写？

makefile 如何检测本地计算机中是否有命令可用？

determine_sum = \
        sum=; \
        for x in sha1sum sha1 shasum 'openssl dgst -sha1'; do \
          if type "$${x%% *}" >/dev/null 2>/dev/null; then sum=$$x; break; fi; \
        done; \
        if [ -z "$$sum" ]; then echo 1>&2 "Unable to find a SHA1 utility"; exit 2; fi

checksums.dat: FORCE
    $(determine_sum); \
    $$sum *.org

Answer 1

这是一个 POSIX shell 变量替换功能：

${var%Pattern} Remove from $var the shortest part of $Pattern that matches the back end of $var.
${var%%Pattern} Remove from $var the longest part of $Pattern that matches the back end of $var.

因此，如果 var="abc def ghi jkl"

echo "${var% *}" # will echo "abc def ghi"
echo "${var%% *}" # will echo "abc"