zsh运行存储在变量中的命令？

Question

在shell脚本(在.zshrc中)我试图执行一个命令,该命令作为字符串存储在另一个变量中.网络上的各种消息来源说这是可能的,但我没有得到我期望的行为.也许这是~在命令的开头,或者也许是使用sudo,我不确定.有任何想法吗？谢谢

function update_install()
{
    # builds up a command as a string...
    local install_cmd="$(make_install_command $@)"
    # At this point the command as a string looks like: "sudo ~some_server/bin/do_install arg1 arg2"
    print "----------------------------------------------------------------------------"
    print "Will update install"
    print "With command: ${install_cmd}"
    print "----------------------------------------------------------------------------"
    echo "trying backticks"
    `${install_cmd}`
    echo "Trying \$()"
    $(${install_cmd})
    echo "Trying \$="
    $=install_cmd
}

输出:

Will update install
With command: sudo ~some_server/bin/do_install arg1 arg2

trying backticks
update_install:9: no such file or directory: sudo ~some_server/bin/do_install arg1 arg2
Trying $()
update_install:11: no such file or directory: sudo ~some_server/bin/do_install arg1 arg2
Trying $=
sudo ~some_server/bin/do_install arg1 arg2: command not found

Answer 1

用途eval:

eval ${install_cmd}

Answer 2

正如上文3.1节"为什么$var这里var="foo bar"没有做什么,我期待？" 在Z-Shell FAQ中,您可以使用shwordsplitshell选项告诉zsh您希望它按空格分割变量并将它们视为多个单词.同一页面还讨论了您可能想要考虑的替代方案.