zsh运行存储在变量中的命令？

Question

在shell脚本中（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。网络上的各种消息称这是可能的，但我没有得到我期望的行为。也许这是在命令开始时的~，或者可能是sudo的使用，我不确定。有任何想法吗？由于

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

使用eval：

eval ${install_cmd} 

Answer 2

如§3.1 "Why does $var where var="foo bar" not do what I expect?" of the Z-Shell FAQ解释的，可以使用shwordsplit外壳选项告诉zsh，你希望它拆分变量了以空格为单位并将其视为多个单词。该页面还讨论了您可能需要考虑的替代方案。