zsh運行存儲在變量中的命令？

Question

在shell腳本中（在.zshrc中）我試圖執行一個命令，該命令作爲字符串存儲在另一個變量中。網絡上的各種消息稱這是可能的，但我沒有得到我期望的行爲。也許這是在命令開始時的~，或者可能是sudo的使用，我不確定。有任何想法嗎？由於

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 
} 

輸出：

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解釋的，可以使用shwordsplit外殼選項告訴zsh，你希望它拆分變量了以空格爲單位並將其視爲多個單詞。該頁面還討論了您可能需要考慮的替代方案。