zsh 运行存储在变量中的命令? [英] zsh run a command stored in a variable?

问题描述

在 shell 脚本(在 .zshrc 中)中，我试图执行一个命令，该命令以字符串形式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.也许是命令开头的 ~ ，或者是 sudo 的使用，我不确定.有任何想法吗?谢谢

In a shell script (in .zshrc) I am trying to execute a command that is stored as a string in another variable. Various sources on the web say this is possible, but I'm not getting the behavior i expect. Maybe it's the ~ at the beginning of the command, or maybe it's the use of sudo, I'm not sure. Any ideas? Thanks

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

推荐答案

使用eval:

eval ${install_cmd}

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