bash脚本中的嵌套大小写 [英] Nested case in bash script

问题描述

我已经在bash脚本中编写了以下函数，但是它不起作用. 我缺少明显的东西吗?

I have written the following function in a bash script but it is not working. Am I missing something obvious?

main_menu() {
dialog \
    --title "Sim Gateway Infomation Utility" \
    --menu "What do you want to do?" 12 60 5 \
    Summary "View overall summary" \
    Details "View details of a sim bank" \
    Modify "Modify used minutes of a sim" \
    Exit "Exit" \
    2>$tempfile

retval=$?
case retval in
    0)

        choice=`cat $tempfile`
        case $choice in
            Summary) summary;;
            Details) details;;
            Modify) modify;;
            Exit) clean_up;;
        esac
        ;;
    1)
        confirm_exit;;
    255)
        confirm_exit;;
esac

}

推荐答案

此文章讨论了dialog;我没有经验.

This article discusses dialog; I'm not experienced with it.

您的"case retval in"必须为"case $retval in"(或"case "$retval" in").

Your 'case retval in' needs to be 'case $retval in (or 'case "$retval" in).

[ @Idelic指出，我的原始答案比必要的更为保守.]

字符串"retval"与您在大写语句中列出的所有选项都不匹配(使用"*"选项可检测到意外内容).如果$retval包含空格，则双引号可防止发生意外.通常，在case "$var" in语句(以及大多数其他地方)中的变量周围使用双引号是个好主意.在这种特殊情况下，这无关紧要；退出状态始终是数字.在'case "$choice" in'语句中，我对变量的引号会比较满意-但您可能仍然很安全(我需要阅读有关dialog的更多信息，以确保其作用以及是否对其起作用.会产生空间-甚至不​​会产生任何空间.

The string 'retval' matches none of the options you list in your outer case statement (use a '*' option to detect the unexpected). The double quotes prevent accidents if $retval ever contained spaces; in general, it is a good idea to use double quotes around the variable in case "$var" in statements (and most other places too). In this particular case, it would not matter; the exit status is always a number. In the 'case "$choice" in' statement, I'd be more comfortable with the quotes around the variable - but you may still be safe (I'd need to read more about dialog to be sure of what it does and whether it ever generates spaces - or nothing even).

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