bash 脚本中的嵌套大小写
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Nested case in bash script
提问by Morchuboo
I have written the following function in a bash script but it is not working. Am I missing something obvious?
我在 bash 脚本中编写了以下函数，但它不起作用。我错过了一些明显的东西吗？
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
}
}
回答by Jonathan Leffler
This articlediscusses dialog; I'm not experienced with it.
此文章讨论dialog; 我没有这方面的经验。
Your 'case retval in' needs to be 'case $retval in(or 'case "$retval" in).
您的 ' case retval in' 必须是 ' case $retval in（或 ' case "$retval" in）。
[@Idelic notes that my original answer was more conservative than necesssary.]
[ @Idelic 指出，我最初的回答比必要的更保守。]
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 $retvalever contained spaces; in general, it is a good idea to use double quotes around the variable in case "$var" instatements (and most other places too). In this particular case, it would not matter; the exit status is always a number. In the 'case "$choice" in' statement, I'd be more comfortable with the quotes around the variable - but you may still be safe (I'd need to read more about dialogto be sure of what it does and whether it ever generates spaces - or nothing even).
字符串 'retval' 不匹配您在外部 case 语句中列出的任何选项（使用 '*' 选项来检测意外情况）。如果$retval包含空格，双引号可以防止发生意外；一般来说，最好在case "$var" in语句中的变量周围使用双引号（以及大多数其他地方）。在这种特殊情况下，这无关紧要；退出状态始终是一个数字。在 ' case "$choice" in' 语句中，我会更喜欢变量周围的引号 - 但你可能仍然是安全的（我需要阅读更多关于dialog它的功能以及它是否会产生空格 - 或者什么都不产生甚至）。
