string bash字符串与多个正确值进行比较
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bash string compare to multiple correct values
提问by eagle00789
i have the following piece of bashscript:
我有以下一段 bashscript：
function get_cms {
echo "input cms name"
read cms
cms=${cms,,}
if [ "$cms" != "wordpress" && "$cms" != "meganto" && "$cms" != "typo3" ]; then
get_cms
fi
}
But no matter what i input (correct and incorrect values) it never calls the function again, because I only want to allow 1 of those 3 inputs.
I have tried it with || with [ var != value ] or [ var != value1 ] or [ var != value1 ]but nothing works.
Can someone point me in the right direction?
但是无论我输入什么（正确和不正确的值），它都不会再次调用该函数，因为我只想允许这 3 个输入中的 1 个。我用 || 试过了 与[ var != value ] or [ var != value1 ] or [ var != value1 ]但没有作品。有人可以指出我正确的方向吗？
采纳答案by devnull
Instead of saying:
而不是说：
if [ "$cms" != "wordpress" && "$cms" != "meganto" && "$cms" != "typo3" ]; then
say:
说：
if [[ "$cms" != "wordpress" && "$cms" != "meganto" && "$cms" != "typo3" ]]; then
You might also want to refer to Conditional Constructs.
您可能还想参考Conditional Constructs。
回答by Edgar Grill
If the main intent is to check whether the supplied value is not found in a list, maybe you can use the extended regular expression matching built in BASH via the "equal tilde" operator (see also this answer):
如果主要目的是检查是否在列表中找不到提供的值，也许您可以通过“等号”运算符使用 BASH 中内置的扩展正则表达式匹配（另请参阅此答案）：
if ! [[ "$cms" =~ ^(wordpress|meganto|typo3)$ ]]; then get_cms ; fi
Have a nice day
祝你今天过得愉快
回答by Alfe
Maybe you should better use a casefor such lists:
也许您应该更好地将 acase用于此类列表：
case "$cms" in
wordpress|meganto|typo3)
do_your_else_case
;;
*)
do_your_then_case
;;
esac
I think for long such lists this is better readable.
我认为长期这样的列表这是更好的可读性。
If you still prefer the ifyou can do it with single brackets in two ways:
如果你仍然喜欢if你可以通过两种方式用单括号做到这一点：
if [ "$cms" != wordpress -a "$cms" != meganto -a "$cms" != typo3 ]; then
or
或者
if [ "$cms" != wordpress ] && [ "$cms" != meganto ] && [ "$cms" != typo3 ]; then
回答by Renich
Here's my solution
这是我的解决方案
if [[ "${cms}" != +(wordpress|magento|typo3) ]]; then
