在while循環中不記住修改變量

Question

我試着問人們哪些接口用作iptables上的公共接口： while循環接縫不記住變量，我真的不知道爲什麼。

#!/bin/sh 
### Show all available interfaces and configure to use them in while-loop: 
# Here is the answer of the function: 
# "${PUB_IF}" != "eth0" -a "${PUB_IF}" != "eth1" -a "${PUB_IF}" != "vmbr0" 
function good_interfaces() { 
ip a | grep '^[0-9]:' | grep -v 'lo' | awk '{ print $2 }' | sed -e 's/^/"/' -e 's/:$/"/' -e '/$/ i\-a "${PUB_IF}" !=' | sed '$!N; s/\n/ /g' | tr '\n' ' ' | awk 'sub("^...", "")' 
} 

### Show available interfaces 
# Here is the answer of "echo ${IF_LIST}" 
# eth0? eth1? vmbr0? 
IF_LIST=$(ip a | grep '^[0-9]:' | grep -v 'lo' | awk '{ print $2 }' | sed 's/:/\?/g' | tr '\n' ' ') 

while [ $(good_interfaces) ]; do 
# When i replace $(good_interfaces) by its answer, the loop works: 
# while [ "${PUB_IF}" != "eth0" -a "${PUB_IF}" != "eth1" -a "${PUB_IF}" != "vmbr0" ] --> without the variable it works but isn't personalized for all machines 
    echo "${IF_LIST}" 
    read -p "Enter the interface to use: " PUB_IF 
done; 

這裏是ip a答案：在你的代碼

1: lo: <LOOPBACK,UP,LOWER_UP> mtu 65536 qdisc noqueue state UNKNOWN group default 
    link/loopback 00:00:00:00:00:00 brd 00:00:00:00:00:00 
    inet 127.0.0.1/8 scope host lo 
     valid_lft forever preferred_lft forever 
    inet6 ::1/128 scope host 
     valid_lft forever preferred_lft forever 
2: eth0: <BROADCAST,MULTICAST,UP,LOWER_UP> mtu 1500 qdisc mq master vmbr0 state UP group default qlen 1000 
    link/ether xx:xx:xx:xx:xx:xx brd ff:ff:ff:ff:ff:ff 
    inet x.x.x.x/x brd x.x.x.x scope global eth0 
     valid_lft forever preferred_lft forever 
3: eth1: <NO-CARRIER,BROADCAST,MULTICAST,UP> mtu 1500 qdisc mq state DOWN group default qlen 1000 
    link/ether xx:xx:xx:xx:xx:xx brd ff:ff:ff:ff:ff:ff 

Answer 1

首先，一些評論：

• #!/bin/sh
  拆卸空間上的這條線（和使用bash）。目前該行只評
• ip a | grep '^[0-9]:' | grep -v 'lo' | a lot more
  對不起，該命令是越來越複雜的。當你看不到輸入時，你認爲你可以在兩個月內自己理解嗎？
• IF_LIST=$(ip a | grep '^[0-9]:' | grep -v 'lo' | and more)
  一些代碼與功能共享good_interfaces
• while [ $(good_interfaces) ]; do
  while循環想要一個真/假的表情，你希望你的函數返回的字符串將被視爲一個表達。

那麼應該如何實現一個讀取PUB_IF的構造並在有效時打破？

把所有有效的接口放在一個帶清晰分隔符的字符串中（我將使用空格）。 我會在字符串前後添加一個空格，以便稍後進行匹配。
要求PUB_IF並檢查。

valid_interfaces=" eth0 eth1 vmbr0 " 
PUB_IF="" 
# Edit: changed "while [ true ]; do" into "while true; do" 
while true; do 
    read -p "Enter the interface to use: " PUB_IF 
    if [[ "${valid_interfaces}" = *;${PUB_IF};* ]]; then 
     echo "${PUB_IF} is valid" 
     break 
    else 
     echo "Nope, please enter something like${valid_interfaces}" 
    fi 
done 

的另一個問題是如何找到有效的接口。使用剪切：

valid_interfaces=" $(ip a | grep ": <" | grep -v "lo" | cut -d: -f2 | tr '\n' ' ')