bash - 比较两个变量并回显缺失值

Question

我有一个如下变量：

Variable1="PanicA0 PanicA1"
Variable2="PanicA0"

我必须比较 variable1 和 variable2 并且应该将输出回显为PanicA1ie "PanicA1" is not in Varaiable2。我如何使用 shell 脚本来实现这一点？

Answer 1

这是脚本

varone=`echo "PanicA0 PanicA1" | cut -d' ' -f1-`
vartwo=`echo "PanicA0" | cut -d' ' -f1-`

for i in $varone; do
    for j in $vartwo; do
        if [[ $i = $j ]]; then 
            echo "Matched: $i"
            break
        else
            echo "$i is not in vartwo"
        fi
    done
done

Answer 2

• 第 1 步：将第一个变量拆分为每个单词的数组
• 第二步：遍历数组
• 第 3 步：在双括号内使用模式匹配

.

Variable1="PanicA0 PanicA1"
variable2="PanicA0"

varArr=($Variable1)
for word in "${varArr[@]}"
do
   [[ $variable2 == *${word}* ]] || echo "$word is not in varaiable2"
done