bash - 获取网页内容并进行比较

Question

有人可以帮助/纠正我编写一个 shell 脚本吗，这就是我想要做的

• 获取网页内容
• 比较内容
• 如果内容相同则返回0
• 如果不返回 2

这是我的代码

for cluster in $CLUSTERS
do
    for applis in $(eval echo \$${cluster}_APPLIS)
    do
       CONTENT=$(wget -q -O - "http://$server/$applis")
       if [ "$CONTENT" -eq 1 ]
       then
       exit_code=0
       else
       exit_code=2
       fi
     done
done

[[ -z "$error_server" ]] && error_server="aucune"

case $exit_code in
    "2")
            echo "CRITICAL - App Version Mismatch"
            exit 2
            ;;
    "1")
            echo "WARNING - instance(s) indisponible(s)"
            exit 1
            ;;
    "0")
            echo "OK - All apps have the save version"
            exit 0
            ;;
    *)
            echo "CRITICAL - there's something wrong with this script ..."
            exit 2
            ;;
esac

任何帮助或建议将不胜感激

问候工厂

Answer 1

您可以使用 diff 来比较文件的两个版本。第一个变量用于获取网页的第一个版本。我不确定 first 应该在第一个循环内部还是外部初始化。

first=1
for cluster in $CLUSTERS
do
     for applis in $(eval echo \$${cluster}_APPLIS)
     do
        if [ $first -eq 1 ];then
             first=0
             PREV_CONTENT=$(wget -q -O - "http://$server/$applis")
        else
             CONTENT=$(wget -q -O - "http://$server/$applis")
             diff $PREV_CONTENT $CONTENT
             PREV_CONTENT=$CONTENT
             result=$?
        fi
        if [ "$result" -eq 0 ]
        then
           exit_code=0
        else
           exit_code=2
       fi
    done

完毕