Shell 脚本 bash：基于月份移动文件迭代

Question

我对 shell 脚本知之甚少，但不幸的是我必须编写一个。我想问一下bash脚本迭代移动文件，我需要移动按月排序的日志文件，这些文件将由cronjob执行。计划是将 mtime +30（1 个月前）文件移动到几个文件夹中，cronjob 将每天执行，例如：

前

/home/Work/LogFiles/20131200012.log
/home/Work/LogFiles/thisLogIsDifferent.log
/home/Work/LogFiles/20120322222.log 
/home/Work/LogFiles/20140100011.log
/home/Work/LogFiles/thisLogIsDifferent2.log

后

/home/Work/LogFiles/thisLogIsDifferent.log
/home/Work/LogFiles/thisLogIsDifferent2.log
/home/Work/LogFiles/2013/DEC/20131200012.log
/home/Work/LogFiles/2012/MAR/20120322222.log 
/home/Work/LogFiles/2014/JAN/20140100011.log

我对我必须使用的方法一无所知。所以这是我糟糕的 shell 脚本：

BASE_DIR=/home/Work/LogFiles
REPORT_DIR_YEAR=$BASE_DIR/`date +%Y`
REPORT_DIR=$REPORT_DIR_YEAR/`date +%b`

NOW=$(date +"%Y%m")

if ! [ -d $REPORT_DIR_YEAR ]; then
    mkdir $REPORT_DIR_YEAR

    if ! [ -d $REPORT_DIR ]; then
        mkdir $REPORT_DIR
    fi
fi

#THIS PART NEED TO BE RE-ARRANGED
#What I expect is not date=NOW; BUT SOME KIND LIKE date %m-1? but I still don't have any ideas about modify date function.

for file in find $BASE_DIR -maxdepth 1 -type f -mtime +30 -name '*$NOW*'
do

 month=$(ls -l $file | awk '{ print $6 }')
    case "$month" in
      "Jan") mv $file $REPORT_DIR_YEAR/$month/$file echo "$file moved to $REPORT_DIR/$file";;
      "Feb") mv $file $REPORT_DIR_YEAR/$month/$file echo "$file moved to $REPORT_DIR/$file";;
      "Mar") mv $file $REPORT_DIR_YEAR/$month/$file echo "$file moved to $REPORT_DIR/$file";;
      "Apr") mv $file $REPORT_DIR_YEAR/$month/$file echo "$file moved to $REPORT_DIR/$file";;
      "May") mv $file $REPORT_DIR_YEAR/$month/$file echo "$file moved to $REPORT_DIR/$file";;
      "Jun") mv $file $REPORT_DIR_YEAR/$month/$file echo "$file moved to $REPORT_DIR/$file";;
      "Jul") mv $file $REPORT_DIR_YEAR/$month/$file echo "$file moved to $REPORT_DIR/$file";;
      "Aug") mv $file $REPORT_DIR_YEAR/$month/$file echo "$file moved to $REPORT_DIR/$file";;
      "Sep") mv $file $REPORT_DIR_YEAR/$month/$file echo "$file moved to $REPORT_DIR/$file";;
      "Oct") mv $file $REPORT_DIR_YEAR/$month/$file echo "$file moved to $REPORT_DIR/$file";;
      "Nov") mv $file $REPORT_DIR_YEAR/$month/$file echo "$file moved to $REPORT_DIR/$file";;
      "Dec") mv $file $REPORT_DIR_YEAR/$month/$file echo "$file moved to $REPORT_DIR/$file";;
          *) echo " Do nothing " ;;
    esac

done

是的，这种情况$month不适用于之前的 for 循环$file。为什么？我不知道。我只是从各种来源、论坛、for 循环中的示例中复制，但它不起作用。

Answer 1

首先，解析 的输出绝不是一个好主意，ls因为它会导致各种问题。获取文件年龄的更好方法是stat. 例如：

$ ls -l 20120322222.log 
-rw-r--r-- 1 terdon terdon 0 Jan  1  2012 20120322222.log
$ stat -c %y 20120322222.log 
2012-01-01 00:00:00.000000000 +0100

所以，现在我们知道如何获取文件的年龄，问题是如何将其转换为三个字母的月份名称。最简单的是使用date：

 $ date -d "2012-01-01" +"%b"
Jan

结合这两个命令给出：

$ date -d "$(stat -c %y 20120322222.log)" +"%b"
Jan

因此，考虑到这一点，您可以将脚本编写为：

#!/usr/bin/env bash
BASE_DIR=/home/Work/LogFiles


## Find those files that are older than a month
find "$BASE_DIR" -maxdepth 1 -mtime +30 -type f -name "20*" | 
 while IFS= read -r file; do
    ## Get the file's modification year
    year="$(date -d "$(stat -c %y "$file")" +%Y)"
    ## Get the file's modification month
    month="$(date -d "$(stat -c %y "$file")" +%b)"

    ## Create the directories if they don't exist. The -p flag
    ## makes 'mkdir' create the parent directories as needed so
    ## you don't need to create $year explicitly.
    [[ ! -d "$BASE_DIR/$year/$month" ]] && mkdir -p "$BASE_DIR/$year/$month"; 

    ## Move the file
    mv "$file" "$BASE_DIR/$year/$month"
done

上面的脚本假设您想获得文件的真实修改日期，而不是解析名称。如果您想解析名称，请告诉我，我会相应地进行修改。