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Shell 腳本 bash：根據月份移動檔案迭代

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Shell 腳本 bash：根據月份移動檔案迭代

我對shell腳本知之甚少，但不幸的是我必須寫一個。我想詢問有關 bash 腳本迭代移動文件的問題，我需要移動按月份排序的日誌文件，這些文件將由 cronjob 執行。計劃是將 mtime +30（1 個月前）檔案移至幾個資料夾中，並且 cronjob 將每天執行，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

後

/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

我對我必須使用的方法一無所知。這是我糟糕的 shell 腳本：

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，這個範例$month不適用於先前的 for 迴圈$file。為什麼？我不知道。我只是從各種來源、論壇、for 循環中複製範例，但它不起作用。

答案1

首先，它是解析輸出不是一個好主意ls因為它會導致各種各樣的問題。取得文件年齡的更好方法是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，現在我們知道如何獲取文件的年齡，問題是如何將其轉換為三個字母的月份名稱。最簡單的是使用date：

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

結合這兩個命令給出：

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

因此，考慮到這一點，您可以將腳本編寫為：

#!/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

上面的腳本假設您想要取得檔案的真實修改日期，而不是解析名稱。如果您想解析名稱，請告訴我，我將進行相應修改。

答案2

感謝 terdon：我設法獲取他/她的腳本並對其進行修改，以便它可以在 OS X 中運行。

#!/usr/bin/env bash
BASE_DIR=/Users/user/


## Find those files that are older than a month
find "$BASE_DIR" -maxdepth 1  -type f |
 while IFS= read -r file; do
    ## Get the file's modification month
    month="$(stat -f '%Sm' -t '%m' "$file")"
    ## Get the file's modification day
    day="$(stat -f '%Sm' -t '%d' "$file")"
    ## Get the file's modification hour
    hour="$(stat -f '%Sm' -t '%H' "$file")"


    ## 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/$month/$day/$hour" ]] && mkdir -p "$BASE_DIR/$month/$day/$hour";

    ## Move the file
    mv "$file" "$BASE_DIR/$month/$day/$hour"
done

答案3

這是我的解決方案：

#!/bin/bash

BASE_DIR="${1}"

if [ -z "${BASE_DIR}" ]; then
  BASE_DIR="/home/Work/LogFiles"
fi

if [ ! -d "${BASE_DIR}" ]; then
  echo "Error: '${BASE_DIR}' does not exists." >2
  exit 1
fi

declare -a MONTH_NAMES
MONTH_NAMES[1]='JAN'
MONTH_NAMES[2]='FEB'
MONTH_NAMES[3]='MAR'
MONTH_NAMES[4]='APR'
MONTH_NAMES[5]='MAY'
MONTH_NAMES[6]='JUN'
MONTH_NAMES[7]='JUL'
MONTH_NAMES[8]='AUG'
MONTH_NAMES[9]='SEP'
MONTH_NAMES[10]='OCT'
MONTH_NAMES[11]='NOV'
MONTH_NAMES[12]='DEC'


find "${BASE_DIR}" -maxdepth 1 -mtime +30 -type f -name '*.log' \
  | grep -e '/[0-9]*.log$' \
  | while read FILE; do

  FILENAME="$(basename "${FILE}")"

  FILE_YEAR="$(echo "${FILENAME}" | cut --bytes=1-4)"
  FILE_MONTH="$(echo "${FILENAME}" | cut --bytes=5-6)"

  FILE_MONTH_NAME="${MONTH_NAMES[${FILE_MONTH}]}"

  REPORT_DIR="${BASE_DIR}/${FILE_YEAR}/${FILE_MONTH_NAME}"
  test -d "${REPORT_DIR}" || mkdir -p "${REPORT_DIR}"

  mv "${FILE}" "${REPORT_DIR}"

  echo "'${FILENAME}' moved to '${REPORT_DIR}'"
done

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