Question

我想检查最近2天的备份文件，并比较当前文件（今天）的大小大于另一个文件（昨天）的大小，不同的备份文件名。

我有10台使用MySQL的服务器，每天都从数据库创建每台服务器的备份，然后发送到备份服务器。

由于某些服务器上有多个数据库，因此服务器在同一天创建不同的备份文件，然后发送到备份服务器。一世需要确定是否今天进行了备份以及文件（今天）是否为比旧文件大（昨天）？

55M 25. Mai 03:45 /home/backup/server1/SW1_20190525.sql.gz

48M 25. Mai 03:45 /home/backup/server1/SW2_20190525.sql.gz

39M 25. Mai 03:45 /home/backup/server1/SW3_20190525.sql.gz

35M 25. Mai 03:45 /home/backup/server1/SW4_20190525.sql.gz

42M 25. Mai 03:45 /home/backup/server1/SW5_20190525.sql.gz

57M 26. Mai 03:45 /home/backup/server1/SW1_20190526.sql.gz

51M 26. Mai 03:45 /home/backup/server1/SW2_20190526.sql.gz

20K 26. Mai 03:45 /home/backup/server1/SW3_20190526.sql.gz

45M 26. Mai 03:45 /home/backup/server1/SW4_20190526.sql.gz

48M 26. Mai 03:45 /home/backup/server1/SW5_20190526.sql.gz

我希望检查：

SW1_20190526.sql.gz> SW1_20190525.sql.gz

SW2_20190526.sql.gz> SW2_20190525.sql.gz

SW3_20190526.sql.gz> SW3_20190525.sql.gz

SW4_20190526.sql.gz> SW4_20190525.sql.gz

SW5_20190526.sql.gz> SW5_20190525.sql.gz

确定时说0，确定

何时不说1错误（例如=> SW3_20190526.sql.gz）

Answer 1

只需将〜/ testing / *更改为path / to / your / files * 记得把*

注意：此版本的脚本假定您在所需的日期运行它。例如（日期命令和最新文件条目匹配）

# Set directory to loop through
dir=~/testing/*

# Todays date used to grep for today's file
today=$(date +%Y%m%d)

# Yesterday's date used to grep for yesterday's file
yesterday=$(date -d "yesterday" '+%Y%m%d')

# Start point to loop through each SW? files
num=1

# Start looping through directory
for file in $dir; do

    # Get today's file by date
    file_td=$(ls SW[$num]* | grep $today)

    # Get yesterday's file by date 
    file_yd=$(ls SW[$num]* | grep $yesterday)

    # Get size of today's file
    size_td=$(stat -c %s $file_td)

    # Get size of yesterday's file
    size_yd=$(stat -c %s $file_yd)

    # Test if file size is equal or greater than
    if (($size_td >= $size_yd)); then
        echo "$file_td is equal or bigger that $file_yd"
    else
        echo "$file_yd is bigger $file_td"
    fi

    # Limit SW? to 5 so its doesn't attempt the script on none existing file
    # You change the 5 to how ever many SW's there are in the directory
    ((num++)) && [[ $num > 5 ]] && exit

done

注意：这个版本的脚本不做任何假设，只是获取目录接受的最后两个条目。这样可以模拟“今天”和“昨天”，而无需您在最后一天运行脚本。

# File Directory to loop through
dir=~/testing/*

# Starting point for file names SW1 - SW?
num=1

# Begin looping through directory
for file in $dir; do

    # Set the latest file in directory SW? to today
    file_td=$(ls SW[$num]* | sort -u -r | sed '1q;d')        

    # Set the second latest file in directory SW? to yesterday
    file_yd=$(ls SW[$num]* | sort -u -r | sed '2q;d')

    # Get size of today's file
    size_td=$(stat -c %s $file_td)

    # Get size of yesterday's file
    size_yd=$(stat -c %s $file_yd)

     # Test if file size is equal or greater than
    if (($size_td >= $size_yd)); then
        echo "$file_td is equal or bigger that $file_yd"
    else
        echo "$file_yd is bigger $file_td"
    fi

    # Limit SW? to 5 so its doesn't attempt the script on none existing files
    # You change the 5 to how ever many SW's there are in the directory 
    ((num++)) && [[ $num > 5 ]] && exit

done

如何按创建日期和文件大小比较不同的文件？

1 个答案: