linux实现自动删除最旧的几个文件详解

#!/bin/bash
#不同的项目有不同的路径
array[0]='project1'
array[1]='project2'
array[2]='com/project3'
array[3]='com/phase/project4'
array[4]='project5'
array[5]='com/stor/sproject6'
#项目的主干目录是相同的
RELEASE="/opt/devapps/nexus/sonatype-work/nexus/storage/release/"

for path in ${array[@]};
do
 #拼接文件路径
 releasepath=${RELEASE}${path}
 cd $releasepath
 #判断是否存在该目录
 if [ $? -eq 0 ];
 then
  echo $releasepath
  echo "Contains file:"
  #输出所有的内容
  echo *
  num=`ls -l | grep '^d' | wc -l`;
  #判断文件夹的数量是否超过5个（我只想保留最新的5个文件夹）
  if [$num -gt 5 ];
  then
   #计算超过5个多少
   num=`expr $num - 5`
   clean=`ls -tr | head -$num | xargs`
   echo "will delete file:"
   echo ${clean}
   #-n1 每次处理1个文件
   ls -tr | head -$num | xargs -i -n1 rm -rf {}
  fi
 fi
done

#!/bin/bash
array[0]='project1'
array[1]='project2'
array[2]='com/project3'
array[3]='com/phase/project4'
array[4]='project5'
array[5]='com/stor/sproject6'
RELEASE="/opt/devapps/nexus/sonatype-work/nexus/storage/release/"

#清空备份文件
BACKUP="/tmp/storage/"
cd $BACKUP
if [ $? -eq 0 ];
then
 rm -rf *
fi

#清除超过5个文件之外最老的那些文件
for path in ${array[@]};
do
 releasepath=${RELEASE}${path}
 cd $releasepath
 if [ $? -eq 0 ];
 then
  echo $releasepath
  echo "Contains file:"
  echo *
  num=`ls -l | grep '^d' | wc -l`;
  if [$num -gt 5 ];
  then
   num=`expr $num - 5`
   clean=`ls -tr | head -$num | xargs`
   echo "will delete file:"
   echo ${clean}
   #把文件移动到备份文件夹更安全
   ls -tr | head -$num | xargs -i -n1 mv {} $BACKUP
  fi
 fi
done

crontab -e 

0 0 1 * * /opt/project/removecode.sh > /opt/project/remove.log 2>&1 &