Question

shell删除保留最新10个文件

如下图:每个名称的zip包N个,需要删除zip包,每个(aaa,bbb,ccc,ddd)名称只保留最新的十个zip包

Answer 1

最新按照哪个时间来, 创建时间, 访问时间, 还是修改时间?  命令参考如下, 可以写成脚本定时调度.

1. 按照文件生成时间:

   stat --printf="%W,%n\n" aaa* | sort -r | tail +11 | awk -F"," '{print "rm "$2";"}' | sh

   stat --printf="%W,%n\n" bbb* | sort -r | tail +11 | awk -F"," '{print "rm "$2";"}' | sh

   stat --printf="%W,%n\n" ccc* | sort -r | tail +11 | awk -F"," '{print "rm "$2";"}' | sh

2. 按照文件修改时间:

   stat --printf="%Y,%n\n" aaa* | sort -r | tail +11 | awk -F"," '{print "rm "$2";"}' | sh

   stat --printf="%Y,%n\n" bbb* | sort -r | tail +11 | awk -F"," '{print "rm "$2";"}' | sh

   stat --printf="%Y,%n\n" ccc* | sort -r | tail +11 | awk -F"," '{print "rm "$2";"}' | sh

3. 按照文件访问时间:

   stat --printf="%X,%n\n" aaa* | sort -r | tail +11 | awk -F"," '{print "rm "$2";"}' | sh

   stat --printf="%X,%n\n" bbb* | sort -r | tail +11 | awk -F"," '{print "rm "$2";"}' | sh

   stat --printf="%X,%n\n" ccc* | sort -r | tail +11 | awk -F"," '{print "rm "$2";"}' | sh