1、将/etc/issue文件中的内容转换为大写后保存至/tmp/issue.out文件中
[[email protected] ~]# cat /etc/issue | tr ‘a-z‘ ‘A-Z‘ > /tmp/issue.out [[email protected] ~]# cat /tmp/issue.out \S KERNEL \R ON AN \M HTTP://WWW.MAGEDU.COM TTY IS \L HOSTNAME IS \N CURRENT TIME IS \T
2、将当前系统登录用户的信息转换为大写后保存至/tmp/who.out文件中
[[email protected] ~]# who | tr ‘a-z‘ ‘A-Z‘ > /tmp/who.out [[email protected] ~]# cat /tmp/who.out ROOT PTS/0 2016-07-29 09:28 (10.1.250.82) ROOT PTS/1 2016-07-29 09:48 (10.1.250.82) ROOT PTS/2 2016-07-29 10:52 (10.1.250.82)
3、一个linux用户给root发邮件，要求邮件标题为”help”，邮件正文如下：
Hello, I am 用户名,the system version is here,pleasehelp me to check it ,thanks!
操作系统版本信息
[[email protected] ~]$ mail -s help root < mail.txt [[email protected] ~]$ cat mail.txt Hello, I am xiaoshui,the system version is here,pleasehelp me to check it ,thanks! LSB Version: :core-4.1-amd64:core-4.1-noarch:cxx-4.1-amd64:cxx-4.1-noarch:desktop-4.1-amd64:desktop-4.1-noarch:languages-4.1-amd64:languages-4.1-noarch:printing-4.1-amd64:printing-4.1-noarch Distributor ID: CentOS Description: CentOS Linux release 7.2.1511 (Core) Release: 7.2.1511 Codename: Core
刚开始以为可以使用` `进行命令引用，然后发现使用后发现还是原来的字符
4、将/root/下文件列表，显示成一行，并文件名之间用空格隔开
[[email protected] /]# ls / | xargs bin boot dev etc f1 home lib lib64 media mnt opt proc root run sbin srv sys test testdir tmp usr var
5、file1文件的内容为：”1 2 3 4 5 6 7 8 9 10” 计算出所有数字的总和
[[email protected] ~]# cat file1 | tr ‘ ‘ ‘+‘ | bc 55 [[email protected] ~]# let s=0;for i in `cat file1` ;do s=$[$i+$s] ;done ; echo "sum=$s" sum=55
网上还看到一种是用echo 1 2 3 4 5 6 7 8 9 10| xargs -n1 | echo $[ $(tr ‘\n‘ ‘+‘) 0 ]的，结果是正确的，最后一点tr ‘\n‘ ‘+‘不明白什么意思
6、删除Windows文本文件中的‘^M‘字符
[[email protected] ~]# cat m.txt | tr -d ‘\r\n‘ > newm.txt
7、处理字符串“xt.,l 1 jr#!$mn2 c*/fe3 uz4”，只保留其中的数字和空格
[[email protected] ~]# tr -d "[[:punct:]]|[[:alpha:]]" < grep.txt 1 2 3 4 [[email protected] ~]# cat grep.txt xt.,l 1 jr#!$mn2 c*/fe3 uz4 [[email protected] ~]# grep -E -o "[[:space:]]|[[:digit:]]" grep.txt 1 2 3 4 [[email protected] ~]# grep -E -o "[[:space:]]|[[:digit:]]" grep.txt | cat -A $ 1$ $ 2$ $ 3$ $ 4$
使用grep但是会出现换行的情况，但是确实也按要求提取到了
8、将PATH变量每个目录显示在独立的一行
[[email protected] ~]# echo $PATH | tr ‘:‘ ‘\n‘ /usr/lib64/qt-3.3/bin /usr/local/sbin /usr/local/bin /usr/sbin /usr/bin /root/bin [[email protected] ~]# echo $PATH | xargs -d: -n1 /usr/lib64/qt-3.3/bin /usr/local/sbin /usr/local/bin /usr/sbin /usr/bin /root/bin [[email protected] ~]# echo $PATH | grep -E -o "/?[[:alnum:]]*/?[[:alnum:]]*/[[:alnum:]]+\-?[[:alnum:]].?[[:alnum:]]+/[[:alnum:]]+:?" | tr -d ‘:‘ /usr/lib64/qt-3.3/bin /usr/local/sbin /usr/local/bin /usr/sbin /usr/bin /root/bin
第三种同样使用正则表达式，最后去掉冒号还是用的tr命令，因为刚刚预习到正则，虽然麻烦，就当做练习吧
9、删除指定文件的空行
[[email protected] ~]# tr -s ‘\n‘ < emout.txt hello nihao hahah how are you how old are you [[email protected] ~]# cat emout.txt hello nihao hahah how are you how old are you [[email protected] ~]# sed -i ‘/^$/‘d emout.txt [[email protected] ~]# cat emout.txt hello nihao hahah how are you how old are you
tr -s 只是去掉空行显示出来，但是并没有真正的删除空行，使用sed却是可以
10、将文件中每个单词（字母）显示在独立的一行，并无空行
[[email protected] ~]# grep -E -o "\<[[:alpha:]]+\>" al.txt xiao shui nihao ni haha ni txt j moring father mother [[email protected] ~]# cat al.txt xiao shui nihao+++ni haha ni.txt j 3 2 num23mo moring father mother
刚开始也想用tr命令将空格转换为\n，但是数字也会出现，所以就没有使用
时间： 2024-10-15 20:16:08