我正在Linux上工作,我正在运行以下程序:
$ ls -t postgresq*.log | head -n1 | xargs grep "ALTER USER"
< pg_user 2021-07-15 05:03:41.609 EDT > LOG: statement: ALTER USER username WITH ENCRYPTED PASSWORD 'JUly@#12' valid until '2021-07-20';
但是,我只想在下面的字符串中进行grep:
ALTER USER username WITH
我的文件包含以下几行
File.txt
Unix is good
Linux and unix is different?
Linux is also good, then what about unix?
这里我要输出
(1st line blank)
Linix and
Linux is also good, then what about
在这里vi命令或任何其他命令都会给出这个输出?搜索特定的单词,如果这是对的话,然后删除那个词,然后删除该行中的所有单词。
对于grep特定的关键字和它的内容,我有一个问题。这是示例文件,实际文件比这个文件大。
示例
user@linux:~$ cat url.txt
abcrandomtextdef another random text blablabla
another random iwantthis text abcrandomtextdef url=https://www.google.com ghirandomtextjkl
ghirandomtextjkl another random text yadayada
wxyz iwantthis abcdef url=yahoo.com yaday
这是我的样本文件
user@linux:~$ cat file.txt
Line 1
Line 2
Line 3
Line 4
Line 5
user@linux:~$
我可以用grep -A2 'e 2' file.txt打印第2-4行。
user@linux:~$ grep -A2 'e 2' file.txt
Line 2
Line 3
Line 4
user@linux:~$
我也可以用grep -n打印出行号。
user@linux:~$ grep -nA2 'e 2' file.txt
2:Line 2
3-Line 3
4
例如,如果我的Linux系统上的文件中有以下文本:
10-02-2020
given as file name) for lines containing a match to the given PATTERN. By default, grep prints the matching lines.
In addition, two variant programs egrep and fgrep are available. egrep is the same as grep -E. fgrep is the same as grep -F
16-02-2020
The top progr
如何使用linux命令将.txt文件中的分隔符从当前逗号(,)更改为分号(;)?
这是我的ME_1384_DataWarehouse_*.txt文件:
Data Warehouse,ME_1384,Budget for HW/SVC,13/05/2022,10,9999,13/05/2022,27,08,27,08
Data Warehouse,ME_1384,Budget for HW/SVC,09/05/2022,10,9999,09/05/2022,45,58,45,58
Data Warehouse,ME_1384,Budget for HW/SVC,25/05/2022,10,999