过滤 Linux 命令输出
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Filtering Linux command output
提问by Raajkumar
I need to get a row based on column value just like querying a database. I have a command output like this,
我需要根据列值获取一行，就像查询数据库一样。我有这样的命令输出，
Name ID Mem VCPUs State
Time(s)Domain-0 0 15485 16 r----- 1779042.1
prime95-01 512 1
-b---- 61.9
名称 ID Mem VCPUs State
Time(s)域-0 0 15485 16 r----- 1779042.1
prime95-01 512 1
-b---- 61.9
Here I need to list only those rows where state is "r". Something like this,
在这里，我只需要列出 state 为“r”的那些行。像这样的东西，
Domain-0 0 15485 16
r----- 1779042.1
域-0 0 15485 16
r----- 1779042.1
I have tried using "grep" and "awk" but still I am not able to succeed.
我曾尝试使用“grep”和“awk”，但仍然无法成功。
Please help me on this issue.
请帮助我解决这个问题。
Regards, Raaj
问候， 拉吉
回答by trojanfoe
try:
尝试：
awk ' ~ /^r.*/ { print }'
Like this:
像这样：
cat file | awk ' ~ /^r.*/ { print }'
回答by dtech
There is a variaty of tools available for filtering.
有多种工具可用于过滤。
If you only want lines with "r-----" grep is more than enough:
如果您只想要带有“r-----”的行，grep 就足够了：
command | grep "r-----"
Or
或者
cat filename | grep "r-----"
Quotes are optional but might prevent grep tripping over the -'s
引号是可选的，但可能会防止 grep 被 - 绊倒
回答by Patrick Echterbruch
grep solution:
grep解决方案：
command | grep -E "^([^ ]+ ){4}r"
What this does (-E switches on extended regexp):
这是做什么的（-E 打开扩展正则表达式）：
The first caret (^) matches the beginning of the line. [^ ] matches exactly one occurence of a non-space character, the following modifier (+) allows it to also match more occurences.
第一个插入符号 (^) 匹配行的开头。[^ ] 匹配非空格字符的一次出现，以下修饰符 (+) 允许它也匹配更多次出现。
Grouped together with the trailing space in ([^ ]+ ), it matches any sequence of non-space characters followed by a single space. The modifyer {4} requires this construct to be matched exactly four times.
与 ([^ ]+ ) 中的尾随空格组合在一起，它匹配后跟单个空格的任何非空格字符序列。修改器 {4} 要求此构造精确匹配四次。
The single "r" is then the literal character you are searching for.
单个“r”就是您要搜索的文字字符。
In plain words this could be written like "If the line starts <^> with four strings that are followed by a space <([^ ]+ ){4}> and the next character is , then the line matches."
简而言之，这可以写成“如果该行以 <^> 开头的四个字符串后跟一个空格 <([^ ]+ ){4}> 并且下一个字符是 ，则该行匹配。”
A very good introduction into regular expressions has been written by Jan Goyvaerts (http://www.regular-expressions.info/quickstart.html).
Jan Goyvaerts (http://www.regular-expressions.info/quickstart.html) 写了一篇非常好的正则表达式介绍。
回答by Johnsyweb
grepcan handle this for you:
grep可以为您处理：
yourcommand | grep -- 'r-----'
It's often useful to save the (full) output to a file to analyse later. For this I use tee.
将（完整）输出保存到文件以供以后分析通常很有用。为此，我使用tee.
yourcommand | tee somefile | grep 'r-----'
If you want to find the line containing "-b----" a little later on without re-running yourcommand, you can just use:
如果您想稍后在不重新运行的情况下找到包含“-b----”的行yourcommand，您可以使用：
grep -- '-b----' somefile
No need for cathere!
cat这里不需要！
I recommend putting --after your call to grepsince your patterns contain minus-signs and if the minus-sign is at the beginning of the pattern, this would look like an option argument to greprather than a part of the pattern.
我建议--在您调用 to 之后放置，grep因为您的模式包含减号，并且如果减号位于模式的开头，则这看起来像是 to 的选项参数grep而不是模式的一部分。
回答by Anoop Kumar
Filtering by awk cmd in linux:-
在 linux 中通过 awk cmd 过滤：-
Firstly find the column for this cmd and store file2 :-
首先找到这个 cmd 的列并存储 file2 :-
awk '/Domain-0 0 15485 /' file1 >file2
awk '/Domain-0 0 15485 /' 文件 1 > 文件 2
Output:-
输出：-
Domain-0 0 15485 16 r----- 1779042.1
域-0 0 15485 16 r----- 1779042.1
after that awk cmd in file2:-
在文件 2 中的 awk cmd 之后：-
awk '{print $1,$2,$3,$4,"\n",$5,$6}' file2
awk '{打印 $1,$2,$3,$4,"\n",$5,$6}' file2
Final Output:-
最终输出：-
Domain-0 0 15485 16
域-0 0 15485 16
r----- 1779042.1
r----- 1779042.1
