How can I use "sed" command to get the last n lines of a huge text file (e.g. A.txt) and copy them into a new text file(e.g. B.txt)? I do not want to remove that lines from A.txt.
如何使用“sed”命令获取大文本文件的最后n行(例如A.txt)并将它们复制到新的文本文件(例如B.txt)中?我不想从A.txt中删除那些行。
3 个解决方案
#1
25
You don't. You use tail -n NUMLINES for that.
你没有。你使用tail -n NUMLINES。
tail -n 100 A.txt > B.txt
#2
10
Here's how to use sed to print the last 10 lines of a file:
以下是使用sed打印文件的最后10行的方法:
sed -e :a -e '$q;N;11,$D;ba'
You should probably only use this if you're planning on executing more sed commands on these lines. Otherwise the tail command is designed for this job.
如果你计划在这些行上执行更多的sed命令,你应该只使用它。否则tail命令是为此作业设计的。
#3
2
Using GNU sed, here's how to get the last 10 lines:
使用GNU sed,这里是如何获得最后10行:
(For n lines, replace 11, with n+1)
(对于n行,替换11,用n + 1)
sed -ne':a;$p;N;11,$D;ba' A.txt > B.txt
sed -ne':a; $ p; N; 11,$ D; ba'At.txt> B.txt
Note: On my Mac, with MacPorts, GNU sed is invoked as gsed. To use Apple's sed you would have to separate the label: sed -ne':a' -e'$p;N;11,$D;ba'*
注意:在我的Mac上,使用MacPorts,GNU sed被调用为gsed。要使用Apple的sed,你必须将标签分开:sed -ne':a'-e'$ p; N; 11,$ D; ba'*
Explanation:
sed -ne' invoke sed without automatic printing pattern space
sed -ne'调用没有自动打印模式空间的sed
:a label for looping
:循环标签
$p on last line, print pattern space, then quit
$ p在最后一行,打印模式空间,然后退出
N slurp the next line
N啜饮下一行
11,$D on line 11 through last line, remove the first line from pattern space (i.e. [^\n]*\n)
11,第11行到最后一行的$ D,从模式空间中删除第一行(即[^ \ n] * \ n)
ba' loop to :a
ba'循环到:a
Further
Because of the loop, sed continuously appends the next line to pattern space. After line 11 and through the last line (11,$D) sed begins removing the first line from pattern space. At the last line the pattern space is printed (`$p'), which contains the last 10 most recently slurped lines (the last 10 lines of file A.txt).
由于循环,sed不断地将下一行附加到模式空间。在第11行和最后一行(11,$ D)之后,sed开始从模式空间中删除第一行。在最后一行打印模式空间(`$ p'),其中包含最近10个最近的slurped行(文件A.txt的最后10行)。
#1
25
You don't. You use tail -n NUMLINES for that.
你没有。你使用tail -n NUMLINES。
tail -n 100 A.txt > B.txt
#2
10
Here's how to use sed to print the last 10 lines of a file:
以下是使用sed打印文件的最后10行的方法:
sed -e :a -e '$q;N;11,$D;ba'
You should probably only use this if you're planning on executing more sed commands on these lines. Otherwise the tail command is designed for this job.
如果你计划在这些行上执行更多的sed命令,你应该只使用它。否则tail命令是为此作业设计的。
#3
2
Using GNU sed, here's how to get the last 10 lines:
使用GNU sed,这里是如何获得最后10行:
(For n lines, replace 11, with n+1)
(对于n行,替换11,用n + 1)
sed -ne':a;$p;N;11,$D;ba' A.txt > B.txt
sed -ne':a; $ p; N; 11,$ D; ba'At.txt> B.txt
Note: On my Mac, with MacPorts, GNU sed is invoked as gsed. To use Apple's sed you would have to separate the label: sed -ne':a' -e'$p;N;11,$D;ba'*
注意:在我的Mac上,使用MacPorts,GNU sed被调用为gsed。要使用Apple的sed,你必须将标签分开:sed -ne':a'-e'$ p; N; 11,$ D; ba'*
Explanation:
sed -ne' invoke sed without automatic printing pattern space
sed -ne'调用没有自动打印模式空间的sed
:a label for looping
:循环标签
$p on last line, print pattern space, then quit
$ p在最后一行,打印模式空间,然后退出
N slurp the next line
N啜饮下一行
11,$D on line 11 through last line, remove the first line from pattern space (i.e. [^\n]*\n)
11,第11行到最后一行的$ D,从模式空间中删除第一行(即[^ \ n] * \ n)
ba' loop to :a
ba'循环到:a
Further
Because of the loop, sed continuously appends the next line to pattern space. After line 11 and through the last line (11,$D) sed begins removing the first line from pattern space. At the last line the pattern space is printed (`$p'), which contains the last 10 most recently slurped lines (the last 10 lines of file A.txt).
由于循环,sed不断地将下一行附加到模式空间。在第11行和最后一行(11,$ D)之后,sed开始从模式空间中删除第一行。在最后一行打印模式空间(`$ p'),其中包含最近10个最近的slurped行(文件A.txt的最后10行)。