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This topic created in 2972 days ago, the information mentioned may be changed or developed.
sed 中有多行处理的命令：N、D、P
现有文件 test.txt ，内容如下：
line one
line two
line three
line four
分别对 test.txt 执行下面的命令：
[normal@centos7-server tmp]$ sed -n 'N; /two/P' test.txt
line one
[normal@centos7-server tmp]$ sed -n 'N; /one\nline/P' test.txt
line one
[normal@centos7-server tmp]$ sed 'N; /two/D' test.txt
line three
line four
[normal@centos7-server tmp]$ sed 'N; /one\nline/D' test.txt
line two
line three
line four
N 用于把下一行数据追加到 pattern space，如果没有下一行，则退出循环
P 打印 pattern sapce 中的第一行数据
D 删除 pattern space 中的第一行数据，如果 pattern space 中仍然有数据，继续从头执行（不跳到下一个循环）
上面的例子中两条 P 命令的输出结果一致，我的问题是，为什么两条 D 命令的结果不一致，请问造成这种差异的原因是什么？
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1
jasonyang9 Mar 21, 2018  1
你加个`=;`在表达式前面能看清楚点。
``` sed '=; N; /two/D' test.txt 1 2 3 line three line four ``` ``` sed '=; N; /one\nline/D' test.txt 1 2 line two line three 4 line four ``` |
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2
yyai3 Mar 21, 2018 1
我猜是因为 D 命令是从匹配的字符开始删除到换行符，所以 one 开始删除到\n
https://docstore.mik.ua/orelly/unix/sedawk/appa_03.htm D [ address1 [, address2 ]] D Delete first part (up to embedded newline) of multiline pattern space created by N command and resume editing with first command in script. If this command empties the pattern space, then a new line of input is read, as if the d command had been executed. |
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