Question

我有一个偶尔会有分割线的文件。

分裂由两个连续的具有字母字符的行组成。


5

00:00:00,000 --> 00:00:00,000

Alphabetic characters
Alphabetic characters

6

00:00:00,000 --> 00:00:00,000

Alphabetic characters

7

00:00:00,000 --> 00:00:00,000

Alphabetic characters
Alphabetic characters

8

00:00:00,000 --> 00:00:00,000

Alphabetic characters

.....

我想加入拆分线：


5

00:00:00,000 --> 00:00:00,000

Alphabetic characters Alphabetic characters

6

00:00:00,000 --> 00:00:00,000

Alphabetic characters

7

00:00:00,000 --> 00:00:00,000

Alphabetic characters Alphabetic characters

8

> 00:00:00,000 --> 00:00:00,000

Alphabetic characters Alphabetic characters
.....

使用sed。我不清楚如何加入前一行。有什么建议吗？

Answer 1

Food(String food) {
    // need `this` to refer to instance variable food since there's 
    // scope overlap.
    this.food = food;
}

背靠背匹配两条线，满足第一条线以字母字符或空格开头的条件，第二条线以相同的方式开始。加入他们的空间。

Answer 2

sed的另一种方法：

sed '/^[[:alpha:]]/{N;/\n[[:alpha:]]/s/\n/ /}' file

当找到以字母字符开头的行时，使用N命令将下一行添加到模式空间。然后在后跟带空格的字母字符时替换换行符。

Answer 3

sed用于单个行上的简单替换，即全部。对于其他任何你应该使用awk：

$ awk '/[[:alpha:]]/{ if (buf=="") {buf=$0; next} else {$0=buf OFS $0; buf=""} } 1' file

5

00:00:00,000 --> 00:00:00,000

Alphabetic characters Alphabetic characters

6

00:00:00,000 --> 00:00:00,000


7

00:00:00,000 --> 00:00:00,000

Alphabetic characters Alphabetic characters

8

00:00:00,000 --> 00:00:00,000

Alphabetic characters Alphabetic characters

.....

上述功能可以在所有兼容POSIX的UNIX系统上稳健，便携，高效地工作。

sed：根据第四个加入线

3 个答案: