查找并将字符串移动到行尾

Question

我需要将一个特定的字符串移到行尾，但我完全不知道如何。这是一个例子：

:88:Some text/Some text/Some text/**TEXT TO MOVE**/Some textSome text/Some text/Some textSome text/

:29:Some textSome text/**TEXT TO MOVE**/Some textSome text/Some text/Some text/

...

每条线的长度和＆＃34;有些文字＆＃34;斜线之间的每一行都不同。 每行以＆＃34; /＆＃34;结尾。始终以＆＃34;：xx：＆＃34;开始其中x - 一些数字。

有没有人可以帮助我？

Answer 1

假设您**TEXT TO MOVE**的字段数与相同，您可以这样做：

$ echo -e ':88:Some text/Some text/Some text/**TEXT TO MOVE**/Some textSome text/Some text/Some textSome text/
:29:Some textSome text/**TEXT TO MOVE**/Some textSome text/Some text/Some text/' \
| awk -F'/' '
    {
      for (i = 1; i <= NF - 5; i++) { 
        printf("%s/", $i)
      }
      printf("%s/%s/%s/", $(NF - 3), $(NF - 2), $(NF - 1));
      print $(NF - 4) "/"
}'
:88:Some text/Some text/Some text/Some textSome text/Some text/Some textSome text/**TEXT TO MOVE**/
:29:Some textSome text/Some textSome text/Some text/Some text/**TEXT TO MOVE**/

Answer 2

无论在移动的字符串中包含哪些RE元字符，这都会有效，因为它适用于字符串，而不是RE：

$ cat file
Some text/Some text/**TEXT TO MOVE**/Some text/Some text/
Some text/**TEXT TO MOVE**/Some textSome text/

$ awk -v str='**TEXT TO MOVE**' 's=index($0,str){$0 = substr($0,1,s-1) substr($0,s+length(str)) str }1' file
Some text/Some text//Some text/Some text/**TEXT TO MOVE**
Some text//Some textSome text/**TEXT TO MOVE**