如何在文件中添加新列
我有这样的文件，其中包含 5 列。但我的文件中的某些行错误地有 4 列，例如在该文件的第 5 行和第 6 行中，您可以看到第二列丢失了。我想用空格替换丢失的第二列，而不用字段分隔符“，”干扰文件中的其他行。
11111,5323,6296,29-May-2010,1 22222,5323,6296,24-May-2010,1 33333,5323,6296,24-Jun-2010,1 44444,5323,6296,24-Jun-2010,1 55555,8061,15-Jul-2010,1 66666,6296,29-May-2010,1 77777,5323,6296,29-May-2010,1 88888,6296,29-May-2010,1 99999,6296,27-May-2010,1 10101,5323,6296,29-May-2010,1
输出我需要的。
11111,5323,6296,29-May-2010,1 22222,5323,6296,24-May-2010,1 33333,5323,6296,24-Jun-2010,1 44444,5323,6296,24-Jun-2010,1 55555,,8061,15-Jul-2010,1 66666,,6296,29-May-2010,1 77777,5323,6296,29-May-2010,1 88888,,6296,29-May-2010,1 99999,,6296,27-May-2010,1 10101,5323,6296,29-May-2010,1
I have file like this which contains 5 columns. But some rows in my file have 4 columns by mistake, for example in this file at the 5th and 6th rows, you can see that the second column is missing. I want to replace the missing 2nd column with a blank space without disturbing other rows in my file with field separator ",".
11111,5323,6296,29-May-2010,1 22222,5323,6296,24-May-2010,1 33333,5323,6296,24-Jun-2010,1 44444,5323,6296,24-Jun-2010,1 55555,8061,15-Jul-2010,1 66666,6296,29-May-2010,1 77777,5323,6296,29-May-2010,1 88888,6296,29-May-2010,1 99999,6296,27-May-2010,1 10101,5323,6296,29-May-2010,1
Output that I need.
11111,5323,6296,29-May-2010,1 22222,5323,6296,24-May-2010,1 33333,5323,6296,24-Jun-2010,1 44444,5323,6296,24-Jun-2010,1 55555,,8061,15-Jul-2010,1 66666,,6296,29-May-2010,1 77777,5323,6296,29-May-2010,1 88888,,6296,29-May-2010,1 99999,,6296,27-May-2010,1 10101,5323,6296,29-May-2010,1
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更短的路
shorter way
使用 Vim，我会在 Ex-Mode 中输入以下行：
:%s/^(\d+),(\d+,\d+-\w+-\d+,\d)$/\1, ,\2/g
Using Vim I'd enter the following line in Ex-Mode:
:%s/^(\d+),(\d+,\d+-\w+-\d+,\d)$/\1, ,\2/g