在bash文件的每一行中添加一个数字
[英]Add a number to each line of a file in bash
问题描述
I have some files with some lines in Linux like: 
我在Linux中有一些带有某些行的文件，例如：
2013/08/16,name1,,5000,8761,09:00,09:30
2013/08/16,name1,,5000,9763,10:00,10:30
2013/08/16,name1,,5000,8866,11:00,11:30
2013/08/16,name1,,5000,5768,12:00,12:30
2013/08/16,name1,,5000,11764,13:00,13:30
2013/08/16,name2,,5000,2765,14:00,14:30
2013/08/16,name2,,5000,4765,15:00,15:30
2013/08/16,name2,,5000,6765,16:00,16:30
2013/08/16,name2,,5000,12765,17:00,17:30
2013/08/16,name2,,5000,25665,18:00,18:30
2013/08/16,name2,,5000,45765,09:00,10:30
2013/08/17,name1,,5000,33765,10:00,11:30
2013/08/17,name1,,5000,1765,11:00,12:30
2013/08/17,name1,,5000,34765,12:00,13:30
2013/08/17,name1,,5000,12765,13:00,14:30
2013/08/17,name2,,5000,1765,14:00,15:30
2013/08/17,name2,,5000,3765,15:00,16:30
2013/08/17,name2,,5000,7765,16:00,17:30
My column separator is " , " and in the third column (currently ,, ), I need the entry number within the same day. 我列分隔符为“ ”并且在第三列（目前, ,, ），我需要在同一天内的条目号。 For example, with date 2013/08/16 I have 11 lines and with date 2013/08/17 7 lines, so I need add the numbers for example: 例如，日期为2013/08/16的我有11行，日期为2013/08/17的我有7行，因此我需要添加数字，例如：
2013/08/16,name1,1,5000,8761,09:00,09:30
2013/08/16,name1,2,5000,9763,10:00,10:30
2013/08/16,name1,3,5000,8866,11:00,11:30
2013/08/16,name1,4,5000,5768,12:00,12:30
2013/08/16,name1,5,5000,11764,13:00,13:30
2013/08/16,name2,6,5000,2765,14:00,14:30
2013/08/16,name2,7,5000,4765,15:00,15:30
2013/08/16,name2,8,5000,6765,16:00,16:30
2013/08/16,name2,9,5000,12765,17:00,17:30
2013/08/16,name2,10,5000,25665,18:00,18:30
2013/08/16,name2,11,5000,45765,09:00,10:30
2013/08/17,name1,1,5000,33765,10:00,11:30
2013/08/17,name1,2,5000,1765,11:00,12:30
2013/08/17,name1,3,5000,34765,12:00,13:30
2013/08/17,name1,4,5000,12765,13:00,14:30
2013/08/17,name2,5,5000,1765,14:00,15:30
2013/08/17,name2,6,5000,3765,15:00,16:30
2013/08/17,name2,7,5000,7765,16:00,17:30
I need do it in bash. 我需要用bash来做。 How can I do it? 我该怎么做？
1 个解决方案
解决方案1
4 2013-08-29 17:36:50
This one's good too: 这也很好：
awk -F, 'sub(/,,/, ","++a[$1]",")1' file
Output: 输出：
2013/08/16,name1,1,5000,8761,09:00,09:30
2013/08/16,name1,2,5000,9763,10:00,10:30
2013/08/16,name1,3,5000,8866,11:00,11:30
2013/08/16,name1,4,5000,5768,12:00,12:30
2013/08/16,name1,5,5000,11764,13:00,13:30
2013/08/16,name2,6,5000,2765,14:00,14:30
2013/08/16,name2,7,5000,4765,15:00,15:30
2013/08/16,name2,8,5000,6765,16:00,16:30
2013/08/16,name2,9,5000,12765,17:00,17:30
2013/08/16,name2,10,5000,25665,18:00,18:30
2013/08/16,name2,11,5000,45765,09:00,10:30
2013/08/17,name1,1,5000,33765,10:00,11:30
2013/08/17,name1,2,5000,1765,11:00,12:30
2013/08/17,name1,3,5000,34765,12:00,13:30
2013/08/17,name1,4,5000,12765,13:00,14:30
2013/08/17,name2,5,5000,1765,14:00,15:30
2013/08/17,name2,6,5000,3765,15:00,16:30
2013/08/17,name2,7,5000,7765,16:00,17:30
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