Bash：从文件中提取日期

Question

I am new to bash and I have a file like this :

2012-11-22 11:36:55,909 1353551815756 1353551815909 0 true myapi 10 203051 203051:ShopDb:ShopDb
2012-11-22 11:37:00,292 1353551820146 1353551820292 0 true myapi 10 201897 201897:ShopDb:ShopDb
2012-11-22 11:38:01,824 1353551881672 1353551881824 0 true myapi 10 203051 203051:ShopDb:ShopDb

In a loop,line I want to extract date part (2012-11-22 11:36:55) convert to time-stamp and assign to a variable(or simply dis How to achieve this in bash ?

Answer 1

You can also use date command, as one of the way to do it.

while IFS=, read x y; do
    date --date "$x" +%s
done < file.txt

+%s converts it to the timestamp as you want.

Answer 2

$ while IFS=" ," read f1 f2 f3
> do
>  echo $f1 $f2
> done < file
2012-11-22 11:36:55
2012-11-22 11:37:00
2012-11-22 11:38:01

By setting the IFS to a space and comma makes it easy to extract the 1st 2 fields in the while loop.

Answer 3

You can use GNU awk to get the timestamps and capture them using a bash array:

array=($(awk -F, '{ print mktime(gensub(/[:-]/," ","g",$1))}' file.txt))

Then you can iterate through them:

for i in "${array[@]}"; do echo "$i"; done

Results:

1353548215
1353548220
1353548281

Or simply select one to print; for example, to echo the second element:

echo "${array[1]}"

Results:

1353548220

Answer 4

以下将为您提取日期：

awk 'BEGIN{FS="[ ,]"; OFS=" ";} {print $1, $2}' input_file

Answer 5

Use the cut command:

cut -c1-19 file

This will extract characters 1 through to 19 from each line.

To just retrieve the first line:

head -n 1 file | cut -c1-19

and to assign that to a bash variable:

myTime=$(head -n 1 file | cut -c1-19)