awk：将日期字符串的格式设置为YYYYMMDD至YYYY-MM-DD

Question

I have a CSV file which I parse using awk because I don't need all columns. The problem I have is that one column is a date but in the format YYYYMMDD but I need it in YYYY-MM-DD and I don't know how to achieve that.

I already tried with split($27, a) but it doesn't split it - so a[0] returns the whole string.

Answer 1

You could use substr :

printf "%s-%s-%s", substr($27,0,4), substr($27,5,2), substr($27,7,2)

Assuming that the 27 ^th field was 20140318 , this would produce 2014-03-18 .

Answer 2

Use your awk output as input to date -d , eg

$ date -d 20140918 +'%Y-%m-%d'
2014-09-18