How to convert one column to date format in awk script?

Question

``` cat test.txt | awk 'BEGIN { FS = "	" } ;{ print $44=strftime("%d %b %Y",$44) "	" $44 }' ``` I have received: ``` 31 12 1969 2012-09-27 ``` But I have expected: ``` 27 Sep 2012 2012-09-27 ``` Where is the error? Thanks.

Accepted Answer

if you really want to convert the format with awk build-in functions, this example may help you:

```
kent$  echo "2012-09-27" |awk '{t=$0;gsub("-"," ",t);print strftime("%d %b %Y", mktime(t" 00 00 00"))}' 
27 Sep 2012
```

read the description of following functions in man page:

```
 mktime(datespec)
 strftime([format [, timestamp[, utc-flag]]])
```

edit try to fix you old command, but no input file, cannot test it...

```
 awk 'BEGIN {FS=OFS="	"}{t=$44;gsub("-"," ",t); print strftime("%d %b %Y", mktime(t" 00 00 00")),$44}' test.txt
```

How to convert one column to date format in awk script?

Solution

Problem