On 17-Feb-11 13:30:53, Brett Parker wrote:

On 17 Feb 12:51, Richard Parsons wrote:

...

Hi everyone

I have some data like this:

something:01/01/1980:something something-else:12/05/1993:foobar another-something:32/01/2012:barfoo etc

In other words, it's a colon delimited file in which the second field is a date in the form dd/mm/yyyy.

How do I sort the whole file in date order by the second field? I can see that 'sort -k' allows me to pick the positional part of the fields and -t allows me to pick the delimiter. But I can't work out how to combine them so that the whole file is sorted.

sed -e 's#:([0-9][0-9])/([0-9][0-9])/([0-9][0-9][0-9][0-9]):#:\3-\2-\1:#' file | sort -d ":" -k 2

Should work. Note that's entirely untested and may eat your cat though.

It will eat the cat. There needs to be a pipe into a further 'sed' to undo the reversal of dd/mm/yyyy to yyyy/mm/dd:

sed -e 's#:([0-9][0-9])/([0-9][0-9])/([0-9][0-9][0-9][0-9]):#:\3-\2-\1:#' file | sort -d ":" -k 2 | sed -e 's#:([0-9][0-9])/([0-9][0-9])/([0-9][0-9][0-9][0-9]):#:\3-\2-\1:#'

Ted. (PS: The spillover of "file" is due to my mailer line length)

-------------------------------------------------------------------- E-Mail: (Ted Harding) ted.harding@wlandres.net Fax-to-email: +44 (0)870 094 0861 Date: 17-Feb-11 Time: 14:40:39 ------------------------------ XFMail ------------------------------

On 17 Feb 14:40, Ted Harding wrote:

On 17-Feb-11 13:30:53, Brett Parker wrote:

...

On 17 Feb 12:51, Richard Parsons wrote:

...

Hi everyone

I have some data like this:

something:01/01/1980:something something-else:12/05/1993:foobar another-something:32/01/2012:barfoo etc

In other words, it's a colon delimited file in which the second field is a date in the form dd/mm/yyyy.

How do I sort the whole file in date order by the second field? I can see that 'sort -k' allows me to pick the positional part of the fields and -t allows me to pick the delimiter. But I can't work out how to combine them so that the whole file is sorted.

sed -e 's#:([0-9][0-9])/([0-9][0-9])/([0-9][0-9][0-9][0-9]):#:\3-\2-\1:#' file | sort -d ":" -k 2

...

Should work. Note that's entirely untested and may eat your cat though.

It will eat the cat. There needs to be a pipe into a further 'sed' to undo the reversal of dd/mm/yyyy to yyyy/mm/dd:

sed -e 's#:([0-9][0-9])/([0-9][0-9])/([0-9][0-9][0-9][0-9]):#:\3-\2-\1:#' file | sort -d ":" -k 2 | sed -e 's#:([0-9][0-9])/([0-9][0-9])/([0-9][0-9][0-9][0-9]):#:\3-\2-\1:#'

Dates should be in YYYY-MM-DD format anyways - it's better, I deliberately didn't convert it back, it's less ambiguous this way. I was improving the input, obviously!

Sheesh ;)

On 17 Feb 14:40, Ted Harding wrote:

On 17-Feb-11 13:30:53, Brett Parker wrote:

...

On 17 Feb 12:51, Richard Parsons wrote:

...

Hi everyone

I have some data like this:

something:01/01/1980:something something-else:12/05/1993:foobar another-something:32/01/2012:barfoo etc

In other words, it's a colon delimited file in which the second field is a date in the form dd/mm/yyyy.

How do I sort the whole file in date order by the second field? I can see that 'sort -k' allows me to pick the positional part of the fields and -t allows me to pick the delimiter. But I can't work out how to combine them so that the whole file is sorted.

sed -e 's#:([0-9][0-9])/([0-9][0-9])/([0-9][0-9][0-9][0-9]):#:\3-\2-\1:#' file | sort -d ":" -k 2

...

Should work. Note that's entirely untested and may eat your cat though.

It will eat the cat. There needs to be a pipe into a further 'sed' to undo the reversal of dd/mm/yyyy to yyyy/mm/dd:

sed -e 's#:([0-9][0-9])/([0-9][0-9])/([0-9][0-9][0-9][0-9]):#:\3-\2-\1:#' file | sort -d ":" -k 2 | sed -e 's#:([0-9][0-9])/([0-9][0-9])/([0-9][0-9][0-9][0-9]):#:\3-\2-\1:#'

EWRONG!

The last sed there should be...

sed -e 's#:([0-9][0-9][0-9][0-9])-([0-9][0-9])-([0-9][0-9]):#:\3/\2/\1:#'

So there, ner ner ner ner ner ner!