linux - bash sort - how do I sort using timestamp

Question

I need to sort a file using shell sort in linux. The sort needs to be based on timestamp values contained within each of file's rows. The timestamps are of irregular format and don’t specify the leading zeros to months, days, etc, so the sorts I am performing are not correct (i.e. their format is “M/D/YYYY H:MI:S AM”; so so “10/12/2012 12:16:18 PM” comes before “7/24/2012 12:16:18 PM”, which comes before “7/24/2012 12:17:18 AM”).

Is it possible to sort based on timestamps?

I am using the following command to sort my file:

sort -t= -k3 file.txt -o file.txt.sorted

(use equal sign as a separator => -t=; use 3rd column as a sort column => -k3)

A sample file is as follows:

<r id="abcd" t="10/12/2012 12:16:17 AM"><d><nv n="name" v="868" /><nv n="name0" v="73" /><nv n="name1" v="13815004" /></d></r>
<r id="defg" t="7/24/2012 12:16:17 PM"><d><nv n="name" v="0" /><nv n="name0" v="0" /><nv n="name1" v="0" /></d></r>
<r id="abcd" t="7/24/2012 12:16:17 PM"><d><nv n="name" v="0" /><nv n="name0" v="0" /><nv n="name1" v="0" /></d></r>
<r id="zxy" t="7/24/2012 12:16:17 PM"><d><nv n="name" v="0" /><nv n="name0" v="0" /><nv n="name1" v="59542676" /></d></r>
<r id="fghj" t="7/24/2012 12:16:17 PM"><d><nv n="name" v="38" /><nv n="name0" v="0" /><nv n="name1" v="3004537" /></d></r>
<r id="defg" t="7/24/2012 12:16:18 AM"><d><nv n="name" v="177" /><nv n="name0" v="0" /><nv n="name1" v="5888870" /></d></r>

Answer 1

The linux date command does a fine job of parsing dates like this, and it can translate them into more sortable things, like simple unix-time integers.

Example:

cat file | while read line; do
    datestring=$(sed -e 's/^.* t="([^"]*)".*$/1/' <<<"$line")
    echo "$(date -d "$datestring" +%s) $line"
done | sort -n

then you could pass that through the appropriate cut invocation if you want that unix timestamp removed again.