How to pass a date range to awk as a variable

Here is my case. ``` bash ~]# TIME="2012-05-25 06:42:57" bash ~]# echo "2012-05-25 00:16:51,610" | awk -v var=$TIME '{if ($0 < var) print $0}' ``` Then, here is the error message ``` awk: 06:42:57 awk: ^ syntax error ``` I just want to pass a date range to my awk command. How to archive this? Please help. Thanks. Modify the case ``` START_TIME="2012-05-24 00:00:00" END_TIME="2012-05-24 01:00:00" echo "2012-05-24 00:10:10" | awk -v "START=$START_TIME" -v "END=$END_TIME" '{ if ( $0 > START && $0 < END) print $0 }' ``` It seems not working in IF conditions. ``` awk: { if ( $0 < START && $0 > END) print $0 } awk: ^ syntax error ``` After serval trying, seems found the solution with another approach. ``` echo "2012-05-24 00:10:10" | awk '{ if ( $0 > "'"$START_TIME"'" && $0 < "'"$END_TIME"'" ) print $0 }' ``` Not sure how to do it with awk variable "-v". Anyone have idears?