How to match the particular line using regular expression?

I need to match the following line using regular expression.

Single lines contain the following things (example):

2010/11/29 09:开发者_如何学编程37:55 (2768)FMS:600 ERROR> Received SIGWARNING: nonstandard use of escape in a string literal
2010/11/29 09:37:55 (2768)FMS:600 ERROR> Received SIGNOTICE: exp: select * from follow_me_switch.call_from_entries where follow_me_group_id in ( select id from follow_me_switch.follow_me_groups where profile_id =105 and follow_me_switch.call_from_entries.call_from_number !~ \'^s*$\'

From those lines I need to match and remove the date and time stamps until the ERROR> How can we match this using efficient regular expression?

Thanks in advance.

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Ehm, how about:

^2010\/11\/29 09:39:57 \(2786\)Db_Wrapper\.pm:1404 ERROR\>

But somehow I think, you meant something different...

EDIT: This answer applied to an earlier question, that was later edited.

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In case you meant every line that has error in db_wrapper that would be something like

^\d{4}/\d{1,2}/\d{1,2} \d{2}:\d{2}:\d{2} \(\d+\)Db_Wrapper\.pm:\d+ ERROR>

or much easier, just

^.*Db_Wrapper\.pm:\d+ ERROR>

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Does this what you want?

^.*?(?=ERROR)

and replace with an empty string

See it here on Regexr

This will match everything till the first ERROR is ahead.

.*? is a lazy match that only will match as less as possible

(?=ERROR) is a positive look ahead that checks when the string ERROR is next

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One more regexp:

^\d{4}/\d{2}/\d{2}\s\d{2}:\d{2}:\d{2}\s\(\d+\)Db_Wrapper\.pm:\d+\sERROR>