Search results matching tag 'HTTPERR input format'

RE: Plugin log Parsing

LogParser User : Jeff — Wed, 31 Jan 2007 17:25:00 GMT

[QUOTE User="LogParser User : Jeff"]Quote: Is there way to parse this plugin log? Here is the sample line within the log. [17/Jan/2007:00:01:19 -0600] - audit "krsekime6 (win32:4170775546178949265)" 6527 Report queued: from 10.34.106.4...[/QUOTE]

Assuming krsekime6 is the machine name, then the parsing is:

select "extract_token(extract_token(text,1,'"'),0,' ') as computername from <filename>"

RE: Plugin log Parsing

LogParser User : DEinspanjer — Wed, 31 Jan 2007 16:30:00 GMT

Yes, the plugin uses standard C# regex syntax.

RE: Plugin log Parsing

LogParser User : Young — Wed, 31 Jan 2007 13:19:00 GMT

[QUOTE User="LogParser User : Young"]Quote: I don't believe that is a format that LP understands so you will have to define the format in some way. You could use lots of string parsing, or you might try the RegexInputFormat plugin that I wrote...[/QUOTE]

Ok, I will take a look at your regular express pulgin. Here is what I have far has regex expression for the first part. Would this express work with your plugin.

[(.*) .....\] \"....? (\S*)

RE: Plugin log Parsing

LogParser User : DEinspanjer — Wed, 31 Jan 2007 00:52:00 GMT

I don't believe that is a format that LP understands so you will have to define the format in some way. You could use lots of string parsing, or you might try the RegexInputFormat plugin that I wrote and put in the Repository (link at the bottom of this page)

Plugin log Parsing

LogParser User : Young — Tue, 30 Jan 2007 18:14:00 GMT

Is there way to parse this plugin log? Here is the sample line within the log.

[17/Jan/2007:00:01:19 -0600] - audit "krsekime6 (win32:4170775546178949265)" 6527 Report queued: from 10.34.106.44, 23667 bytes, 844 ms

I need the Machine name. I thought about do it this vbscript using regular expression but, I thought logparser might be easier. Here is the syntax I came up with:

logparser -:httperr <filename> "select s-computername from <filename>"

The syntax above doesn't work. Does anyone have ideas? I am new to logparser.

Selective logfile

LogParser User : Smithers — Fri, 15 Apr 2005 17:08:00 GMT

Trying to determine best way to select the httperr.log file in my sql query but having problems with my logic. In trying to select the most current logfiles (i.e. httperr50 -httperr59), in my FROM clause I tried to specify the parameter "httperr5?" and httperr5* but I always get include the httperr5.log file which I do not want. Point me in the right direction!

--Never mind...Answered my own question when I read the help file! Had to specfiy httperr*.log in from and use the minDateMod parameter.