0

i am trying to extract a substring from log message, usually a log entry like this:

time="2023-06-01T00:05:30Z" level=debug msg="a foo is printing something" at=foobar

But when the message is too long it will be truncated like this:

time="2023-06-01T00:05:30Z" level=debug msg="This is a pretty long message

To extract the msg field, i want to use a regex which starts matching from the first double quote, until it sees the enclose double quote, or reaches the end of the string.

I tried msg=\"(.*)(\"|$) but it doesn't work as expected. Thanks!

codewarrior
  • 723
  • 7
  • 22
  • How does whatever's logging these messages handle embedded quotes? – Corvus Jun 01 '23 at 00:49
  • 1
    That looks like XML. If it is, [use a parser](https://stackoverflow.com/a/1732454). – InSync Jun 01 '23 at 00:58
  • Assuming that you're using C#: [`msg=(?"?)(.+?)((?<-q>\1)|$)`](https://regex101.com/r/hckOyc/1) – InSync Jun 01 '23 at 01:24
  • 2
    Two main things, change `(.*)` to `(.*?)` to turn off greedy matching and use the s-flag (single-line) if your regex engine supports it so `.` will include embedded newline characters. Then you'll need to add something to handle whatever syntax for embedding a double-quote you're using. – Chris Maurer Jun 01 '23 at 05:02
  • If the regex is a hardcoded string, then you need to double escape! – Poul Bak Jun 05 '23 at 19:46
  • 1
    Just don’t use the dot: `msg="[^"]*"?` – Holger Jun 13 '23 at 14:55

2 Answers2

0

This would be a good use-case for look aheads, since that's how you're naturally defining when the string should end. Since you want either the end of the string ($) or a doublquote (") - but whichever comes first, you could get away with using something like:

msg="(.*?)(?="|$)

Where the capture group is the contents of the msg field. Chris Maurer's answer is very important -- the .*? vs .* makes the behavior lazy//stop as soon as possible. Otherwise, it would default to continuing to end of line. However, if there are many lines you are looking through, you would either need to make this a multi-line flagged regex (which is slow), or add in a check for end of line rather than end of string. That would look like this:

msg="(.*?)(?="|\n|$)

Checked in Regex101 and it looks to match both cases. Best of luck! To reiderate, though, Chris hit the nail on the head- you not only need to be careful with your greedy/lazy quantifer but also escaping that " out. Most langauages use \" to do that.

EDIT: @chepner brought up a good point. If there is a " in the msg field, it'll throw your regex way off. Instead, we can check for either end of line/string OR just the next tag! Here's what I came up with for that:

msg=\"(.*?)(?=\" [a-z]+=|\n|$)

Assuming all the tags are lowercase a-z's and followed by =" then this should work just fine, as seen here: enter image description here

Sand
  • 198
  • 11
  • 2
    A greedy operator alone won't help with a quoted string that contains an (escaped) quote, like `'... msg="I said \"Hello\""...'`. – chepner Jun 05 '23 at 19:30
  • @chepner Oooh! Very intersting point! In a scenario like this it would be best to avoid relying on " at all. If you knew what the next key was after msg, look for that instead, or end of string - whichever happens first. Good point!!! – Sand Jun 05 '23 at 19:36
0

If I understand the question correctly, if the line is too long it breaks and continues to the next line.

You can use the (?s) syntax to toggle-on, "single-line mode", which will cause the dot, ., character to additionally match new-line delimiters.

(?s)msg=\"(.+?)\"

For example, this will match the following.

time="2023-06-01T00:05:30Z" level=debug msg="a foo is printing something"

And

time="2023-06-01T00:05:30Z" level=debug msg="This is a pretty long message
abc
def
ghi"

Where, capture group 1, will return

a foo is printing something

And

This is a pretty long message
abc
def
ghi
Reilas
  • 3,297
  • 2
  • 4
  • 17