How to extract the end of an string after a certain character with regex/java?

Question

I to parse multiple lines of text that, for example, look like this:

{"Name":"pathology[876]", "cpu":"0.58","mem":"18.39", "vm":"1542.14"}
{"Name":"/opt/pathology/bin/pathology[876]", "cpu":"0.58","mem":"18.39", "vm":"1542.14"}
{"Name":"/usr/sbin/ofonod[760]", "cpu":"0.00","mem":"0.00", "vm":"0.00"}
{"Name":"/opt/networking/bin/network_manager[370]", "cpu":"0.20","mem":"53.43", "vm":"4225.69"}
{"Name":"/usr/bin/dmrouterd[913]", "cpu":"0.00","mem":"0.00", "vm":"0.00"}

I have to extract every process name, but some come alone and as well with their related path which I have to ignore, for example: pathology[876] is that same thing as /opt/pathology/bin/pathology[876]. I have to generalize this process to take the process name indifferently of the path. How could I take the desired string between the last / and the end of the string?

So far I have computed the following regex that treats paths like: /opt/<anything>/bin/<anything> extracting part after bin/, but there is a problem where the path is longer, for example /opt/<anything>/bin/pat/pathology[876] I get pat/pathology[876] while I would want only pathology[876].

"(Name)":("\/opt\/(.*?)\/bin\/(.*?)"|"(.*?)")

This looks like JSON. do you want to iterate over each JSON record? Or do you want to treat them as a single entry like a string? — aitchkhan
– aitchkhan, Commented Jun 15, 2020 at 14:23
I want to treat them like a string, the idea is that this JSON structure was embedded in a longer message, but I only selected it to be more concise with my question. — LiviuIosim
– LiviuIosim, Commented Jun 15, 2020 at 14:36
@LiviuIosim Anything wrong with the answer I gave 10 minutes ago? — MonkeyZeus
– MonkeyZeus, Commented Jun 15, 2020 at 14:42
Can we assume that every entry will look like your excerpt? Something like alphanumerical string followed by a square bracket enclosed process number? — PJProudhon
– PJProudhon, Commented Jun 15, 2020 at 14:49

Felix · Accepted Answer · 2020-06-15 14:59:14Z

2

my steps to create such regex are:

Thinking about which characters are (not) included in my target string? In this case all chars are allowed, but " and / are not allowed: ([^/\"]+)
What is written before my target string? In this case an optional string like /.../.../ which always starts and ends with /. To catch all ../../../ we can write ([^"\/]+\/)* and to catch the first / and make it optional we just extend it to (\/([^"\/]+\/)*)?
What is written after my target string? -> "

The final regex could be:

"Name":"(?:\/(?:[^"\/]+\/)*)?([^/\"]+)"

(Note the syntax (?:X) will group the expression X but will not be captured as a "result group")

I've tested and saved this regex here: https://regex101.com/r/WnSNNk/2

answered Jun 15, 2020 at 14:59

Felix

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2 Comments

MonkeyZeus Over a year ago

Albeit a little hard to decypher, this is much more performant than my answer so +1 :-)

Felix Over a year ago

@MonkeyZeus thanks for your +1. But I like your "simple" regex more. It is easier to understand. I just didn't saw it.

MonkeyZeus · Accepted Answer · 2020-06-15 14:26:40Z

2

This would do it for you:

[^\/"]+(?=", "cpu")

In English:

Per line, find everything that's not a forward slash nor double quote leading up to ", "cpu"

https://regex101.com/r/u3rhUf/1/

answered Jun 15, 2020 at 14:26

MonkeyZeus

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1 Comment

LiviuIosim Over a year ago

I am sorry for being inactive yesterday, this works really well, thank you for your input.

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