grep question

newest_newbie has asked for the wisdom of the Perl Monks concerning the following question:

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Re: grep question by Roy Johnson (Monsignor) on May 26, 2005 at 21:53 UTC
I think your confusion is in what tool extracts from what. `grep` extracts elements from a list; regex capture extracts pieces from a string. You're using both in your example, when you only need to use a regex: `perl -e 'print "param=value" =~ /param=(.)/'` On the plus side, at least nobody had to edit your post to add code tags. Caution:* Contents may have been coded under pressure.	[reply] [d/l] [select]
Re^2: grep question by newest_newbie (Acolyte) on May 27, 2005 at 00:29 UTC
I just assumed that grep would behave the same way as split does. I was wrong. `perl -e'print split /param=(.*)/, "param=value"'` [download] returns 'value', right? Thanks	[reply] [d/l]
Re^3: grep question by Roy Johnson (Monsignor) on May 27, 2005 at 00:41 UTC
It does, but that's not the normal usage of split, either. You're saying that you've got a series of records separated by `param=`anything and that, in addition to returning the things that are separated by that, you also want to return the part in the parentheses. You can use the `=~` operator to apply a regex to a string (which was what I showed you in my example). That is the straightforward way to do what you want. If you have a list or array of strings to apply, you can use map or foreach to set $_ to each of them in turn, and then you don't need the `=~` operator, because a regex applies to $_ by default. Caution: Contents may have been coded under pressure.	[reply] [d/l] [select]
Re^4: grep question by newest_newbie (Acolyte) on May 27, 2005 at 00:59 UTC
Re: grep question by japhy (Canon) on May 26, 2005 at 23:25 UTC
Most people have been showing `map { /param=(.)/ ? $1 : () }`, but if you're using list context (like print() supplies), you can just write `map { /param=(.)/ }`. Jeff `japhy` Pinyan, P.L., P.M., P.O.D, X.S.: Perl, regex, and `perl` hacker How can we ever be the sold short or the cheated, we who for every service have long ago been overpaid? ~~ Meister Eckhart	[reply] [d/l] [select]
Re: grep question by tlm (Prior) on May 26, 2005 at 21:45 UTC
Why `grep`? The purpose of `grep` is to filter a list according to some criterion, which doesn't sound at all like what you want to do. Of course, if one is hellbent on using `grep` for this, Perl will oblige with WTDIs galore, but this hardly qualifies as an argument in favor of using `grep` here. the lowliest monk	[reply]
Re^2: grep question by fishbot_v2 (Chaplain) on May 26, 2005 at 22:33 UTC
Of course, if one is hellbent on using `grep` for this, Perl will oblige with WTDIs galore... One such solution: `perl -e 'print grep { s/param=// } @{[ "param=value" ]};'` [download] ...but this hardly qualifies as an argument in favor of using `grep` here Agreed. Other than as a proof-of-concept, don't do this. I'd personally use `map` for the OPs exact stated desire. I suspect, though, if the OP gave the larger context of the problem, better solutions might be apparent.	[reply] [d/l] [select]
Re^3: grep question by Tanktalus (Canon) on May 26, 2005 at 23:26 UTC
Personally, I'd use both map and grep for what the OP is doing: `return map { /memtype=(.*)/i; $1 } grep /memtype=/, $self -> collect_d +ata ('all');` [download] I realise that this probably doesn't speed things up, maybe even slows it down. But it does what it says much more obviously to me. I'm collecting all the data. Then I'm extracting just the ones I want (memtype), and then I'm grabbing the stuff on the right of the equals sign. Just seems more straightforward to me. As for brian d foy's concern about using $1 when you haven't checked if it matched, I hope this is ok - I agree with him, and I think this is ok. ;-)	[reply] [d/l]
Re: grep question by trammell (Priest) on May 26, 2005 at 21:26 UTC
One possible solution: `perl -le 'print map {/param=(.*)/;$1} "param=value";'` [download] I just hope you're not planning on using this to process CGI arguments...	[reply] [d/l]
Re^2: grep question by demerphq (Chancellor) on May 26, 2005 at 21:31 UTC
I think you meant: `perl -le 'print map { /param=(.*)/ ? $1 : () } "param=value";'` [download] $1 is only set if the match is succesful. (merlyn i beat you to it ;-) --- $world=~s/war/peace/g	[reply] [d/l]
Re^3: grep question by trammell (Priest) on May 26, 2005 at 22:06 UTC
No, the OP describes the input as a 'param=value' list. I stand by my post.	[reply]
Re^4: grep question by brian_d_foy (Abbot) on May 26, 2005 at 22:27 UTC
Re^5: grep question by trammell (Priest) on May 26, 2005 at 23:26 UTC
Re^2: grep question by newest_newbie (Acolyte) on May 26, 2005 at 22:26 UTC
Thanks Thanks for all the advises I actually had tried something like `map { /param=(.)/; return $1; }` which did not work. I have modified my code to be like this `return map { /memtype=(.?)"/i ? $1 : (); } $self -> collect_data ('all');` as suggested by you monks! PS: I hope at least the reply post looks better !!	[reply] [d/l] [select]
Re^3: grep question by Joost (Canon) on May 26, 2005 at 23:21 UTC
`return`ing from a grep actually returns from the underlying subroutine (not just the grep statement), so you probably don't want that (unless you want to return the first match in some perverse way) Update: ps: please use `<code>` tags for code (not pre) and use regular HTML for the rest (except links: use `[link name or id://xxxxx]` for them) see writeup formatting tips "What should it profit a man, if he should win a flame war, yet lose his cool?"	[reply] [d/l] [select]