World: r3wp

Perhaps a good starting point is to think that when you "parse" something, 
you are asking does it conform to the rules I have supplied. If it 
does parse returns true, if it doesn't parse returns false.

>> parse "abcdefghi" ["abcdefghi"]

== true

>> parse "abcdefghi" ["abcde"]    

== false

You're already familiar with to,thru,  end and to end: 

>> parse "abcdefghi" ["a" to "i"]
 
== false
>> parse "abcdefghi" ["a" thru "i"]
 
== true
>> parse "abcdefghi" ["abcde" to end]

== true

Perhaps the second thing to realise is a rule can be split into "sub-rules':

>> parse "abcdefghi" ["abcde" "fghi"]

== true

Each "subrule" can have some Rebol code executed if it is "triggerred":


>> parse "abcdefghi" ["abcde" (print "abcde found") "fghi"(print 
"fghi found")]

abcde found

fghi found

== true

>> parse "abcdefghi" ["abcde" (print "abcde found") "xyz"(print "xyz 
found")] 
 
abcde found

== false

(What Gregg referred to was a much more sophisticated way of getting 
the parse to show you what happened).

Sub-rules can be optional by using the | (or) and enclosing the options 
in a block:


>> parse "abcdefghi" ["abcde" (print "abcde found") ["fghi" (print 
"fghi found") | "xyz"(print "xyz found")]]                       
                                                   
abcde found
fghi found
== true

Sorry about the formatting, It's one of the big problems with AltME 
under Mac OS X.

Skip tells parse to move to the next item:

>> parse "abcdefghi" ["abcdefgh" skip]
== true   ;; because the skip took us to the end

We can specify that a sub-rule should be repeated:


>> parse "abcdefghi" ["abcde" 4 skip] 

== true

A better example of repetition:

>> parse "aaa" [4 "a"]

== false

Some and Any are forms of repetition, this shows the difference:

>> parse "aaa1000" [some "a" to end]

== true

>> parse "aaa1000" [any "a" to end] 

== true

>> parse "bbb1000" [any "a" to end]   

== true

>> parse "bbb1000" [some "a" to end]

== false

I apologise if you are already happy with these basic concepts, in 
which case I hope you don't mind the refresher.

Maybe these are some variations of what you are looking for


parse/all "fd doixx s x x x oie    x } " [some [copy d   "x" (print 
d) | skip]]


parse/all "fd doixx s x x x oie    x } " [some [copy d 1 2  "x" (print 
d) | skip]]


parse/all "fd doixx s x x x oie    x } " [some [copy d  2  "x" (print 
d) | skip]]


parse/all "fd doixx s x x x oie    x } " [some [copy d   "xx" (print 
d) | skip]]


parse/all "fd doixx s x x x oie    x } " [some [[copy d  "x"  copy 
e  "x" (print [e d]) ] | skip]]


parse/all "fd doixx s x x x oie    x } " [some [ (g: copy "" ) 2 
[copy d  "x"  (append g d)  ]  (print g )  | skip]]

or you are looking for the pairs

 parse/all "fd doixx s x x x oie    x } "  [ some [  [ (g: copy "" 
 ) 2 [ copy d "x"  (append g d ) any notx  | skip  ] (if not empty? 
 g [print g]) ]  ] ]

I forgot notx

notx: complement charset "x"

parse/all "fd doixx s x x x oie    x } "  [ some   [ (g: copy "" 
) 2 [ copy d "x"  (append g d ) any notx  | skip  ] (if not empty? 
g [print g]) ]  ]

Thanks PeterWood. I like to think I am ok with the most basic concepts, 
so now I am trying to learn things that will help me some my real 
life probelms in a better way.  I use parse pretty much every day 
& always have a rebol console up on my work PC, but ANY SOME & OPT 
& |  I do not understand in context.  I understand them in abstract 
terms, but not how to apply them in conjuction with [] . I do understand 
your examples of some & any (these examples are usefull to me). skipping 
an un-known number of chars to get to the next match is the bit I 
find hard to understand how to construct, paticularly if it needs 
to be done in the context of a previous match.

sqlab, I dont know about this syntax at all. I dont think I understand 
what is happening here.

copy to "x"  & copy thru "x"  I understand, but copy "x"  I didn't 
expect to see.

In the parens you use the COPY function, not the PARSE copy operation. 
Is that what you meant?

The compliment syntax & the    to 1 3 digit   where digit is a charset 
seems to be "unreliable" as far as I can understand.

this is what I dont expect.

parse/all "fd doixx s x x x oie    x } " [some [copy d   "x" (print 
d) | skip]]

I dont think I have ever seen the PARSE copy operation documented. 
  I will have  a hunt for it.

have you ever read the parse documentation in the old RT publisehd 
rebol 2.3 pdf  ?  its a good reference... there are only minor changes 
from that version up to the latest... I don't think any of the examples 
would fail in the current parse.

chpter 15 or the Rebol Core Manual  http://www.rebol.com/docs/core23/rebolcore-15.html
 may have a use of this syntax in a complicated example, but no description 
of what is happening exactly.

yep, that's the online version of it.

OK, here's what happens: The next recognized pattern is COPY/part'ed 
and assigned to the variable. If the length of the matched pattern 
is 0, #[none] is assigned to the variable.

Yes, I have read it a lot, but it seems more of a reference for people 
who already know, rather then an explanation of Parse operations.

Thanks BrianH, I was sort of guessing it must be like a variation 
of copy thru "x" that does not skip like thru...  I think I get that 
now. Thanks.

Note that the assignement to the variable happens *after* the pattern 
is recognized, so any code inside the pattern that references the 
value of the variable will get the old value. Like this:
>> x: "old"
== "old"
>> parse "new" [copy x ["new" (print x)] (print x)]
old
new
== true

The same goes for the set operation of block parsing.

That is pretty important! I had not realised that before & this copuld 
account for some of the unpredictable behaviour I get.. I thought 
the patern was complete at your first print statement.   These [] 
have lots of subtle influence.

[ and ] are a grouping construct.

This is my nemisis. I can't understand how this prints XXXX then 
XX , not XX  three times.  It seems to have a will of its own.

parse/all { X X  XX X X} [some[[copy x "X" (prin x) [copy y "X" (print 
y) | skip] | skip]]]    
I have been stuck on this (in various forms) for over a week now

Well first of all, you have an extra [ ] in there, just after the 
some.

My thinking is that I expect the inner copy to be executed after 
the first "X" is found, then come back out of the inner bit when 
the next "X" is found.

oops.. my bad with the extra [ ]    I keep trying all sorts & that 
got left behind.

parse/all { X X  XX X X} [some [copy x "X" (prin x) [copy y "X" (print 
y) | skip] | skip]]

Character at a time:
- the outer skip
- copy x "X" (prin x)
- the inner skip
- copy x "X" (prin x)
- the inner skip
- the outer skip
- copy x "X" (prin x)
- copy y "X" (print y)
- the outer skip
- copy x "X" (prin x)
- the inner skip
- copy x "X" (prin x)
- the outer skip

Try this:

>> parse/all { X X  XX X X} [some [copy x "X" (prin x) [copy y "X" 
(print y) | skip (prin "i")] | skip (prin "o")]]
oXiXioXX
oXiXo== true

Now that last outer skip seems to me that it should be an inner skip, 
but I am clearly wrong :(

Ah. that is a good trick to print i & o with the skips, this will 
help a lot, thanks.

Isee from your analysis that my error is to expect the inner skip 
to skip back to the beginning of that loop...  not sure why I expected 
that, but it is clearly wrong. I dont know if there is a way to make 
the inner loop behave like that, at an earlier point I did hve OPT 
in front of it...  I will be able to make progress with my experiments 
now armed with the new trick you have taught me, but for now I have 
to go & get some sleep I am afraid.  Thanks again.

>> parse/all { X X  XX X X} [(prin 'a) some [(prin 'b) "X" (prin 
'c) [(prin 'd) "X" (print 'e) | (prin 'f) skip (prin 'g)] (prin 'h) 
| (prin 'i) skip (prin 'j)] (prin 'k)]
abijbcdfghbcdfghbijbcde
hbijbcdfghbcdfijbik== true

Now the fij is a bug in parse - it should be fgh.

Excuse, if I did confuse you with my examples-

I just tried to show you that you can get loops with an definite 
loop counter in parse.

regarding 

parse/all "fd doixx s x x x oie    x } " [some [copy d   "x" (print 
d) | skip]]
what did you expect?

If you know what you are looking for you can extend it to


parse/all "fd doixx s x x x oie    x } " [some [copy d   ["x"   | 
"y" | "z" ]    (print d) | skip]]
and you will get your searched values.

But maybe I just don't understand the problem.

Thanks Brian, I am finaly getting it I think. the prin 'a etc is 
better than debugging techniquies I have tried because it is small 
and dosen't add too much to the complexity, I can see I could reduce 
this method further with something like a: does [prin 'a]

Thanks sqlab, no need to excuse yourself please, your examples are 
great & I learnt a new use for COPY in PARSE. This has made your 
examples clear to me now, so thanks for spending your time helping 
me.  


The problem I have set myself is purely to understand parse more 
clearly so I have enough know-how to write any scripts I need without 
spending all day doing it. That is why I start off anking one question, 
then jump to another question if I don't fully understand the help 
I get.  


I have used parse a fair bit all ready, but limited myself to very 
simple concepts. see http://www.rebol.org/script-information.r?script-name=cisco-extract.r
and marvel that it even works ;-)   Thanks.

I have been reminded that   |skip   is a word  | skip   is OR skip. 
 it is so easy to miss the space between | & skip

Right, I would say that the following snippit is the most educational 
thing I have done with PARSE.  It shows me a lot of things about 
what is happening & validates the construction and use of charsets 
& whatever the 'address block is called.     Thanks everyone for 
your help.

digit: charset [#"0" - #"9"]
address: [1 3 digit "." 1 3 digit "." 1 3 digit "." 1 3 digit]

a: does [prin 'a] b: does [prin 'b] c: does [prin 'c] d: does [prin 
'd] e: does [prin 'e] f: does [prin 'f]
parse/all {1 23 4.5.6.12 222.1.1.1 7 8} [some[

 (a) copy x address (prin x) some[ (b) copy y address break | skip 
 (c)] (print y) | skip (d)
]]
adadadadada4.5.6.12bcb222.1.1.1

why not use block parsing?

parse [ 1 23 4.5.6.12 222.1.1.1 7 8 ] [ integer! integer! tuple! 
tuple! integer! integer! ]

Because he's parsing Cisco config files.