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sundarkudos
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Joined: 16 Oct 2008
Posts: 39
Location: Chennai
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Hi,
A requirement goes like this.
ABC 1ABC 2ABC 3ABC
If the first word in the line is 'ABC', then i need to find the starting position of the 2nd word.It can have any number of spaces in between 1st and 2nd word.
Is there any function in PLI to make use for it???
Please help.
Thanks,
Sundar. |
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dick scherrer
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Joined: 23 Nov 2006
Posts: 19244
Location: Inside the Matrix
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Hello,
What is the "second word" to you? 1ABC or the second occurrence of ABC? |
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sundarkudos
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Joined: 16 Oct 2008
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Hi Dick,
The position of '1' in the word '1ABC'.
Thanks,
Sundar. |
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donateeye
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Joined: 01 Jun 2007
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S=SUBSTR(SOURCE,1,3);
IF SUBSTR(S,1,3) = 'ABC' THEN
DO;
T=SUBSTR(SOURCE,LENGHT(S)+1,LENGTH(SOURCE)-LENGTH(S));
S1=TRIM(T,' ');
K=INDEX(T,SUBSTR(S1,1,1);
K=K+3;
END;
K IS YOUR ANSWER
Example :
ABC......ABCD...DEF
S = ABC
T = ......ABCD...DEF
S1 = ABCD...DEF
K = 7
K = 7 + 3 + 10
FINALLY K = 10 WHICH IS NOTHING BUT THE START OF THE SECOND WORD IN THE SOURCE (ABCD) |
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enrico-sorichetti
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Joined: 14 Mar 2007
Posts: 10873
Location: italy
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the code posted does not satisfy the request...
it depends on the string "ABC" in the source
a correct logic would keep a word count and alternating the search for a blank and non blank chars
pseudo logic ...
initialize the <word> count, position, length
position the string pointer at the beginning of the string
find the first non blank char ( identifies the beginning of the next <word>)
update the word count and position
find the first blank ( identifies the end of the current <word>)
process the current <word> ( store the length)
keep looping |
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donateeye
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S=SUBSTR(SOURCE,1,3);
IF S = 'ABC' THEN
DO;
T=SUBSTR(SOURCE,LENGHT(S)+1,LENGTH(SOURCE)-LENGTH(S));
S1=TRIM(T,' ');
K=INDEX(T,SUBSTR(S1,1,1);
K=K+3;
END;
the above code should work (apologize, there was a typo error in my previous post)
Please advice me why the code does not satisfy the request.? |
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enrico-sorichetti
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Joined: 14 Mar 2007
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looks like You did not care to understand the original request and my comment ...
ABC 1ABC and so on are just place holders
and as said before Your code relies on the ABC <thing> being in the input string
so do not insist on posting wrong code and meditate on my <pseudo> code |
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donateeye
Warnings : 2
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Yes, I was considering his example, here is the generic code
THE CODE :
M = MATCHING WORD
S=SUBSTR(SOURCE,1,LENGTH(M));
IF M=S THEN
DO;
T=SUBSTR(SOURCE,LENGHT(S)+1,LENGTH(SOURCE)-LENGTH(S));
S1=TRIM(T,' ');
K=INDEX(T,SUBSTR(S1,1,1);
K=K+LENGTH(S);
END;
LETS CONSIDER THE FOLLOWING :
M is the matching word, lets say it is 'APPLE'
Lets have Source as 'APPLE....BOY.....CAN' (... ARE SPACES)
How it works :
M = APPLE 'AS GIVEN'
S = APPLE
M IS EQUAL TO S, SO GOES INTO THE LOOP
T = ....BOY......CAN
S1 = BOY......CAN
K = 5
K = K + 5
FINALLY K = 10 WHICH IS NOTHING BUT THE START OF THE SECOND WORD IN THE SOURCE |
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enrico-sorichetti
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Joined: 14 Mar 2007
Posts: 10873
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I think that You still miss the point...
why do You insist in match the string about a known <thing>
consider a string of arbitrary words with an arbitrary number of spaces
use SUBSTR to find out the starting position of the next word ( first non blank )
use INDEX to find the end of a word ( first blank ) |
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sundarkudos
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Joined: 16 Oct 2008
Posts: 39
Location: Chennai
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I'm screwed.  Please help.
Thanks,
Sundar. |
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Robert Sample
Global Moderator
Joined: 06 Jun 2008
Posts: 8696
Location: Dubuque, Iowa, USA
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| Code: |
IF UPPERCASE(SUBSTR(TRIM(STRINGVAR),1,3)) = 'ABC'
THEN DO;
ABCLOC = INDEX(UPPERCASE(STRINGVAR),'ABC') ;
FIRSTCHAR = SUBSTR(TRIM(SUBSTR(STRINGVAR,ABCLOC+3)),1,1) ;
SECONDWORDLOC = INDEX(SUBSTR(STRINVAR,4),FIRSTCHAR)+3 ;
END; |
I haven't tested this, so beware. You need to verify that the first word is ABC. Once that is done, find its location in the string (leading blanks may mean it does not start in the first position). Find the first character of the second word. Find the location of that character in the string -- SECONDWORDLOC is that starting position. |
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dick scherrer
Moderator Emeritus
Joined: 23 Nov 2006
Posts: 19244
Location: Inside the Matrix
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Hello,
| Quote: |
| I'm screwed. Please help. |
Possibly because the requirement as posted has confused more than it has aided in getting a solution. . .
Suggest you show multiple input lines with differing values and explain the "answer" you want when each set is processed. Use the Code tag to preserver alignment and a "ruler" to show data positions:
| Code: |
| ----+----1----+----2----+----3----+----4----+----5 |
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enrico-sorichetti
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Joined: 14 Mar 2007
Posts: 10873
Location: italy
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all depends on who is holding the screwdriver  |
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donateeye
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Sundarkudos....
tell me whether you tried the code that I posted ???? and let me know what do you expect and what did my code deliver ??? for your reference, this is the code I am referring to :
M = MATCHING WORD /* in your example, this was 'ABC' */
S=SUBSTR(SOURCE,1,LENGTH(M));
IF M=S THEN
DO;
T=SUBSTR(SOURCE,LENGHT(S)+1,LENGTH(SOURCE)-LENGTH(S));
S1=TRIM(T,' ');
K=INDEX(T,SUBSTR(S1,1,1);
K=K+LENGTH(S);
END;
Please try this out and let me if you are not getting what you expected..? |
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enrico-sorichetti
Superior Member
Joined: 14 Mar 2007
Posts: 10873
Location: italy
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the code is wrong in my way of reading the topic because it relies on the presence of a matching as You call it word
try to run it without change on two different strings
| Code: |
"word1 word2 wrord3 ..."
"someotherword1 adifferentword2 asimpleword3" |
I just reread the whole topic and ....
maybe there is a language/phylosophical/way of reading things barrier here,
mostly phylosophical I guess
given a string find the starting position of the second/any word
the algorithm is general an useful to everybody
the two things are examples and the "ABC" is a placeholder
I guess that this way is the most reasonable ( but still my opinion )
otherwise the TS should have used an additional adjective like VERBATIM
to specify that the data to be processed was EXACTLY and ONLY the data posted
in this is the case Your algorithm might work
but I feel that my interpretation is the right one because the TS said
| Quote: |
| A requirement goes like this |
and he was trying to explain to us dummies what he mean by <word> 
I have some personal urgencies right now ... NO IS NOT WHAT YOU THINK
I' ll post later a code snippet which given a string will parse it and build trhee arrays
words array
length array
poisition array |
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donateeye
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You are right... Perhaps, the ideas werent communicated perfectly.
According to sundarkudos, he has a character value stored in a variable (lets consider M as the variable name, and the character value is any group of characters for eg, 'apple' or 'childhood' etc)
Also, he has another variable 'S' where group of words (like a sentence) is stored. eg; S has the value 'apple tree is good' OR it might have the value 'orange tree is good'.
Now, all he have is these variables :
M (in our example, M is assigned the character value 'apple')
S (in our example, S is assigned the character value 'apple tree is good')
His requirement is :
IF the value in M is the first word in S, THEN find the starting position of the second word (in our example, find the position of 't' in the variable S)
ELSE ignore everything.
The code which I posted first checks whether M is the first word in S (assuming we declare M as char varying, so the whole of M should be the first word in S, inorder for the process to continue)
then does the rest.
so, if we have 'orange' in M, then this would go to the ELSE part and does nothing.
If we have 'apple' in M, and if we had different number of spaces between the first and second word in S, then too this code would give the position of where the second word begins.
Sundarkudos....!! hope you are there reading this. Please correct me, and also try the code, I couldnt see you popping in for long time. |
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enrico-sorichetti
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Joined: 14 Mar 2007
Posts: 10873
Location: italy
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I will not have access to a PL/I compiler for a while..
here is a prototype written in rexx...
the PL/I coding might follow the same
| Code: |
str = "someword anotherword againaword"
wc = 0 /* word count */
ws = 0 /* word start , works also as an indicator of a word found */
len = length(str)
do i = 1 to len
if ws = 0 then do
if substr(str,i,1) = " " then ,
iterate
ws = i
wc = wc + 1
end
else do
if substr(str,i,1) \= " " then ,
iterate
wl = i - ws
say wc wl substr(str,ws,wl)
ws = 0
end
end
if ws > 0 then do
wl = i - ws
say wc wl substr(str,ws,wl)
end
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enrico-sorichetti
Superior Member
Joined: 14 Mar 2007
Posts: 10873
Location: italy
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and here is another prototype
using the POS/INDEX builtin function for the terminating blank
instead of substringing char by char
| Code: |
str = "someword anotherwor againaword "
wc = 0
ws = 0
len = length(str)
i = 1
do while ( i < len )
if substr(str,i,1) = " " then do
i = i + 1
iterate
end
ws = i
wc = wc + 1
we = pos(" ",str,i)
if we = 0 then
we = len + 1
wl = we - ws
say wc wl substr(str,ws,wl)
ws = 0
i = we
end
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enrico-sorichetti
Superior Member
Joined: 14 Mar 2007
Posts: 10873
Location: italy
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and here is a quick and dirty PLI for it
CODE:
| Code: |
STRPARS: PROC OPTIONS(MAIN);
DCL S CHAR(80) ;
DCL (I,L) FIXED BIN(31) ;
DCL (WC, WS, WE, WL ) FIXED BIN(31);
S = ' thisisword1 someword2 anotherword3' ;
L = LENGTH(S) ;
PUT SKIP LIST(L,S) ;
WC = 0 ;
WS = 0 ;
I = 1 ;
DO WHILE ( I < L );
IF SUBSTR(S,I,1) = ' ' THEN
I = I + 1 ;
ELSE DO ;
WS = I ;
WC = WC + 1 ;
WE = INDEX(SUBSTR(S,WS),' ') ;
IF WE = 0 THEN WE = L + 1 ;
ELSE WE = WE + WS ;
WL = WE - WS - 1 ;
PUT SKIP LIST (WC, WS, WL, SUBSTR(S,WS,WL)) ;
WS = 0 ;
I = WE ;
END ;
END ;
END STRPARS;
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RESULT:
| Code: |
80 thisisword1 someword2 anotherword3
1 7 11 thisisword1
2 19 9 someword2
3 29 12 anotherword3 |
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prino
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Joined: 07 Feb 2009
Posts: 1306
Location: Vilnius, Lithuania
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Why doesn't anyone mention "VERIFY", now with the option of starting in the middle of the string?
Sheesh... |
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dbzTHEdinosauer
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Joined: 20 Oct 2006
Posts: 6966
Location: porcelain throne
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donateeye,
your code presupposes the knowledge of the first word in the string.
My interpretation of Sundardudos's requirement is:
find the position of the second word in any string,
I find that enrico's solutions provide that info
Actually, they provide a complete answer to an unasked question:
what are the starting and ending positions of all words in a string.
If Sundardudos only wants to know the attributes of the second word,
he can modify enrico's code.
using your code, you would have to add a routine to run previous that
would extract the first word.
Now, analyzing the complete string (all the words),
is something that is often done -
but, if for some reason, the TS only wants the 2nd,
truncating enrico's code will provide the info. |
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enrico-sorichetti
Superior Member
Joined: 14 Mar 2007
Posts: 10873
Location: italy
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Hi Robert
| Quote: |
| Why doesn't anyone mention "VERIFY", now with the option of starting in the middle of the string? |
because the manual version I was looking at did not have the 3 args version
and I would have had anyway to use a trick |
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donateeye
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Yes, Sundarkudos has clear said
If the first word in the source is xxxxxx then...
which clearly means that he should be aware of the first word...
considering that, the code just needs to find only the starting position of the second word.
Interesting is that, the man who started this topic is no where here to read all the suggestions and codes for him.... |
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dbzTHEdinosauer
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Joined: 20 Oct 2006
Posts: 6966
Location: porcelain throne
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donateeye,
you are pounding sand.
one assumes the program is to handle more than one input,
thus the first word is not always known.
enrico and the other seniors do not write code for simple accept/display cobol programs.
their examples are an attempt to drag those
with little experience
out of the dark chasm of ignorance that they seem to reside
and provide insight to the world of professional programming. |
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donateeye
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| well.... you are right.... My apologize to enrico.... |
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