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Using awk to find next word available

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Top Forums Shell Programming and Scripting Using awk to find next word available
# 15  
Old 02-08-2010
try using perl as below:-

Code:
perl  -wnl -e ' /Start/ ... /[a-zA-Z]/ and /\d+\s+\d+/ and print  ;'  infile.txt

BR

SmilieSmilieSmilie

---------- Post updated at 18:37 ---------- Previous update was at 18:33 ----------

or even better

Code:
perl  -wnl -e ' /Start/ ... /^\D+/ and /\d+\s+\d+/ and print  ;'  infile.txt

SmilieSmilieSmilie
# 16  
Old 02-08-2010
NickC, in you sample the first character of the lines that need to be printed are starting with a digit. I am testing for that: /^[a-zA-Z]/ and /^[^0-9]/ mean testing for the case that the first character is a letter or not a digit respectively.

IMO, the construct that you suggest would not work since it would print the labels as well and not every occurence.
# 17  
Old 02-08-2010
ahmad.diab: the perl script worked well, but only printed out the first instance of the number string after /Start/

Scrutinizer: I'm sorry - what I failed to mention was that when the number string (that includes the exponents) was finished, the next word does not always start with the first character. There might be one or two blank characters before the letters begin. I don't know if your code is looking for something as the first character.

This is why I thought it might be easier to start the search with /Start/ and stop the next time awk encounters a line that does NOT contain the string "E+" in it - and then print out all the lines in between! Smilie And then repeat this throughout the file.

For more clarity, I generated some output files. So here are a couple of samples (in red color are the lines I want printed):

Code:
               EPSILON              EXTERNAL WORK      START
                1         -5.1077400E-06          2.3338035E+03
                2         -1.2286651E-05          2.6901846E+03
                3         -2.4254409E-06          4.0875334E+03   
 *** USER INFORMATION MESSAGE 4114 (OUTPX2)
     DATA BLOCK OUGV1    WRITTEN ON FORTRAN UNIT 12, TRL =

Or

Code:
         EPSILON              EXTERNAL WORK      START
               1         -4.3371111E-09          8.6807975E+05 
1    OUTPUT FILES                  PAGE    22

0     BASELINE                                                                                   
 *** SYSTEM INFORMATION MESSAGE 6916 (DFMSYN)
     DECOMP ORDERING
 *** USER INFORMATION MESSAGE 5293 (SSG3A)
    FOR DATA BLOCK KLL
 EPSILON              EXTERNAL WORK      START
                2         -3.1694217E-10          2.4290496E+07 
1    OUTPUT FILES                         PAGE    23

0     BASELINE                                                                            
 *** SYSTEM INFORMATION MESSAGE 6916 (DFMSYN)
     DECOMP ORDERING 
 *** USER INFORMATION MESSAGE 5293 (SSG3A)
    FOR DATA BLOCK KLL
 EPSILON              EXTERNAL WORK      START
                3         -1.1569729E-09          1.5256892E+07 
1    OUTPUT FILES            PAGE    24

0     BASELINE                                                                            
 *** SYSTEM INFORMATION MESSAGE 6916 (DFMSYN)
     DECOMP ORDERING


It is the differing formats that made it difficult for me. Initially, I had it working with a simple awk script that just searched between /START/ and /USER/ and printed the lines in between. But then I came across an output format like the second example! Thanks again for everyone's patience! Smilie
Last edited by NickC; 02-08-2010 at 02:05 PM..
# 18  
Old 02-08-2010
That is quite different from your original input spec.

Try this:
Code:
awk '$1=="START" {flag=1;next} !/^  / {flag=0} flag' infile

try this to remove the spacing:
Code:
awk '!/^  /{p=0}{$1=$1}p;$1=="START"{p=1}' infile

# 19  
Old 02-08-2010
Yeah, sorry - when I posted the first thread I didn't think it made such a big difference but now I know! Live and learn...

I tried those two commands but I'm getting no output at all. Not sure if the !/^ sequence depends on the version of UNIX (like I said above, I'm using AIX and ksh.

Going back to my question above - would it be too complicated or difficult to start the search at START - print out every line that has E+ in it, and stop when a line without E+ is encountered? From what I can tell, that sequence should capture what I'm trying to get. I was trying something like :

awk '/START/ {flag=1;next} /^E\+/ {flag=0} flag {print}' <filename>

hoping that would work - where the E\+ would mean look for "E+" without attaching special character to +. And the ^ would say until E+ is no longer found. But I think you said above that this command wouldn't work. And you're right that it didn't! Smilie But I'm not sure why.
# 20  
Old 02-08-2010
Try this one:
Code:
awk '/START/{p=1;next}p && $2 !~ /E-..$/{p=0} p' file

# 21  
Old 02-08-2010
Got it - that works!! Smilie Now I have to figure out just what you did! Smilie

Thanks again all of you for your patience and help!
Last edited by NickC; 02-08-2010 at 04:45 PM..
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