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sed command not getting the right output

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# 8  
Old 01-13-2016
Yes, "^" in a bracket expression is a negation. man regex:
Quote:
A bracket expression is a list of characters enclosed in "[]". It normally matches any single character from the list (but see below). If the list begins with '^', it matches any single character (but see below) not from the rest of the list.
The "g" flag is needed as we (hopefully) have multiple (2) matches that all need to be replaced.
# 9  
Old 01-13-2016
Quote:
Originally Posted by RudiC
Yes, "^" in a bracket expression is a negation. man regex:The "g" flag is needed as we (hopefully) have multiple (2) matches that all need to be replaced.
Thank you, but do you mind taking a look at my code and tell me whats wrong with it and how id fix it ?
# 10  
Old 01-13-2016
As MadeInGermany said, it's the greedy .* that makes it output the LAST combination of -<>: chars. On top, in the bracket expression, there's a "," too many. Your (modified) code snippet works:
Code:
sed -ne 's/[^<>:-]*\([-<>:]*\).*/\1/;p' file
<-:::
<-::
<-:::
->:::
->::

# 11  
Old 01-13-2016
Quote:
Originally Posted by RudiC
As MadeInGermany said, it's the greedy .* that makes it output the LAST combination of -<>: chars. On top, in the bracket expression, there's a "," too many. Your (modified) code snippet works:
Code:
sed -ne 's/[^<>:-]*\([-<>:]*\).*/\1/;p' file
<-:::
<-::
<-:::
->:::
->::

Got it. Do you mind looking at this ? its probably easy for you to solve (thread-related).

This code :
Code:
sed -e 's/[^-><:]*\([-<>]*[:]*\).*\[-<>]*[:]*/\1/;h' datasets/q13data.txt

Gives me this output :
Code:
"Oct<-:::NULL<-:: phi
SunSat<-::3.1415<-:::`Nov
WedBlue<-::Mar->:::OctJun

You see, there are 2 [<>-:] patterns here, one in the begining and one close to the EOL.

How would I filter those two out ? Id like to use grouping here if possible

EDIT: I changed the code to :
Code:
sed -ne 's/[^-><:]*\([-<>]*[:]*\)[^-><:]*\([-<>]*[:]\)[^-><:]*/\1\2/;p' datasets/q13data.txt

and get this output

Code:
->:::<-::: `
->::->::2.8chi
<-:::->:::3.14
<-:::<-:: phi
<-::<-:::`Nov
<-::->:::OctJun

How to filter out the last part :P?
Last edited by jozo95; 01-13-2016 at 06:30 PM..
# 12  
Old 01-13-2016
Try
Code:
sed 's/[^<-]*\([-<>:]*\)[^<-]*\([-<>:]*\).*/\1 \2/' file
<-::: <-:::
<-:: <-:::
<-::: <-:::
->::: <-:::
->:: ->:::

This User Gave Thanks to RudiC For This Post:
jozo95
# 13  
Old 01-14-2016
Quote:
Originally Posted by RudiC
Try
Code:
sed 's/[^<-]*\([-<>:]*\)[^<-]*\([-<>:]*\).*/\1 \2/' file
<-::: <-:::
<-:: <-:::
<-::: <-:::
->::: <-:::
->:: ->:::

wow... perfect my friend, thank you very much!!!!
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