CODEWITHSUNDEEP
regexbashsedcygwin
GoofyBall
GoofyBall

Reputation: 411

sed - Return characters before pattern

 $  echo "ABC XYZ 12/123/52/ ABBDBDAD 562.4224.32 02381831522" | sed 's/[^a-zA-Z]//g' raw.tmp

Using the above, I am trying to extract ABC XYZ from a line (spaces preserved). My regex returns ABCXYZABBDBDAD: I am a noob at regex and still have a lot to learn.

In summary, how do I get the substring ABC XYZ from a line before a number with whitespace preceding it?

Upvotes: 1

Views: 440

Answers (3)

Anshul
Anshul

Reputation: 730

You need to write following

echo "ABC XYZ 12/123/52/ ABBDBDAD 562.4224.32 02381831522" | sed 's/.*\(ABC XYZ\).*/\1/g'

Output

ABC XYZ

Point is - I believe you are trying to extract 'ABC XYZ' (exactly). So you extract that and substitute entire line with it

Edit I think i missed the point. You basically want 'Str1 Str2 '

In that case following works

echo "ABC XYZ 12/123/52/ ABBDBDAD 562.4224.32 02381831522" | sed 's/\([a-zA-Z ][a-zA-Z ]*\).*/\1/g'

Upvotes: 0

fedorqui
fedorqui

Reputation: 290045

This can make it:

$ echo "ABC XYZ 12/123/52/ ABBDBDAD 562.4224.32 02381831522" | sed -n 's/\([A-Z]* [A-Z]*\) [0-9]*.*/\1/p' 
ABC XYZ

Explanation:

sed -n 's/\([A-Z]* [A-Z]*\) [0-9]*.*/\1/p' 

  \([A-Z]* [A-Z]*\)  == catch WORD + space + WORD
       [0-9]*.*      == some number + space + rest of string
  /\1/p              == print catched string

Upvotes: 3

konsolebox
konsolebox

Reputation: 75548

Perhaps this one

echo "ABC XYZ 12/123/52/ ABBDBDAD 562.4224.32 02381831522" | sed -n 's/^\([a-zA-Z ]\+\).*/\1/gp' > raw.tmp

Or more accurately

echo "ABC XYZ 12/123/52/ ABBDBDAD 562.4224.32 02381831522" | sed -n 's/^\([a-zA-Z][a-zA-Z ]\+[a-zA-Z]\).*/\1/gp'

Which restricts characters that begin with letters and ends up with letters as well.

Upvotes: 2

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