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06-26-2013

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How to use regular exp to replace the character in the file?

Hi Unix Gurus,

yesterday I asked a question and got answer, it works fine.
I have one more thing need to help
in the code

Code:

awk '{print substr($0,1,3)"xxx"substr($0,7)}' file

If I have 50 charactor's need to be replaced, is there any easy way to use reg exp or I have to input 50 XXXXx......

Thanks in advance

ken6503

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06-26-2013

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You could do something like this:

Code:

awk -v x=$(printf 'x%.s' {1..50}) '{print substr($0,1,3) x substr($0,7)}' f

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06-26-2013

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You can do it all in awk:

Code:

awk 'BEGIN { 
  c = "x"; n = 50
  for (i = 0; ++i <= n;) s = s c
  }
{
  print substr($0, 1, 3) s substr($0, 7)
  }' infile

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06-26-2013

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Quote:

Originally Posted by radoulov

You can do it all in awk:

Code:

awk 'BEGIN { 
  c = "x"; n = 50
  for (i = 0; ++i <= n;) s = s c
  }
{
  print substr($0, 1, 3) s substr($0, 7)
  }' infile

The script works, would you please explain a little bit about "s = s c", what this is doing?

Thanks

---------- Post updated at 12:42 PM ---------- Previous update was at 12:08 PM ----------

Quote:

Originally Posted by Subbeh

You could do something like this:

Code:

awk -v x=$(printf 'x%.s' {1..50}) '{print substr($0,1,3) x substr($0,7)}' f

Thanks for your reply. the script works.
one more question, If I have two parts of string need to be replaced. how could I achieve this?
for example:
I have original string like

Code:

abcdererdfd1233345r3434sdswewe

I need replace from 3rd to 5th with 20 'X', from 8th 10th with 40 'Y'

Code:

ab"XXXX... 20X"re"YYYY.... 40Y"d1233345r3434sdswewe

thanks in advance

---------- Post updated at 12:49 PM ---------- Previous update was at 12:42 PM ----------

Quote:

Originally Posted by radoulov

You can do it all in awk:

Code:

awk 'BEGIN { 
  c = "x"; n = 50
  for (i = 0; ++i <= n;) s = s c
  }
{
  print substr($0, 1, 3) s substr($0, 7)
  }' infile

one more question, If I have two parts of string need to be replaced. how could I achieve this?
for example:
I have original string like

Code:

Code:

abcdererdfd1233345r3434sdswewe

I need replace from 3rd to 5th with 20 'X', from 8th 10th with 40 'Y'

Code:

Code:

ab"XXXX... 20X"re"YYYY.... 40Y"d1233345r3434sdswewe

thanks in advance

ken6503

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06-26-2013

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Something like this:

Code:

awk 'BEGIN {
  s = ARGV[1]
  cx = 20; cy = 40
  for (i = 0; ++i <= cy;) {
    i <= cx && x = x "x"
    y = y "y"
    }  
  print substr(s, 1, 2) x substr(s, 6, 2) y substr(s, 11)    
  }' <your_string>

Code:

% awk 'BEGIN {
  s = ARGV[1]
  cx = 20; cy = 40
  for (i = 0; ++i <= cy;) {
    i <= cx && x = x "x"
y = y "y"
}
  print substr(s, 1, 2) x substr(s, 6, 2) y substr(s, 11)
  }' abcdererdfd1233345r3434sdswewe
abxxxxxxxxxxxxxxxxxxxxreyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyd1233345r3434sdswewe

radoulov

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Try also

Code:

awk 'BEGIN {X=sprintf ("%50s"," "); gsub (" ","x", X); print X}'
xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

You can do the same for Y...

RudiC

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06-26-2013

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Another one with GNU awk:

Code:

% awk 'BEGIN {
  FIELDWIDTHS = "2 3 2 3 99"
  $0 = ARGV[1]
  cx = 20; cy = 40
  for (i = 0; ++i <= cy;) {
    i <= cx && x = x "x"
y = y "y"
}
  print $1 x $3 y $5             
  }'   abcdererdfd1233345r3434sdswewe
abxxxxxxxxxxxxxxxxxxxxreyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyyd1233345r3434sdswewe

And Perl (I've just realized, that you want X and Y, not x and y

):

Code:

perl -le '($_ = shift) =~ 
  s/(..)...(..)...(.*)/$1.("X" x 20).$2.("Y" x 40).$3/e;
  print ' abcdererdfd1233345r3434sdswewe
abXXXXXXXXXXXXXXXXXXXXreYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYd1233345r3434sdswewe

Last edited by radoulov; 06-26-2013 at 05:43 PM..

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