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unix command to replace anything between between two delimiter positions

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.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;

-1

Please help me with a unix command to replace anything between two delimiter positions.
For ex: I have multiple files with below header data and I want replace the data between * delimiters at 9th and 10th position

ISA*00* *00* *ZZ*80881 *ZZ*TNC0022 *190115*1237*^*00501*000320089*0*P*|~

My output should like this:

ISA*00* *00* *ZZ*80881 *ZZ*TNC0022 *190327*1237*^*00501*000320089*0*P*|~

linux

share|improve this question

edited Mar 27 at 16:43

ceving

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asked Mar 27 at 16:39

kiran kiran

1

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  You should provide some simpler sample input/output that you will elaborate for this very particular case. This way, it will be more accurate for others users when searching the same response and you will be able to transpose the logic to your needs.
  
  – Gilles Quenot
  Mar 27 at 17:01
  
  

  
  
  
  

add a comment | 

-1

Please help me with a unix command to replace anything between two delimiter positions.
For ex: I have multiple files with below header data and I want replace the data between * delimiters at 9th and 10th position

ISA*00* *00* *ZZ*80881 *ZZ*TNC0022 *190115*1237*^*00501*000320089*0*P*|~

My output should like this:

ISA*00* *00* *ZZ*80881 *ZZ*TNC0022 *190327*1237*^*00501*000320089*0*P*|~

linux

share|improve this question

edited Mar 27 at 16:43

ceving

11k44 gold badges6262 silver badges108108 bronze badges

asked Mar 27 at 16:39

kiran kiran

1

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  You should provide some simpler sample input/output that you will elaborate for this very particular case. This way, it will be more accurate for others users when searching the same response and you will be able to transpose the logic to your needs.
  
  – Gilles Quenot
  Mar 27 at 17:01
  
  

  
  
  
  

add a comment | 

Please help me with a unix command to replace anything between two delimiter positions.
For ex: I have multiple files with below header data and I want replace the data between * delimiters at 9th and 10th position

ISA*00* *00* *ZZ*80881 *ZZ*TNC0022 *190115*1237*^*00501*000320089*0*P*|~

My output should like this:

ISA*00* *00* *ZZ*80881 *ZZ*TNC0022 *190327*1237*^*00501*000320089*0*P*|~

linux

share|improve this question

edited Mar 27 at 16:43

ceving

11k44 gold badges6262 silver badges108108 bronze badges

asked Mar 27 at 16:39

kiran kiran

1

ISA*00* *00* *ZZ*80881 *ZZ*TNC0022 *190115*1237*^*00501*000320089*0*P*|~

ISA*00* *00* *ZZ*80881 *ZZ*TNC0022 *190327*1237*^*00501*000320089*0*P*|~

share|improve this question

edited Mar 27 at 16:43

ceving

11k44 gold badges6262 silver badges108108 bronze badges

asked Mar 27 at 16:39

kiran kiran

1

share|improve this question

edited Mar 27 at 16:43

ceving

11k44 gold badges6262 silver badges108108 bronze badges

asked Mar 27 at 16:39

kiran kiran

1

edited Mar 27 at 16:43

ceving

11k44 gold badges6262 silver badges108108 bronze badges

edited Mar 27 at 16:43

ceving

11k44 gold badges6262 silver badges108108 bronze badges

asked Mar 27 at 16:39

kiran kiran

1

asked Mar 27 at 16:39

kiran kiran

1

1 Answer
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-1

Try this:

perl -pe 's/^((?:[^*]**)9)([^*]+)(.*)/$1190327$3/'

The regexp searches for 9 occurences 9 of anything but not being a star [^*] followed by a star * and stores all in the first capture group. The second capture is at least one character not being a star [^*]+. And the third capture is the rest of the line.

A matching line gets replaced by the first part $1, your new value 190327 and the third part $3.

share|improve this answer

answered Mar 27 at 17:01

cevingceving

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-1

Try this:

perl -pe 's/^((?:[^*]**)9)([^*]+)(.*)/$1190327$3/'

The regexp searches for 9 occurences 9 of anything but not being a star [^*] followed by a star * and stores all in the first capture group. The second capture is at least one character not being a star [^*]+. And the third capture is the rest of the line.

A matching line gets replaced by the first part $1, your new value 190327 and the third part $3.

share|improve this answer

answered Mar 27 at 17:01

cevingceving

11k44 gold badges6262 silver badges108108 bronze badges

add a comment | 

-1

Try this:

perl -pe 's/^((?:[^*]**)9)([^*]+)(.*)/$1190327$3/'

The regexp searches for 9 occurences 9 of anything but not being a star [^*] followed by a star * and stores all in the first capture group. The second capture is at least one character not being a star [^*]+. And the third capture is the rest of the line.

A matching line gets replaced by the first part $1, your new value 190327 and the third part $3.

share|improve this answer

answered Mar 27 at 17:01

cevingceving

11k44 gold badges6262 silver badges108108 bronze badges

add a comment | 

Try this:

perl -pe 's/^((?:[^*]**)9)([^*]+)(.*)/$1190327$3/'

The regexp searches for 9 occurences 9 of anything but not being a star [^*] followed by a star * and stores all in the first capture group. The second capture is at least one character not being a star [^*]+. And the third capture is the rest of the line.

A matching line gets replaced by the first part $1, your new value 190327 and the third part $3.

share|improve this answer

answered Mar 27 at 17:01

cevingceving

11k44 gold badges6262 silver badges108108 bronze badges

perl -pe 's/^((?:[^*]**)9)([^*]+)(.*)/$1190327$3/'

share|improve this answer

answered Mar 27 at 17:01

cevingceving

11k44 gold badges6262 silver badges108108 bronze badges

answered Mar 27 at 17:01

cevingceving

11k44 gold badges6262 silver badges108108 bronze badges

answered Mar 27 at 17:01

cevingceving

11k44 gold badges6262 silver badges108108 bronze badges