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How to fix this awk

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Top Forums Shell Programming and Scripting How to fix this awk
# 1  
Old 06-17-2010
How to fix this awk
I have a script which will mask the 9th and 15th column in a record starting with BPR.

The record looks like below before my script
Code:
BPR*C*160860.04*C*ACH*CTX*01*072000326*DA*1548843*3006968523**01*071000013*DA*5529085*100323*VEN

The record will be masked after my script parses this record
Code:
BPR*C*160860.04*C*ACH*CTX*01*072000326*DA*XXX8843*3006968523**01*071000013*DA*XXX9085*100323*VEN*

I do not want the '*' character at the end of the record at any point.
Below is my script
Code:
function chgrec(rec){
  debug=1
  z=split(rec,rarray,"*")
  element=10
  unChanged = rarray[element]
  lUnChg = length(unChanged)-3
  lastFour = substr(unChanged,lUnChg)
  initValue = rarray[element]
  gsub(/[0-9A-Za-z]/,"X",initValue)
  preFix = substr(initValue,1,lUnChg-1)
  element2=16
  unChanged2 = rarray[element2]
  lUnChg2 = length(unChanged2)-3
  lastFour2 = substr(unChanged2,lUnChg2)
  initValue2 = rarray[element2]
  gsub(/[0-9A-Za-z]/,"X",initValue2)
  preFix2 = substr(initValue2,1,lUnChg2-1)
  rec=""
  for (i=1;i<=z;i++){
#    printf("-->1:%s-->2:%s\n", i, rarray[i])> "/tmp/fftt";
     if (debug=0)
       {
        if (i==element)
          rec=rec sprintf("-->%s%s<--(%s)*", preFix, lastFour,rarray[element])
        else
        if (i==element2)
          rec=rec sprintf("-->%s%s<--(%s)*", preFix2, lastFour2,rarray[element2])
        else
            if (i==z)
            rec=rec sprintf("%s*", rarray[i])
       }
     else
       {
        if (i==element)
          rec=rec sprintf("%s%s*", preFix, lastFour,rarray[element])
        else
        if (i==element2)
          rec=rec sprintf("%s%s*", preFix2, lastFour2,rarray[element2])
        else
          rec=rec sprintf("%s*", rarray[i])
       }
  }

# 2  
Old 06-17-2010
Something like this?
Code:
awk -F\* '/^BPR/{$10="XXX" substr($10,4);$16="XXX" substr($16,4)}1' OFS="*" file

# 3  
Old 06-17-2010
Is there a way where you can find the problem in my script.
And also I should show only the last 4 characters in the 9th column and I should the mask the rest. You code works good but it masks only the 1st the 3 characters.
# 4  
Old 06-17-2010
One way:
Code:
awk -F\* 'BEGIN{X="XXXXXXXXXX"}
/^BPR/{
  n=length($10)-4
  $10=substr(X,1,n) substr($10,n+1)
  n=length($16)-4
  $16=substr(X,1,n) substr($16,n+1)
}1' OFS="*" file

# 5  
Old 06-22-2010
Thanks a lot Frank.. It worked.. I wanted to include this in my script and I did as shown below. But I get the syntax error as shown below. Any advise.

Below is the script awkMask
Code:
/./{
  type=substr($0,1,3)
  rec = $0
awk -F\* 'BEGIN{X="XXXXXXXXXX"}
/^BPR/{
n=length($10)-4
$10=substr(X,1,n) substr($10,n+1)
n=length($16)-4
$16=substr(X,1,n) substr($16,n+1)
}1' OFS="*" rec > output
}

I am trying to call this from the other script as shown below,
Code:
awk -f $base/scripts/awkMask $InputFile

I get the below error in my log file

Code:
  7834   syntax error The source line is 58.
  7835   The error context is
  7836          awk >>>  -F\ <<< * 'BEGIN{X="XXXXXXXXXX"}
  7837   awk: The statement cannot be correctly parsed.
  7838   The source line is 58.
  7839   syntax error The source line is 63.

# 6  
Old 06-22-2010
Is this a snippet of an awk program?
Code:
/./{
  type=substr($0,1,3)
  rec = $0

You can't include the code in another awk program this way.
# 7  
Old 06-22-2010
What would be the ideal way to have this in my script..
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