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# 8  
Old 04-01-2008
a bit more paint:
Code:
BEGIN {
  FS=OFS=","
}
{
  split($3, a, "/")
  $2=a[3] sprintf("%02d",a[1]) sprintf("%02d", a[2])

  split($NF, a, "/")
  $NF=a[3] sprintf("%02d", a[1]) sprintf("%02d", a[2])
  print
}

# 9  
Old 04-01-2008
That's nice Smilie
# 10  
Old 04-02-2008
hi

Quote:
Originally Posted by danmero
Just a touch-up
Code:
#cat char.awk 
BEGIN {
  FS=OFS=","
}
{
  split($3, a, "/")
   if (a[1]<10) a[1]=0a[1]
   if (a[2]<10) a[2]=0a[2]
  $3=a[3] a[1] a[2]

  split($NF, a, "/")
   if (a[1]<10) a[1]=0a[1]
   if (a[2]<10) a[2]=0a[2]
  $NF=a[3] a[1] a[2]
  print
}

file
E102,0,1/23/1994,0,1/9/1993,1/9/1993,A,6/4/1994,10450,60.288462
E104,0,6/4/1994,0,E001,E003,A,6/4/1994,7916.667,45.673077
1/23/1994,0,7/30/1993,0,6/4/1994,E003,A,87000,7250,41.826923
E106,0,1/9/1993,0,1/23/1994,E003,A,1/23/1994,3766.667,6/4/1994


i need the logic for converting a file to yyyymmdd formate...

I may have the date value 1 or 2 and 3 ,or contiounsly 45678 columns.......
it should convert into that formate.........the above code is manipluating at location 3 and last location.......

I may have field like effdate,birthdate,deathdate,hiredate,rehiredate in a file and i am not using constant file i am get a file with different date location and my code should identify the date pattern and according it should modify.............please let me know this........it is very urgent.....

Thanks in advance
# 11  
Old 04-02-2008
char.awk:
Code:
BEGIN {
  FS=OFS=","
  SEPslash="/"
}
{
  for(i=1; i<=NF; i++) {
     if (split($i, a, SEPslash) != 3 || match($i, "[^0-9/]"))
        continue;
     split($i, a, "/")
     $i=a[3] sprintf("%02d",a[1]) sprintf("%02d", a[2])
  }
  print
}

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