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Replace Date field in Unix File

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Top Forums Shell Programming and Scripting Replace Date field in Unix File
# 1  
Old 07-19-2010
Replace Date field in Unix File
I have a data file having first 19 characters having the date in the below format-

2010-04-29-00.00.00

I have to check this date field ( first 19 characters) against some specific dates and if date is not in 3 valid dates ( business date available to me , business date - 1 , businessdate -2 ) I have to replace with business date.

For example if business date is 2010-04-29-00.00.00

1) I need to check the date field ( first 19 characters) in the file
2) If the date is not from valid dates (2010-04-29-00.00.00 , 2010-04-28-00.00.00, 2010-04-27-00.00.00) I have to replace it with business date (2010-04-29-00.00.00)

Can somebody suggest the way to achieve this using awk or sed.
# 2  
Old 07-19-2010
You need provide the sample input file first.
# 3  
Old 07-20-2010
File ouput and problem further explained
The file looks like this-
Code:
2010-04-29-00.00.00TSCN00000171585               *         **   ADDR
2010-04-27-00.00.00TSCN00000171585               *         **   ADDR
2010-04-25-00.00.00TSCN00000171585               *         **   ADDR
2010-04-21-00.00.00TSCN00000171585               *         **   ADDR
2010-04-11-00.00.00TSCN00000171585               *         **   ADDR

The First 19 characters are the date field.

My program has got 3 possible valid dates. In the above file I want that if the date is not in

2010-04-29-00.00.00
2010-04-28-00.00.00
2010-04-27-00.00.00

it should be replaced with

2010-04-29-00.00.00


So the last 3 dates (first 19 characters in file) should be changed to
2010-04-29-00.00.00

and file should look like this

Code:
2010-04-29-00.00.00TSCN00000171585               *         **   ADDR
2010-04-27-00.00.00TSCN00000171585               *         **   ADDR
2010-04-29-00.00.00TSCN00000171585               *         **   ADDR
2010-04-29-00.00.00TSCN00000171585               *         **   ADDR
2010-04-19-00.00.00TSCN00000171585               *         **   ADDR

# 4  
Old 07-20-2010
Hi,
try this,
Code:
awk  -F'-00' '{if(/(2010-04-29|2010-04-28|2010-04-27)/)  {print $0}else {gsub($1,"2010-04-29",$0);print $0}}' infile

# 5  
Old 07-20-2010
Code:
awk  '! match($0,/^2010-04-2[7-9]-00.00.00/) {s=substr($0,20,length-19); $0="2010-04-29-00.00.00"s}1' urfile

# 6  
Old 07-21-2010
Thanks , further clarification required
Thanks for your replies. The awk command is working as standalone.

However my 3 valid dates are available in 3 variables-

var1=2010-04-29-00.00.00
var2=2010-04-28-00.00.00
var3=2010-04-27-00.00.00

I have to fit this into the command suggested by you-

Code:
awk  -F'-00' '{if(/(2010-04-29|2010-04-28|2010-04-27)/)  {print $0}else {gsub($1,"2010-04-29",$0);print $0}}'  file1

# 7  
Old 07-21-2010
Code:
var1=2010-04-29-00.00.00
var2=2010-04-28-00.00.00
var3=2010-04-27-00.00.00

awk  '! match($0,/'"$var1"'|'"$var2"'|'"$var3"'/) {s=substr($0,20,length-19); $0="2010-04-29-00.00.00"s}1' urfile

awk  -F'-00' '{if (/'"$var1"'|'"$var2"'|'"$var3"'/)  {print $0} else {gsub($1,"2010-04-29",$0);print $0}}' urfile
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