Search for a string and replace the searched string in the same position in samefile


 
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# 1  
Search for a string and replace the searched string in the same position in samefile

Hi All,

My requisite is to search for the string "0108"(which is the year and has come in the wrong year format) in a particular column say 4th column in a tab delimited file and then replace it with 2008(the correct year format) in the same position where 0108 was found in the same file..The issue is the last two digits in "0108" tells the year..The file may contain "0109","0110" and so on...i need to replace the correct year depending on the last two digits of the string..How do i go about??please help me....

eg

00000000 0 +00000000.00 2008/12/29 11:06:04 0108/12/29 10:05:37 00521363

i need to change this to

00000000 0 +00000000.00 2008/12/29 11:06:04 2008/12/29 10:05:37 00521363

Thanks in advance!!!
# 2  
Code:
awk '{sub(/^../,"20",$6); print}' infile > tmpfile
mv tmpfile infile

Replaces the 1st two didigts always in field 6 against "20". Write the changed output to tmpfile and then moves tmpfile to infile, overwriting it. Maybe this is an option for you.
# 3  
THANKS A MILLION ZAXXON!!!!SmilieSmilie
worked fine....!!!!
# 4  
awk '{sub(/^../,"20",$6); print}' infile > tmpfile

In the above code how does awk identify column no. 6? Based on which delimiter?
# 5  
The default delimiter in awk is space... as Zaxxon replied it searches for 6th column and replaces first two digits with 20...
# 6  
Thanks White...Can I specify a delimiter in the above command? Suppose my infile is a tab delimited file....This command generates tmpfile with delimiter as space. Can I also have the output file to be tab delimited?
# 7  
Code:
$ cat tab.txt
00000000 0 +00000000.00 2008/12/29 11:06:04 0108/12/29 10:05:37 00521363
 
$ awk '{sub(/^../,"20",$6); print}' OFS="\t" tab.txt
00000000        0       +00000000.00    2008/12/29      11:06:04        2008/12/29      10:05:37        00521363

$ cat samp.txt
00000000        0       +00000000.00    2008/12/29      11:06:04        0108/12/29

$ awk '{sub(/^../,"20",$6); print}' samp.txt
00000000 0 +00000000.00 2008/12/29 11:06:04 2008/12/29

$ awk '{sub(/^../,"20",$6); print}' OFS="\t" samp.txt
00000000        0       +00000000.00    2008/12/29      11:06:04        2008/12/29


Hope you can understand the above code with sample examples.....
 

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