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awk to select rows based on condition on column

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# 1  
Old 11-04-2009
Question awk to select rows based on condition on column
I have got a file like this
003ABC00281020091005000100042.810001
003ABC01281020091005000100042.810001
003DEF00281020091005000100044.180001
003DEF01281020091005000100044.180001
003GHI00281020091005000100046.130001
003GHI01281020091005000100046.130001
003DKK00281020091005000100009.370001

i want output like this

003ABC00281020091005000100042.810001

003DEF00281020091005000100044.180001

003GHI00281020091005000100046.130001
003DKK00281020091005000100009.370001
i mean i want to select rows only when 8th character of the line is 0... Is it possible to do it using awk...pls help...
(pls ignore the space between lines Smilie
# 2  
Old 11-04-2009
try this ...

Code:
awk '{chaine=substr($0,8,1);if (chaine==0) print $0}' file

# 3  
Old 11-04-2009
If you need keep the line position:

Code:
$ awk '{if (substr($0,8,1)==0) {print $0} else {print ""} }' urfile
003ABC00281020091005000100042.810001

003DEF00281020091005000100044.180001

003GHI00281020091005000100046.130001

003DKK00281020091005000100009.370001

# 4  
Old 11-04-2009
Code:
nawk ' (substr($0,8,1) == "0"){printf("%s\n\n", $0)} ' file

Last edited by steadyonabix; 11-04-2009 at 05:39 AM..
# 5  
Old 11-04-2009
Try to grep
Code:
grep -E "^.{7}0" inFile

# 6  
Old 11-04-2009
Try this one to
Code:
cat yourfile | awk -F "" '{ if ($8==0)  print $0 }'

Where yourfile is the input file
# 7  
Old 11-04-2009
Quote:
Originally Posted by oky
Try this one to
Code:
cat yourfile | awk -F "" '{ if ($8==0)  print $0 }'

Where yourfile is the input file
Avoid the usage of "cat" here.

Code:
awk -F "" '{ if ($8==0)  print $0 }' yourfile
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