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grep and modify

 
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# 1  
Old 03-16-2010
grep and modify
if input1 has an expression nouse covert allcolumns in to hashes except 6th column.
ex: abc100 has no use in 3rd row and 4th row and became hashes.

BUT if input1 1st column keys-1st row has no words in 3rd column (ex: def200 (key) and 1st row has no words in 3rd column (no x/x or y/y or others) except nouse. make the whole line into has in this case.
on the other hand abc100 (key-1st row) has words in 3rd column and the lines exists.

Hope I didn't confuse you.
Thanx
Ruby

input1
Code:
abc100    abc    x/x    1    2    3
abc100    abd    y/y    2    3    4
abc100    abd    nouse    nouse    nouse    5
abc100    ade    nouse    nouse    nouse    6
def200    ddd    nouse    nouse    nouse    12
def200    def    nouse    nouse    nouse    23

output
Code:
abc100    abc    x/x    1    2    3
abc100    abd    y/y    2    3    4
#    #    #    #    #    5
#    #    #    #    #    6
#    #    #    #    #    #
#    #    #    #    #    #

# 2  
Old 03-17-2010
MySQL
See the following code:

Code:
sed -r '3,$s/([^ \t]* ){4}/#\t/g'  Input_file

Input file:
Code:
abc100    abc    x/x    1    2    3
abc100    abd    y/y    2    3    4
abc100    abd    nouse    nouse    nouse    5
abc100    ade    nouse    nouse    nouse    6
def200    ddd    nouse    nouse    nouse    12
def200    def    nouse    nouse    nouse    23

Output:
Code:
abc100    abc    x/x    1    2    3
abc100    abd    y/y    2    3    4
#       #       #       #       #       5
#       #       #       #       #       6
#       #       #       #       #       12
#       #       #       #       #       23

You are specifying except 6th column. but in your output last two lines are also having hash (#) for 6th column I am having some confusion on that. you can try with above code.
# 3  
Old 03-17-2010
haa
12 and 23 should be hashes because 1st def200 key doesn't have any values in 3rd column. On contrary 1st abc100 has x/x.
Last edited by ruby_sgp; 03-17-2010 at 03:50 AM..
 
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