Extract data based on match against one column data from a long list data


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# 1  
Extract data based on match against one column data from a long list data

My input file:
data_5 Ali 422 2.00E-45 102/253 140/253 24
data_3 Abu 202 60.00E-45 12/23 140/23 28
data_1 Ahmad 256 7.00E-45 120/235 140/235 22
data_4 Aman 365 8.00E-45 15/65 140/65 20
data_10 Jones 869 9.00E-45 65/253 140/253 18
data_50 Pink 785 6.00E-45 12/253 140/253 16
data_6 Twins 133 4.00E-45 192/253 140/253 14
data_9 Orange 165 3.00E-45 150/253 140/253 12
data_7 King 258 1.00E-45 184/253 140/253 80

My output file:
data_50 Pink 785 6.00E-45 12/253 140/253 16
data_6 Twins 133 4.00E-45 192/253 140/253 14
data_9 Orange 165 3.00E-45 150/253 140/253 12
data_7 King 258 1.00E-45 184/253 140/253 80

I would like to extract all the content after it match the column 4 is 6.00E-45 or it other way extract the content after it find out the column 4 is 9.00E-45.
Does anybody got good suggestion about how to get this desired output result? I think awk should be able to do this?
Thanks a lot for sharing Smilie
# 2  
Code:
sed -rn '/([^ \t]* ){3}6.00E-45/,$p' t1

t1 is the file name.
# 3  
Hi, I just try the code that you suggested.
It is not worked Smilie
# 4  
Hi,
Try this ...
egrep '6.00E-45|9.00E-45' hi.txt

the hi.txt should contain your inputs.

Thanks
Sarwan
# 5  
thanks for your suggestion.
It can't work too Smilie
# 6  
how its not working for you? see the following output.

egrep '6.00E-45|9.00E-45' hi.txt
data_10 Jones 869 9.00E-45 65/253 140/253 18
data_50 Pink 785 6.00E-45 12/253 140/253 16
# 7  
Assumimg you want the content of the file after the 4th field matches the pattern:

Code:
awk '$4=="6.00E-45"{p=1}p' file

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