for i in {1..5} do echo $i dlr="$"$i str="{print $dlr}" echo $str awk '$str' <input_file> | while read value do echo $value if [ "$value" == "NULL" ] then echo "inside" cut $i fi done done for i in {1..5} do echo $i dlr="$"$i str="{print $dlr}" echo $str awk '$str' <input_file> | while read value do echo $value if [ "$value" == "NULL" ] then echo "inside" cut $i fi done done for i in {1..5} do echo $i dlr="$"$i str="{print $dlr}" echo $str awk '$str' <input_file> | while read value do echo $value if [ "$value" == "NULL" ] then echo "inside" cut $i fi done done 
How to not display a column that has all NULL values in a file
I have a file that has let's say, 5 columns(obtained by redirecting the results of a Sybase select query). Each column is separated by a tab. I need to filter out the columns that have all NULLs. Any column among the five can be null.
For example, if the columns in the file look like the below:
1000 NULL NULL 2 NULL 7 NULL 1000 2 NULL 7 NULL 1000 2 NULL 7 NULL 1000 2 NULL 7 NULL 1000 2 NULL 7 NULL 1000 2 NULL 7 NULL 1000 2 NULL 7 NULL 1000 2 NULL 7 NULL 1000 2 NULL 7 NULL 1000 2 NULL 7 NULL 1000 2 NULL 7 NULL 1000 2 NULL 7 NULL 1000 2 NULL 7 NULL 1000 2 NULL 7 NULL 1000 2 NULL 7 NULL 1000 2 NULL 7 NULL 1000 2 NULL 7 NULL 1000 2 NULL 7 NULL 1000 2 NULL 7 NULL 1000 2 NULL 7 NULL 1000 2 NULL The output must be(after removing columns 2 and 5), preferably in the same file:
1000 NULL 2 7 1000 2 7 1000 2 7 1000 2 7 1000 2 7 1000 2 7 1000 2 7 1000 2 7 1000 2 7 1000 2 7 1000 2 7 1000 2 7 1000 2 7 1000 2 7 1000 2 7 1000 2 7 1000 2 7 1000 2 7 1000 2 7 1000 2 7 1000 2 I reached:
for i in {1..5} do echo $i dlr="$"$i str="{print $dlr}" echo $str awk '$str' <input_file> | while read value do echo $value if [ "$value" == "NULL" ] then echo "inside" cut $i fi done done value right now, shows all the rows!!! Also, I am not sure how to cut off the column. I am a shell script beginner and unable to proceed further.
Can you please advise how to do this?
default