grep word from column and remove row - shell

Question

I want to grep a word from particular column from a file. Then remove those rows put all remaining rows into another file.
could anyone please help me on shell command to get following output?
I have a file with this format:
1234 8976 897561234 1234 678901234
5678 5678 123456789 4567 123456790
1234 1234 087664566 4567 678990000
1223 6586 212134344 8906 123456789
I want to grep word "1234" in the second column alone and removed those rows alone and put remaining rows in another file. So output should be in this format:
1234 8976 897561234 1234 678901234
5678 5678 123456789 4567 123456790
1223 6586 212134344 8906 123456789
The out should be with 3 rows except 3 row out of 4 rows.
while read value ;do
grep -v ${value:0:10} /tmp/lakshmi.txt > /tmp/output.txt
cp /tmp/output.txt /tmp/no_post1.txt
done < /tmp/priya.txt
Could you please help me to modify this script?

Answer 1

You can use awk for this, if that's good for you:
awk '$2==1234' <file-name>
$2 represents second column and it will return the line:
1234 1234 087664566 4567 678990000
Then you can use sed, grep -v or even awk for further process, either delete this line from current file, or print only the lines that do not match to another file. awk will be much easier and powerful.

Answer 2

Try the following regular expression.
egrep -v "^[[:space:]]*[^[:space:]]+[[:space:]]+1234[[:space:]]+.*$"
Not sure what your intention is, but my best guess is that you want to do the following.
while read value ;do
egrep -v "^[[:space:]]*[^[:space:]]+[[:space:]]+${value:0:10}[[:space:]]+.*$" /tmp/lakshmi.txt > /tmp/output.txt
cp /tmp/output.txt /tmp/no_post1.txt
done < /tmp/priya.txt

Answer 3

For columnar data, awk is often the best tool to use.
Superficially, if your input data is in priya.txt and you want the output in lakshmi.txt, then this would do the job:
awk '$2==1234 { next } { print }' priya.txt > lakshmi.txt
The first pattern detects 1234 (and also 01234 and 0001234) in column 2 and executes a next which skips the rest of the script. The rest of the script prints the input data; people often use 1 in place of { print }, which achieves the same effect less verbosely (or less clearly).
If you want the line(s) with 1234 in another file (filtered.out, say), then you'd use:
awk '$2==1234 { print > "filtered.out"; next } { print }' priya.txt > lakshmi.txt
If the column must be exactly 1234 rather than just numerically equal to 1234, then you would use a regx match instead:
awk '$2 ~ /^1234$/ { next } { print }' priya.txt > lakshmi.txt
The great thing about awk is that it splits the data into fields automatically, and that usually makes it easy to process columnar data with awk. You can also use Perl or Python or other similar scripting languages to do much the same job.

Answer 4

You did not specify the record layout exactly. When an empty first field is replaced by 4 spaces, the clever solutions will fail. Can a field have a space inside?
When your fields have fixed offsets, you might want to check the offset:
grep -v "^.\{9\}1234"
When /tmp/priya.txt has more than 1 line, your while loop becomes ugly:
cp /tmp/lakshmi.txt /tmp/output.txt
while read value ;do
grep -v "^.\{9\}${value}" /tmp/output.txt > /tmp/output2.txt
mv /tmp/output2.txt /tmp/output.txt
done < /tmp/priya.txt
You can also use the -f option of grep:
echo "1234 8976 897561234 1234 678901234
5678 5678 123456789 4567 123456790
1234 1234 087664566 4567 678990000
1223 6586 212134344 8906 123456789" |grep -vf <(sed 's/^/^.\\{9\\}/' /tmp/priya.txt )
or in your case
grep -vf <(sed 's/^/^.\\{9\\}/' /tmp/priya.txt ) /tmp/lakshmi.txt