Question : Bash script to check a column for a specific match

I have the apache logs in the following format:

[Mon Jul 28 00:02:35 2008] [error] [client 99.99.99.99] File does not exist: /home/abc.xyz.com/http/1.gif", referer: http://abc.xyz.com/ggg1/p0002.aspx?p=21sub..,,,

I need to get rid of the "?p=21sub..." which for now  I am doing with the following command:

cut -d? -f1 file.txt

This might not be the most efficient way of doing so.. but it is working ok as long as the line doesn't have a "?" at any other place.

would it be possible to only go to the last field and cut that part

Like I say "if" the last field has a "?" cut everything after that because some lines doesn't have that even or some might have before the last field.

Answer : Bash script to check a column for a specific match

>what i want to do is to make sure that it should go to the last field only and then substract the part " ?p=.* "
sed 's/\( [^ \?]*\)?[^ ]*$/\1/'
This assumes space delimited file .... replace space with any delimiter that you may choose.

>Also, could you help me to code " how to check second-last field is " referer: "
awk ' { print $(NF-2) } '  this would get you second last field
Note that there is a space between referer and URL ... hence technically these are separate fields.