What does the following SED pattern exactly do? - regex

Question

I am working on a CGI script and the developer who worked on this before me has used a SED Pattern.
COMMAND=`echo "$QUERY_STRING" | sed -n 's/^.*com_tex=\([^&]*\).*$/\1/p' | sed "s/%20/ /g"`
Here com_tex is the name of the text box in HTML.
What this line does is it takes a value form the HTML text box and assigns it to a SHELL variable. The SED pattern is apparently (not sure) necessary to extract the value from HTML without the other unnecessary accompanying stuff.
I will also mention the issue what I am asking this. The same pattern is used for a text area where I am entering a command and I need it retrieved exactly as it is. However it's getting jumbled up. Eg. IF I enter the following command in text box:
/usr/bin/free -m >> /home/admin/memlog.txt
The value that gets stored in the variable is:
%2Fusr%2Fbin%2Ffree+-m+%3E%3E+%2Fhome%2Fadmin%2Fmemlog.txt
All of us can get that / is being substituted by %2F, a space by + and the > sign by %3E.
But I just can not figure how this is specified in the above pattern! Will someone please tell me how that pattern works or what pattern should I substitute there so that I would get my entered command instead of the output I am getting?

Answer 1

sed -n
-n switch means "Dont print"
's/
s is for substitutions, / is a delimiter so the command looks like
s/Thing to sub/subsitution/optional extra command
^.*com_tex=
^ means the start of the line
.* means match 0 or more of any character
So it will match the longest string from the start of the line up to com_tex=
\(\)
This is a capture group, whatever is matched inside these brackets is saved and can be used later
[^&]*
[^] When the hat is used inside square brackets it means do not match any characters inside the brackets
* The same as before means 0 or more matches
The capture group combined with this means capture any character except &.
.*$
The same as the first bit except $ means the end of the line, so this matches everything until the end
/\1/p'
After the second / is the substitution. \1 is the capture group from before, so this will substitute everything we matched in the first part(the whole line) with the capture group.
p means print, this must be explicitly stated as the -n switch was used and will prevent other lines from being printed.
|
PIPE
s/%20/ /g
Sub %20 for a space, g means global so do it for every match on the line
HTH :)

Answer 2

This is not performed by any of the patterns. My best guess is that this escaping is performed by the shell or whatever fetches the HTML.
I will try to explain the patterns a little at a time
sed -n
-n specifies that sed should not print out the text to be matched, ie the html, after applying the commands.
The command following is of the form 's/regexp/replacement/flags'
^.*com_tex=\([^&]*\).*$
^ matches the beginning of the line
.* matches zero to many of any character
com_tex= matches the characters literally
\([^&]*\) '\(' specifies the beginning of a group that can later be backreferenced via its index. '[^&]*' matches zero to many characters which are not '&'. '\)' specifies the end of the group.
.* See above
$ matches the end of the line
\1
The above replacement is a backreference to the first (and only) group in the regexp i.e. '[^&]*'. So the replacement replaces the entire line with all characters immediately following 'com_tex=' till the first '&'.
The p flag specifies that if a substitution took place, the current line post substitution should be printed.
sed "s/%20/ /g"
The above is much simpler, it replaces all (not just the first) occurences of '%20' with a space ' '.