以下SED模式到底能做什么？

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 ie '[^&]*'. 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 ' '.