I tried wget, but it simply downloaded webpage, its output wasn't piped to grep. Then I tried curl with example.com: curl -s https://example.com | grep "<a href=\"http" | grep "iana.org" | cut -d "\"" -f 2 Output: https://www.iana.org/domains/example Explanation: curl -s = quiet/silent, no progress meter/error messages grep "<a href=\"http" = grep lines with hyperlinks to URLs, quotation mark is escaped by backslash grep "iana.org" = grep lines with iana.org domain cut -d "\"" -f 2 = output only 2nd field in each grepped line, fields in grepped lines are delimited by quotation marks, quotation mark is escaped by backslash here too

The correct option is B

The correct option is B

should be B

the answer is B

Is the B.

Is the B.

wget | grep "< a href=\*http" | grep "site.com"

correct

Defo B

I prefer B. You will see whole list on your screen, just try it.

wget saves index.html to a file, curl prints out the screen requested web resource, and with commands concatenated with pipes give links inside the web page. The answer is B. You should try in a shell of a linux system to see directly

Probably B

Try this out and see that the answer is B.

By default wget will save page content to a file, so piping to grep will not work. Indeed wget can return page content to standard output, but it requires additional argument flag for that. Even if we accept that wget will return to standard output, the grep command will return only URLs that contain specific domain - not all URLs. Curl will return page to standard output, which can be piped to grep to list only URLs (href tag), and then strip the HTML tags to leave the URLs only

Answer is B. `curl` outputs to stdout which makes it suitable to pipe to grep. `wget` just saves to a file (unless you use flag `-O -`) Tested on Ubuntu 20.04 ``` $ curl -s http://example.com/ | grep '<a href' | cut -d"\"" -f2 https://www.iana.org/domains/example ```

wget | grep "< a href=\*http" | grep "site.com"