How to get the contents of a webpage in a shell variable?

Question

You can use wget command to download the page and read it into a variable as:

content=$(wget google.com -q -O -)
echo $content

We use the -O option of wget which allows us to specify the name of the file into which wget dumps the page contents. We specify - to get the dump onto standard output and collect that into the variable content. You can add the -q quiet option to turn off's wget output.

You can use the curl command for this aswell as:

content=$(curl -L google.com)
echo $content

We need to use the -L option as the page we are requesting might have moved. In which case we need to get the page from the new location. The -L or --location option helps us with this.

There are many ways to get a page from the command line... but it also depends if you want the code source or the page itself:

If you need the code source:

with curl:

curl $url

with wget:

wget -O - $url

but if you want to get what you can see with a browser, lynx can be useful:

lynx -dump $url

I think you can find so many solutions for this little problem, maybe you should read all man pages for those commands. And don't forget to replace $url by your URL :)

Good luck :)

There is the wget command or the curl.

You can now use the file you downloaded with wget. Or you can handle a stream with curl.

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Resources :

• linux.die - man wget
• linux.die - man curl

content=`wget -O - $url`

You can use curl or wget to retrieve the raw data, or you can use w3m -dump to have a nice text representation of a web page.

$ foo=$(w3m -dump http://www.example.com/); echo $foo
You have reached this web page by typing "example.com", "example.net","example.org" or "example.edu" into your web browser. These domain names are reserved for use in documentation and are not available for registration. See RFC 2606, Section 3.