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

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.

Resources :

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