Shell Programming and Scripting

Hi,

I am trying to obtain the HTML code of a PHP page (http:// areferee .com/soccer/test1.php?quiz=50&ran=1&t=5) after the page has been processed; I want pure HTML. Is there a Unix command I can use to do this? I have tried wget, GET, curl but I am getting odd behaviour i.e. it is not completing successfully.
To do this manually, I open this webpage in IE, right-click, View Source and voila I have it. In Chrome and Firefox, it shows me the source code of the page with PHP code.

Is there an alternative?

Do you have lwp-download?

Code:

lwp-download "http://areferee.com/soccer/test1.php?quiz=50&ran=1&t=5" download.html

That's perfect! Thanks.

It's probably due to the & in the URL. The shell will treat it as separate commands, wget LAWS OF THE GAME TEST and ran=1 and t=5. If you want the shell to not split on & like that, you must put the string in double-quotes " ".

IOW, I think it's the quotes that fixed it, not lwp-download...

Any idea what is wrong with this command in bash?

`lwp-download "http://areferee.com/soccer/test.php?quiz=50&ran=1&t=1" test.html`

I'm sure I'm not escaping or something of the sort.

Why are you using back-ticks?

if you just want to run the command, then:-

Code:

lwp-download "http://areferee.com/soccer/test1.php?quiz=50&ran=1&t=5" test.html

Or if you want to get the downloaded size, then:-

Code:

d_size=$( lwp-download "http://areferee.com/soccer/test1.php?quiz=50&ran=1&t=5" test.html )

You were right about the quotes! Silly mistake...

---------- Post updated at 05:28 PM ---------- Previous update was at 05:27 PM ----------

I'm trying to run it in a loop in a bash script so I can iterate the querystring parameter t.