Php include a few php files...

Oni

Hi,

For some reason my small include script is not working.
I've never used the include function before...

<HTML>
<META HTTP-EQUIV="Refresh" CONTENT="40;URL=http://">
<?php include("http://www.b.com/page.php?site=http://www.google.com");?>
<?php include("http://www.b.com/page.php?site=http://www.google.com");?>
<?php include("http://www.b.com/page.php?site=http://www.google.com");?>
<?php include("http://www.b.com/page.php?site=http://www.google.com");?>
<?php include("http://www.b.com/page.php?site=http://www.google.com");?>
<?php include("http://www.b.com/page.php?site=http://www.google.com");?>
<?php include("http://www.b.com/page.php?site=http://www.google.com");?>
<?php include("http://www.b.com/page.php?site=http://www.google.com");?>

</HTML>

Maybe it doesnt work because of the variable set? If so... What can i do?

nemonoman

What in the world are you trying to do? It's hard to guess what that code is meant to do. So very hard to figure out why "it doesn't work".

Try telling us:
What you're trying to accomplish
The code you wrote to do the job
What you expected to happen that didn't happen.

Oni

When I goto "http://www.b.com/page.php?site=http://www.google.com" Something happens while the page loads.

How can I make the same thing happen when the file gets included?
All of the included files output data. That data doesn't get shown in the file that does all the including...

Hope that answers your question?

nemonoman

Sorry, I'm a mystified as before.

I will now GUESS about what you are trying to do.

My guess is that you somehow want to show the output of 'http://www.google.com' on the page being created by your PHP code?

'include' doesn't do what you are trying to do.
'[man]include[/man]' includes and evaluates a file -- it adds the file to the code base that PHP will interpret. 'http://www.google.com' is not going to include any interpretable code.

If you are using PHP 4.3 or greater, you can use this code:

echo file_get_contents("http://www.google.com");

to output the contents of http://www.google.com.

Oni

I tried but It didnt seem to work?

Goto:
http://www.basichyip.com/ptp/test.php?site=http://www.google.com

You will see the content of google.com. Not the source code or anything.
I would like the same content to be displayed on http://basichyip.com/ptp/all1.php Which containts the code that you gave me:


<?php
echo file_get_contents("http://www.basichyip.com/ptp/test.php?site=http://www.google.com");
?>

But when I goto that site, it shows up blank?
I dont just simply copy the code in in test.php because I will add more sites to all1.php.
And the PHP version is 4.4.1 I believe..
Thanks--

nemonoman

for me:

echo file_get_contents("http://www.basichyip.com/ptp/test.php?site=http://www.google.com");

produces the following output:

Google Personalized Home | Sign in Web Images Groups News Froogle Local more » Advanced Search Preferences Language ToolsAdvertising Programs - Business Solutions - About Google©2006 Google

so the code works. Please be more specific about your code (show all of it), and your problem.

Oni

All of my code for all1.php is

<?php
echo file_get_contents("http://www.basichyip.com/ptp/test.php?site=http://www.google.com");
?>

I'll get rid of the "php" after the <? but I don't think it will help?

but the site with that code: http://basichyip.com/ptp/all1.php shows up blank... I use firefox if that matters.

bpat1434

Nemonoman and Oni:

In reference to the remark that include() doesn't do what is expected (display output of google.com on page), it actually should. I quote the manual:

When a file is included, parsing drops out of PHP mode and into HTML mode at the beginning of the target file, and resumes again at the end.

Remote file may be processed at the remote server (depending on the file extension and the fact if the remote server runs PHP or not) but it still has to produce a valid PHP script because it will be processed at the local server. If the file from the remote server should be processed there and outputted only, readfile() is much better function to use.

// Works.
include 'http://www.example.com/file.php?foo=1&bar=2';

So including the URL to google should work. If not properly, try adding the fopen() wrapper and including then.

file_get_contents() should also work for getting the page requested. The difference is that file_get_contents() reads the file into a string, rather than outputting it where the statement is called.

If I understand correctly, you want to include the information from google.com, but while the page is loading, show something? Is that right?
If that's what you want, then I'd say you need to use file_get_contents with ob_start() and ob_flush().... Then you can display what you need as the file is read into a string.

I'm pretty sure you want Output Control. Ajax may even be able to help you here too....

nemonoman

http://dontworry.org/zzz.php

Oni

I can't get it to run??
I think it might be because of webhosts security but I dunno how this would compromise security...
I can do it using fopen but I want to be able to make it be a small line of code.

Like an include or filesgetcontents code but for some reason none of them seem to work...

My webhost admin said he would contact tech about it but I doubt that they would fix it if they had blocked it once.

I just want to run a few scripts by executing only one script... That's all 🙂

nemonoman

Thank Brett/Bpat:
everything you say is true!

nemonoman

http://us2.php.net/manual/en/ref.filesystem.php#ini.allow-url-fopen

This setting may not be enabled on your server.

Oni

Yup, not enabled 🙁

Well, I wont have to use that setting if I can get this piece of code to work.

$fh = fopen("http://www.site.com/url.php?user=Onish", "r");
while (!feof($fh)) :
print fgetss($fh, 1000);
endwhile;
fclose($fh);

$fh = fopen("http://site.com/page.php?userid=1097", "r");
while (!feof($fh)) :
print fgetss($fh, 1000);
endwhile;
fclose($fh);

And the list goes on and on.
I want to be able to access only one page and have the affect like I accessed other sites as well.

The variable repeats, so I don't think it works?
The script neads to show me that data that it has gathered, or it doesn't seem to work...
Can two fopen commands work in the same page together like this?

bpat1434

Reusing the same variable is no big deal. As long as you do it properly, you dhould nove nothing to worry about. Since the allow-fopen-wrappers are off, your code is going to have a few extra calls in it.

Can't say exactly why your code is not working, but I'd try writing all the contents to a variable, and echoing that variable at once. It could also have something to do with multiple <html></html> tags and such (Which would be a malformed document).

Oni

It seems as though the code works great, but only opens the sites on the top of the list.

I put a "set_time_limit(0);" command on the top of the code but it didn't seem to have any effect.

How to make it execute all? It's like a list of 15 sites all listed in the way shown on my post above.

Oni

I've even tried to put them all into one variable. Have the value of the variable change, then have the last changed value be printed.

Didn't seem to work either?!?

$fh = fopen("http://", "r");
while (!feof($fh)) :
$dt = fgetss($fh, 1000);
endwhile;
fclose($fh);

$fh = fopen("http://", "r");
while (!feof($fh)) :
$dt = fgetss($fh, 1000);
endwhile;
fclose($fh);

print $dt;

bpat1434

The code you just posted wouldn't work in the first place, at least, not to get them all. You'd need to append to the variable $dt, not re-write it.

I'm not sure why you can't do all the calls. It could be an execution time thing, it could be a limit from your server.

mvona

Here is some code I put togeather which does what you're trying to do. I am having problems getting it to format however. Would appreciate the help from GURU's out there. :p

<html>
<body>


<?php


function getPage($url)
{
$file = fopen($url,"r") or exit ("Unable to open file!");
echo '<TABLE>'."\n";
echo '<TD>'."\n";
echo $url." contains the following information"                                                                      ;

while (!feof($file))
{
echo fgetss($file, 2054,"<p>");
};
echo '</TD>'."\n";
echo '</table>'."\n";
fclose($file);
};

//define the site
$url = "http://www.MyFirstsite.com/";
//grab the visible text
getPage($url);
$url = "http://www.MySecondsite.com/services/";
getPage($url);
$url = "http://www.MyThirdSite.com/home.htm";
getPage($url);

?>

</body>
</html>

Oni wrote:
Yup, not enabled 🙁

Well, I wont have to use that setting if I can get this piece of code to work.
$fh = fopen("http://www.site.com/url.php?user=Onish", "r");
while (!feof($fh)) :
print fgetss($fh, 1000);
endwhile;
fclose($fh);

$fh = fopen("http://site.com/page.php?userid=1097", "r");
while (!feof($fh)) :
print fgetss($fh, 1000);
endwhile;
fclose($fh);
And the list goes on and on.
I want to be able to access only one page and have the affect like I accessed other sites as well.

The variable repeats, so I don't think it works?
The script neads to show me that data that it has gathered, or it doesn't seem to work...
Can two fopen commands work in the same page together like this?

bpat1434

What's the problem with formatting? What is it you're trying to do exactly with the format of it? Can we see an example of what you have now to compare it to?

mvona

I actually figured out the formatting issues. I was trying to use /n to create a new line but actually needed to embed </br> in double quotes.