How do I access my script from another site ?

[jeddiki]

Hi,
I have my image generator on one website and I want to use
the script on my second website. This scripts tracks the image
downloads to a database so I want to keep the scripts on that server.

Now I don't want to directly refer to the script of the other site in the HTML, so I am trying to access it via an intermediate php page,
so I tried this:

On my second website:

echo "<img src=\"http://www.second_site.com/sys/image_gen.php?url_id=$Db_prod\" 
width='300px' height='200px' 
alt='Thumbnail for $title_sht'  border='2'>";

And then, still on second website, in the image_gen.php

I just have :

if(isset($_GET['url_cd'])){ 
  $prod  = $_GET['url_cd'];
    <img src=\"http://first_web_site.com/gen_the_image.php?id=$prod\">
    }
?>

But this "image redirect" doesn't work

Probably because I am doubling up on the <img src> tags but I don't
know how to do this.

Any ideas on how to make it work ?

[Markus]

Try

echo file_get_contents('http://firstsite');
 

[jeddiki]

That's pretty good, but can I pass the url_cd variable over
with that method ?

I need to pass the ?id=$prod so that the
correct image gets generated.

( the prod code is the url_cd so it is the
same as the ?url_id=$Db_prod

sorry for mixing up the names a bit ! )

[kovik]

Well, if you know that the other page outputs an <img> tag, then why are you wrapping it in an <img> tag? The tag is already made for you. Either only use that tag, or change the script to only output the "src" attribute.

[Markus]

file_get_contents('http://secondsite/?var=1&var2=2')

Edit: Would a header() redirect work?

header("Location: http://secondsite/?var=1&var2=2");

[jeddiki]

Well, if you know that the other page outputs an <img> tag, then why are you wrapping it in an <img> tag? The tag is already made for you. Either only use that tag, or change the script to only output the "src" attribute.

Yes ... but How ?

I tried the following:

<img src=\"http://www.SECOND-SITE.com/sys/webthumb.php?prod_id=$Db_prod\" 
width='300px' height='200px' alt='Thumbnail for $title_sht'>

AND

if(isset($_GET['prod_id'])){ 
  $$Db_prod  = $_GET['prod_id'];
    file_get_contents('http://www.FIRST-SITE.com/image-gen.php?id=$Db_prod');
    }

ALSO

if(isset($_GET['prod_id'])){ 
  $$Db_prod  = $_GET['prod_id'];
    file_get_contents('http://www.FIRST-SITE.com/image-gen.php?id=$Db_prod', FILE_BINARY);
    }

But none of them work :(

Any other suggestions ?



.

[kovik]

file_get_contents returns a string.. You have to echo it. Also, if you are going to inject a variable directly into a string, you need to use double-quotation marks ("), not single-quotation marks ('). Single quotation marks don't parse variables.

[jeddiki]

Thanks for your input,

So I tried this:

if(isset($_GET['prod_id'])){ 
   $$Db_prod  = $_GET['prod_id'];
   echo "file_get_contents(\"http://www.FIRST-SITE.com/image-gen.php?id=$Db_prod\", FILE_BINARY)";
    }

and this:

if(isset($_GET['prod_id'])){ 
   $$Db_prod  = $_GET['prod_id'];
   echo "file_get_contents(\"http://www.FIRST-SITE.com/image-gen.php?id=$Db_prod\")";
    }

But unfortunately neither worked.

[kovik]

...

echo file_get_contents("http://www.FIRST-SITE.com/image-gen.php?id=$Db_prod");

[jeddiki]

Ahh -
now that is more interesting !

OK - I got it to work - thanks

[kovik]

Change this line:

$$Db_prod  = $_GET['prod_id'];

To this:

$Db_prod = $_GET['prod_id'];

Even if it doesn't help, it's likely what you meant to do. Using 2 dollar signs is very different than 1.

[realin]

That should work for you .

http://www.digimantra.com/technology...sing-curl-php/

You can pass the query string in the URL as well.

Thanks
Realin !