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  #1  
Old August 21st, 2002, 07:42 AM
billwj billwj is offline
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parameter transfer in php script

I tried the following way to call another php script:
include "http://.../script1.php?password='i'";

the script1.php is loaded successfully but the parameter "i" is not passed.
Does anyone know the reason?

thanks

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  #2  
Old August 21st, 2002, 07:53 AM
EvilivE EvilivE is offline
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RE: parameter transfer in php script

& is signifigant in passing data. What is the password suppose to equal?

password=pswd

then script1.php will recv password set to pswd.

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Old August 21st, 2002, 08:51 AM
siteworkspro.com siteworkspro.com is offline
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RE: parameter transfer in php script

there is a easier way to pass a variable then that
$password = "'i'"&#65307";

include "http://.../script1.php";

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Old August 22nd, 2002, 05:23 PM
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RE: parameter transfer in php script

and u can make a function to do this, like:

php Code:
Original - php Code
  1.  
  2. function include2($file, $params) {
  3.   expand($params);
  4.   include($file);
  5. }
  6.  
  7. // and call it like this
  8.  
  9. include2("script1.php", array('password'=>'ijklmno...'));

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