rtadams89 Posted June 17, 2008 Share Posted June 17, 2008 I have a php file named "include.php" which contains the following code: <?php echo $_GET['name']; ?> If I go to http://myserver.com/include.php?name=ryan I get a page that contains the text "ryan". I also have another page named "main.php" which has the following code: <?php echo "The name is: "; include ("http://myserver.com/include.php?name=ryan"); echo "."; ?> I would expect the output of this page to be "The name is ryan.". However, I get nothing. I'm pretty sure this is because I am using the include() function incorrectly. What function should I use? Quote Link to comment Share on other sites More sharing options...
The Little Guy Posted June 17, 2008 Share Posted June 17, 2008 You don't need the ?name=ryan in the include. Quote Link to comment Share on other sites More sharing options...
bluejay002 Posted June 17, 2008 Share Posted June 17, 2008 <?php echo "The name is: "; include ("http://myserver.com/include.php?name=ryan"); echo "."; ?> I would expect the output of this page to be "The name is ryan.". However, I get nothing. include ("http://myserver.com/include.php?name=ryan"); does it really work fine? i haven't tried such feat. anyway, why would you do that for? i don't find any good reason for printing a name from another script while passing a parameter inside include() itself. what is it that you are trying to achieve, may i know? Quote Link to comment Share on other sites More sharing options...
rtadams89 Posted June 17, 2008 Author Share Posted June 17, 2008 You don't need the ?name=ryan in the include. I do. Without it, the include.php page won't have anything to process. does it really work fine? i haven't tried such feat. anyway, why would you do that for? i don't find any good reason for printing a name from another script while passing a parameter inside include() itself. It doesn't work fine, that's the problem. The code I posted is a simplified example of what I am trying to do. In reality, the include.php file actually takes an e-mail address as in put, converts it to hex code, and outputs it in the form of a JavaScript document.write command. This is all so that when a robot views the source code of a page, it won't be able to pick out the mailto links and spam them. Quote Link to comment Share on other sites More sharing options...
The Little Guy Posted June 17, 2008 Share Posted June 17, 2008 GET variables come from a URL. so go to: http://mysite.com/main.php?name=ryan then just include "include.php". you don't need the ?name=ryan in the include. Quote Link to comment Share on other sites More sharing options...
bluejay002 Posted June 17, 2008 Share Posted June 17, 2008 GET variables come from a URL. so go to: http://mysite.com/main.php?name=ryan then just include "include.php". you don't need the ?name=ryan in the include. yeah. since you got the name as get in the current page, why not create a function inside "include.php" that do such a feat. then just call it with that parameter and then returns whatever value you desire. Quote Link to comment Share on other sites More sharing options...
PFMaBiSmAd Posted June 17, 2008 Share Posted June 17, 2008 The posted code works (just tested) provided the php.ini settings allow_url_fopen and allow_url_include are on. Is there a reason you are including using a URL instead of directly through a file system path? However, I get nothing.If you are literally getting nothing (a blank page), your actual code probably has a fatal parse or a fatal runtime error. Are you developing and testing your code on a system that has full php error_reporting set (E_ALL) and display_errors turned on? Quote Link to comment Share on other sites More sharing options...
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