How do I include another PHP file that is stored as a variable?

Question

I'm trying to load another page using the include statement but I keep getting the error:

Warning: include(AccountView.php?id=) [function.include]: failed to open stream: No such file or directory in /home/public_html/InsertAccount.php on line 62

$return = "AccountView.php?id=".$ID;
include $return;

I'm not sure what I'm missing to make this perform correctly. Another set of eyes would really be great.

Do you have a file called “AccountView.php?id=” on your server? — Waleed Khan
– Waleed Khan, Commented Feb 8, 2013 at 18:55
As an addendum to @Pekka웃's comment, when you include a PHP file, it will receive all variables in scope at which the include statement is executed. You could use your $ID variable that way. — ajp15243
– ajp15243, Commented Feb 8, 2013 at 18:56
Are you sure you want to do an include and not a redirect? — Sean McSomething
– Sean McSomething, Commented Feb 8, 2013 at 19:00

salezica · Accepted Answer · 2013-02-08 18:56:17Z

1

There's no file with this name:

AccountView.php?id=

The ?id= part is killing you. You're trying to include an URL there, not a filename.

answered Feb 8, 2013 at 18:56

salezica

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1 Comment

user2055499 Over a year ago

Thats correct, sorry. I am trying to include the url with the variable $return. Yes, the file exists and is functioning correctly. I'm trying to return to that page after performing an update to a table.

Kermit · Accepted Answer · 2013-02-08 18:56:56Z

1

Unless you have a file called AccountView.php?id=.php, this won't work.

What you want to do is set the variable ($ID), include your file and then use $ID in the included file:

$ID = '5';
$return = "AccountView.php";
include $return;

answered Feb 8, 2013 at 18:56

Kermit

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2 Comments

user2055499 Over a year ago

Thanks! How would I pass the variable $ID to the next program?

Kermit Over a year ago

@user2055499 It's within the scope of your parent script. You would just call it in AccountView.php

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