[PHP] Using a variable for include statement
Sorry if this is a noob question, I have used PERL, TCL and VB but I am just
getting into PHP. If there is a better place to ask noobie questions then
let me know.
I want to use the include statement but I want to pass the name of the file
rather than hard code it. It seems that PHP needs the file to be in its
defined PATH in order to do this (or am I wrong?).
My code works OK if I use hard coding
Example (works)
include ('somefile.php')
Example (doesn't work)
$TheFile = somefile.php
include ($TheFile)
Am I trying to do something that is impossible?
If it is a path problem then how do I get around this, I can control my
local server config but I only have a local server for development.
Thanks,
Jeff
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Re: [PHP] Using a variable for include statement
On Tue, 2007-09-11 at 14:00 -0300, Jeff Benetti wrote:
Sorry if this is a noob question, I have used PERL, TCL and VB but I am just
getting into PHP. If there is a better place to ask noobie questions then
let me know.
I want to use the include statement but I want to pass the name of the file
rather than hard code it. It seems that PHP needs the file to be in its
defined PATH in order to do this (or am I wrong?).
My code works OK if I use hard coding
Example (works)
include ('somefile.php')
Example (doesn't work)
$TheFile = somefile.php
include ($TheFile)
Am I trying to do something that is impossible?
If it is a path problem then how do I get around this, I can control my
local server config but I only have a local server for development.
If the first works, then the second will work unless you've trimmed out
some useful information for us to see... or maybe it's the lack of a
semi-colon after the assignment (but that's probably just a quick typo
mistake).
Cheers,
Rob.
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Re: [PHP] Using a variable for include statement
you can definately use variables for filenames. i do it all the time.
it's not like XSLT or other languages where it has to be included at
the start or you're screwed.
On 9/11/07, Jeff Benetti [EMAIL PROTECTED] wrote:
Sorry if this is a noob question, I have used PERL, TCL and VB but I am just
getting into PHP. If there is a better place to ask noobie questions then
let me know.
I want to use the include statement but I want to pass the name of the file
rather than hard code it. It seems that PHP needs the file to be in its
defined PATH in order to do this (or am I wrong?).
My code works OK if I use hard coding
Example (works)
include ('somefile.php')
Example (doesn't work)
$TheFile = somefile.php
include ($TheFile)
Am I trying to do something that is impossible?
If it is a path problem then how do I get around this, I can control my
local server config but I only have a local server for development.
Thanks,
Jeff
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To unsubscribe, visit: http://www.php.net/unsub.php
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Re: [PHP] Using a variable for include statement
Jeff Benetti wrote:
Sorry if this is a noob question, I have used PERL, TCL and VB but I am just
getting into PHP. If there is a better place to ask noobie questions then
let me know.
I want to use the include statement but I want to pass the name of the file
rather than hard code it. It seems that PHP needs the file to be in its
defined PATH in order to do this (or am I wrong?).
My code works OK if I use hard coding
Example (works)
include ('somefile.php')
Example (doesn't work)
$TheFile = somefile.php
include ($TheFile)
Am I trying to do something that is impossible?
If it is a path problem then how do I get around this, I can control my
local server config but I only have a local server for development.
Thanks,
Jeff
Jeff, there are a couple ways you can do this.
1) include(/path/to/.$TheFile./);
2) include $TheFile;
Notice the lack of brackets in the last one. Either way has worked for me.
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