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Posted by Jamie Alessio on 01/07/05 19:03
>> This doesn't prove the case either way.
>>
Nice catch on that one but actually there might be another problem with
my example. According to example 16-8 on http://us2.php.net/include/
"Because include() and require() are special language constructs, you
must enclose them within a statement block if it's inside a conditional
block". So, I think I would need to rewrite
if($include)
include('2.php');
to be
if($include) {
include('2.php');
}
in order for it "to work as desired". I'm not exactly sure what it means
to "work as desired" though. Maybe it only applies if you have
additional else/elseif statements in the block?
BUT, also from documentation on http://us2.php.net/include/ it looks
like your example might not hold any water either. "Be warned that parse
error in required file doesn't cause processing halting." Based on that
it appears that introducing the parse error in foo2.php doesn't prove
that the file wasn't included. So, how do we actually prove this one way
or the other? I guess we could just like at the underlying C code. Any
takers?
- Jamie
>[snip]
>You're wrong. The include() and require() statements are only evaluated
>when they are reached in your application code, so there is a big
>difference between your two examples. In you use the second example the
>code will only be included by PHP if the application logic enters the
>case statement that contains the include() statement. Here's a quick
>example to demonstrate this:
>
>////////////
>// 1.php
>///////////
><?php
>$include = 0;
>
>if($include)
> include('2.php');
>
>hello();
>
>?>
>
>////////////
>// 2.php
>///////////
><?
>function hello() {
> print "Hello";
>}
>?>
>
>Save those two snippets into files named 1.php and 2.php. Pull up 1.php
>in your browser and you'll see that the hello() function is undefined.
>[/snip]
>
>This doesn't prove the case either way. It would be similar to writing
>this code:
><?php
>$include = 0;
>
>if($include){
> function hello() {
> print "Hello";
> }
>}
>hello();
>
>?>
>
>Which also throws an error. We looked a while and found nothing
>in the documentation, so we ended up putting an error in the include
>file.
>foo.php
>
><?
>
>$include = "b";
>
>switch ($include){
> case "a":
> include("foo2.php");
> break;
> case "b":
> echo "bar";
> break;
>}
>
>?>
>
>foo2.php
><?
>//note syntax error
>echo "foo"
>?>
>
>Assumably, if includes were processed before the script was
>executed, it would show a syntax error in foo2.php. We ran
>the test and no error was returned, so that pretty much answers
>the question (READ: includes are not processed until they
>are called).
>Just out of curiosity, a lot of people had answers to this
>question, but I couldn't find a shred of evidence in the
>documentation. Did you do similar tests, hear from gurus, or
>something of like? Or did I just miss it somewhere?
>
>Regards,
>Jesse R. Castro
>
>
>
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