ID: 13802
User updated by: [EMAIL PROTECTED]
Reported By: [EMAIL PROTECTED]
Status: Bogus
Bug Type: Documentation problem
Operating System:
PHP Version: 4.0.6
New Comment:
I got it.
Previous Comments:
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[2001-10-23 14:53:32] [EMAIL PROTECTED]
To Robert: It means that the file is included there,
but the code from the file is not executed, because some
condition (eg. an if statement).
if ($foo) {
require "foo.php";
}
This code (as the manual says) will __always__
load in foo.php and will replace the require
statement with the contents of that file, but
this content is only executed if $foo is true.
Goba
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[2001-10-23 14:35:44] [EMAIL PROTECTED]
It is correct the way it is.
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[2001-10-23 14:21:19] [EMAIL PROTECTED]
Quoting the fifth paragraph of the English language version:
"Unlike include(), require() will always read in the target file, even if the line
it's on never executes. If you want to conditionally include a file, use include().
The conditional statement won't affect the require(). However, if the line on which
the require() occurs is not executed, neither will any of the code in the target file
be executed."
I think the last line is contradicting the first line :
FIRST LINE : "...require() will always read in the target file, even if the line it's
on never executes."
LAST LINE : "However, if the line on which the require() occurs is not executed,
neither will any of the code in the target file be executed."
This the same in the Dutch language version.
IMHO the last line should read :
"However, if the line on which the include() occurs is not executed, neither will any
of the code in the target file be executed."
Robert Waarde
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Edit this bug report at http://bugs.php.net/?id=13802&edit=1
