Re: [PHP] version_compare
On Thu, Sep 30, 2010 at 9:43 PM, Brian Smither bhsmit...@gmail.com wrote:
I found this code...
if (version_compare(PHP_VERSION, '5.2.0', '=')) {
It means condition (PHP_VERSION = 5.2.0)
$text=filter_var($text, FILTER_SANITIZE_URL);
}
...to be questionable.
Under what conditions would version_compare() return true, yet the
filter_var() be undefined? Because that's what is happening.
Thank you.
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Re: [PHP] version_compare
On Thu, Sep 30, 2010 at 09:43:22AM -0600, Brian Smither wrote:
I found this code...
if (version_compare(PHP_VERSION, '5.2.0', '=')) {
$text=filter_var($text, FILTER_SANITIZE_URL);
}
...to be questionable.
Under what conditions would version_compare() return true, yet the
filter_var() be undefined? Because that's what is happening.
If your PHP was compiled without support for the filter functions? I
assume when you say that filter_var() is undefined you mean that the
software returns a message that the function itself cannot be found?
Paul
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Re: [PHP] version_compare
Brian Smither wrote:
I found this code...
if (version_compare(PHP_VERSION, '5.2.0', '=')) {
$text=filter_var($text, FILTER_SANITIZE_URL);
}
...to be questionable.
Under what conditions would version_compare() return true, yet the
filter_var() be undefined? Because that's what is happening.
Thank you.
Personally, I would change that to be
if ( function_exists('filter_var') ) {
$text = filter_var($text, FILTER_SANITIZE_URL);
}
Jim
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Re: [PHP] version_compare
Personally, I would change that to be
if ( function_exists('filter_var') ) {
So would I:
*But it's not my code.
*I wish to learn and understand the cause of the problem - not walk around it.
It means condition (PHP_VERSION = 5.2.0)
I understand that. There was a second, more relevant, part to my question.
I have found the version of PHP used as reported in the HTTP header responses.
filter_var() is undefined in PHP/5.2.4_p20070914-pl2-gentoo
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Re: [PHP] version_compare
Brian Smither wrote:
Personally, I would change that to be
if ( function_exists('filter_var') ) {
So would I:
*But it's not my code.
*I wish to learn and understand the cause of the problem - not walk around it.
It means condition (PHP_VERSION = 5.2.0)
I understand that. There was a second, more relevant, part to my question.
I have found the version of PHP used as reported in the HTTP header responses.
filter_var() is undefined in PHP/5.2.4_p20070914-pl2-gentoo
As Paul pointed out, maybe your version of PHP was built without the filter_var
function compiled in.
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[PHP] filter_var (was: Re: [PHP] version_compare)
As Paul pointed out, maybe your version of PHP was built without the filter_var function compiled in. This is what I have learned about PHP with filter_var() as an illustrative point: Many people who provide elaborations on PHP make too many assumptions or are blatently and woefully incomplete. Statements such as PHP 5.2.0 or higher is required for this to work is misleading. The PHP online manual pages for respective functions make no distinction that any particular function or capability is an optional include for a build - considered an extension as opposed to, what?, native code? I have also learned that when coding my own scripts, I must not only read the manual page for that function, but also any metadata for it and its group, such as the Introduction page, Installing/Configuring, etc. And where, when I finally find it, if I can reason out that I should be looking for it, it may say The filter extension is enabled by default as of PHP 5.2.0, that I need to be reasonably cautious because some idiot may build PHP 5.2.0+ without this extension. Are the array_*() functions also an optional extension? -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
