>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 incomple
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.
>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
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
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
>
On Thu, Sep 30, 2010 at 9:43 PM, Brian Smither 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_com
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