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
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
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
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
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
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
6 matches
Mail list logo