Re: Pager : force to display one page ID : SOLVED
| From: | Lorenzo Alberton | Date: | Tue, 21 Dec 2004 18:39:30 +0000 |
| Subject: | Re: Pager : force to display one page ID : SOLVED | ||
| References: | 1 | Groups: | php.pear.general |
| Request: | Send a blank email to pear-general+get-16376@lists.php.net to get a copy of this message | ||
Vincent DUPONT wrote:
Massimiliano, your answer was on the right direction, but we need to user $_REQUEST[] instead of $_GET. The Pager uses the $_REQUEST[$options['urlVar']] to select the page to open. I found this by going into the Pager files.even if that works, I don't reccomend using this method as I don't guarantee it will work in future releases
-----Original Message-----http://pear.php.net/manual/en/package.html.pager.getpagedata.php please notice the optional parameter... -- Lorenzo Alberton http://pear.php.net/user/quipoI would like to force Pager to display a particular page from PHP. By default, the first page is displayed. In the Pager Options, I found nothing like this. Actually, I need to be able to open the page that includes one specific record in the list. But this record may be on any page in the range. With getPageIdByOffset I can have the page ID where my record will be displayed. So I just need to be able to force Pager to open on that page, not on the first one.