Re: Passing variables from a primary web page to a web page that is two levels below
| From: | Ryan Gibson | Date: | Tue, 29 Apr 2003 08:41:25 +0000 |
| Subject: | Re: Passing variables from a primary web page to a web page that is two levels below | ||
| References: | 1 | Groups: | php.general |
| Request: | Send a blank email to php-general+get-145596@lists.php.net to get a copy of this message | ||
It depends on how your getting from page 2 to page 3, if your getting there
using a form then you gust need to put an hidden field in the form, like
you've done for page 3, (can't page 2 & 3 be just one page).
If your not posting the var from page two to page three then you can either:
Hold it in a session var
Hold it in a cookie
Put it in a query string and grab it from $_GET[]
save it to a file...
Etc.
It just depends on how your getting from page 2 to page 3
On 29/4/03 9:09 am, "Ron Allen" <allenro@firsthandcomm.com> wrote:
> This is on the first page starting the query.
>
> <select name="SEARCH" onChange="document.inventorysearch.submit()">
> <OPTION>SELECT
> <option>CAMP
> <option>LIN
> <option>NSN
> <option>SERIAL
> <option>DRAGON
> <option>STATUS
> </select>
>
> Second page
>
> <?PHP
> $SEARCH = $_POST["SEARCH"];
> include 'drop.php';
> $DBName = "******";
> $table = "*****";
> $db = mysql_connect("$DBhost","$DBuser","$DBpass") or
> die("Problem
> connecting");
> mysql_select_db("$DBName") or die("Problem selecting database");
> $query = "SELECT * FROM $table";
> $result = mysql_query($query) or die ("Query failed");
> //let's get the number of rows in our result so we can use it in a for loop
> $numofrows = mysql_num_rows($result);
> ?>
>
> Third page
>
> <input type="hidden" name="SEARCH" value="<? echo $SEARCH;
> ?>">
>
> I have tried $SEARCH = $_POST["SEARCH"]; on the third page as well, but that
> does not work.
>
> Any help would apprciated!
>
>