On Tue, January 17, 2006 4:27 pm, Bing Du wrote:
> Hello everyone,
>
> Here is what I want to accomplish. Query the backend database and
> generate a page listing all the staff members in table format that has
> name, title, phone and office address. Staff name should be a link.
> Clicking the link should query the database again and use the
> information retrieved (e.g. projects, educations, etc.) to generate a
> page for each staff member. A staff list page can be generated pretty
> easily.
>
> Now my question is how to keep the staffID (although it is not shown
> on
> the staff list page, but is queried along with name, title, etc.) so
> that when a staff name is clicked, his/her staffID will then be used
> to
> query the DB for personal information.
>
> Anybody have any handy working examples to give me some hints?
if (isset($_REQUEST['staff_id'])){
$staff_id = (int) $_REQUEST['staff_id']; //crude but effective
filtering / validation
$query = "select * from staff where staff_id = $staff_id";
//next line is BAD CODE, for example purposes only:
$detailed_profile = mysql_query($query) or die(mysql_error());
$row = mysql_fetch_row($detailed_profile);
//this is not so pretty...
echo "<pre>"; var_dump($row); echo "</pre>\n";
}
else{
$query = "select staff_id, name from staff order by name";
//more bad code next line (see above):
$staff = mysql_query($query) or die(mysql_error());
while (list($staff_id, $name) = mysql_fetch_row($staff)){
echo "<a
href=\"$_SERVER[PHP_SELF]?staff_id=$staff_id\">$name</a><br />\n";
}
}
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