RE: Help?
| From: | Don Read | Date: | Mon, 31 Jul 2000 19:35:32 +0000 |
| Subject: | RE: Help? | ||
| References: | 1 | Groups: | php.general |
| Request: | Send a blank email to php-general+get-9257@lists.php.net to get a copy of this message | ||
On 31-Jul-00 Wee Chua wrote:
> Hi everyone,
> I have a field on the form in HTML which is a Selection with 2 options.
> There are Company and Employee. For example,
> Select one: <Select Name="Game">
> <option value"Emp">Employee</option>
> <option value"Comp">Company</option>
> </Select>
>
> Now, I want to use the selection on the form in order to retrieve the data
> from such table in php. Assuming I have all the host, username and password
> are correct and connected, I also have chosen the database I want to run
> with. Here is the SQL statement with php. $result is a variable.
>
> $result = mysql_query("Select * from $Game",$db);
>
> I don't know how to pass the value from the <Select> to implement this SQL
> statement. I also gave me the error message " Supplied argument is not a
> valid MySQL result".
>
> Could anyone tell me how to pass the value from <Select> into SQL statement
> with php. Thanks!
add a bit of debuging ?
if (! isset($db))
echo "No db connection<BR>";
if (! isset($Game))
echo "No Game<BR>";
else
echo "using ".$Game."<BR>";
$result = mysql_query("Select * from $Game",$db) or die(mysql_error());
while ($row=mysql_fetch_result($result)) {
var_dump($row);
echo "<P>";
}
Regards,
--
Don Read dread@texas.net
-- "Stop telling God what to do" - Niels Bohr to A. Einstein