Posted by kirke on 10/05/06 05:27

No, I want to receive the Enginetype(select name) as $Enginetype in the
same file.

then I can use $sql=' ~~~~ where EngineType=$Enginetype ';


cendrizzi wrote:
> Please be more specific. Do you mean you want to combine the files so
> it's posts to itself so you can show that the correct option is
> selected?
>
> On Oct 4, 8:12 pm, "kirke" <hinky...@gmail.com> wrote:
> > Hi. I'm a real beginner of PHP. So if my question is so stupid, plz
> > understand me. Thx.
> >
> > I made a file like following :
> >
> > <form id="form1" name="form1" method="post" action="step2.php">
> > <table >
> > <tr>
> > <select name="EngineType" id="EngineType">
> >
> > <?php
> > require('conf.php')
> > dbconnect();
> > $sql = 'SELECT * FROM dbo_EngineType;
> > $result = mysql_query($sql) or die();
> >
> > while( $row = mysql_fetch_array($result) )
> > {
> > echo '<option
> > value="'.$row['iID'].'">'.strtoupper($row['sDescription']).'</option>';
> > }
> > ?>
> > </select>
> > </tr> </table> </form>
> >
> > And another file have following code :
> >
> > <?php
> > $EngineID = (int)$_POST['EngineType'];
> > ?>
> >
> > At here, How can I combine both file? I mean i want to define $EngineID
> > as 'EngineType' selected value(option). I don't know how can i change
> > $_POST part.
> >
> > Thank you!

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