|
|
Posted by ZeldorBlat on 10/13/52 11:42
Steff wrote:
> Hello every one,
>
> I need your help because I have no idea how to do this...
>
> how can I get the veg_id into the table veg_type according to the vegetable
> name
> chosen in the pick list ?
>
> do I have to create another query in test2.php which select the veg_id
> from the vegetable table ?
>
> I would like to no what's best, I mean what's professional?
>
> I created those two tables for my test:
>
> create table vegetables (
> veg_id number(5),
> veg_name varchar2(30));
>
> insert into vegetables values (1,'carot');
> insert into vegetables values (2,'spinach');
> insert into vegetables values (3,'potato');
> insert into vegetables values (4,'tomato');
> insert into vegetables values (5,'turnip');
>
> create table veg_type (
> type_id number(5));
>
> here is my first php
> page=====================================================
>
> <html>
>
> <head>
>
> <title>test insert</title>
>
> </head>
>
> <body>
>
> <p>test insert into db</p>
>
> <p> </p>
>
> <p> </p>
>
> <form method="POST" action="test.php">
>
> <p><input type="text" name="vegy_name" size="20"><input type="submit"
> value="insert" name="B1"><input type="reset" value="reset" name="B2"></p>
> </form>
>
> <p> </p>
>
> <form method="POST" action="test2.php">
>
> <p><select size="1" name="D1">
> <option selected>Vegy Name</option>
>
> <?php
>
> $c1=oci_connect("stephane","Stef1975",$bdtest05);
>
> $query="select veg_name from vegetables";
>
> $stmt=oci_parse($c1,$query);
>
> oci_execute($stmt);
>
> // here I get to chose what vegetable I want (it works fine)
>
> while ($row = oci_fetch_array($stmt,OCI_NUM))
> {
> print "<option>";
> echo $row[0];
> print "</option>";
> }
>
> oci_free_statement($stmt);
>
> //close connexion
> oci_close($c1);
> ?>
>
> </select><input type="submit" value="insert" name="b3"><input type="reset"
> value="Reset" name="B4"></p>
> </form>
>
> </body>
>
> </html>
>
> and here is my second php
> page========================================================
>
>
> <?php
>
> $veg= $_POST['D1'];
>
>
> $c1=oci_connect("stephane","Stef1975",$bdtest05);
>
> // I have a question mark here because I don't know how to do it
>
> $query="insert into veg_type (id_type) values (?)";
>
> $stmt=oci_parse($c1,$query);
>
> oci_execute($stmt);
>
>
> oci_free_statement($stmt);
>
>
> oci_close($c1);
>
> ?>
>
> // thanks for looking!
>
> //stephane
Change
$query="select veg_name from vegetables";
to the following:
$query = "select veg_id, veg_name from vegetables";
Then, when building your select list, your option tag should look
something like this:
echo '<option value="' . $row[0] . '">' . $row[1] . '</option>';
So now the user will see veg_name ($row[1]) in the drop-down list, but,
when the form is submitted D1 will have the coresponding veg_id
($row[0]) as its value.
Navigation:
[Reply to this message]
|