cascading dropdown menus

[yo123]

Hello everyone:
I need to make a dropdown menu that shows information depending on what the user chooses in the first one. I am using PHP and MySql.
I am new!

I looked for tutorials and I am trying to understand the code... however it is not clear enough. I will say what I understand so you can tell me where I am wrong.
[CODE=php]
<?PHP
require_once('c onnection.php') ;
require_once('o pen_db.php');

$query1=mysql_q uery("SELECT Route_From as Desde FROM route") or die(mysql_error ());
$query2=mysql_q uery("SELECT * FROM route WHERE Route_From=".$_ POST["Desde"]."&quot;") or die(mysql_error ());
?>

<form name="check" method="post">
<!-- First Drop Down starts here -->
<select name="select1" onchange="this. form.submit()">
<?PHP
while ($row = mysql_fetch_ass oc($query1))
{
echo ("<option>{$ row['Route_From']}</option>");
}
?>
</select>

<?PHP
if (isset($_POST["select1"]))
?>
<select name="select2">
<?PHP
while($row = mysql_fetch_ass oc($query2))
{
echo("<option>{ $row['Route_To']}</option>");
}
?>
</select>
</form>
[/CODE]
Ok, so what I understand is:
- I connect to MySql (that works fine)
- I create the queries
- I create the first menu and start filling it up and with the "onchange", when a user selects an option, what he chose is submitted.
- When the first option is submitted, this info is captured with:
<?PHP
if (isset($_POST["select1"]))
?>
- The the second query is done and the second menu is filled.

When I execute the page the error that is shown is the following:
You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '"' at line 1

I hope someone can help me!
Bye

[ak1dnar]

Replace this line and let me know the status.

[PHP]$query2=mysql_q uery('SELECT * FROM route WHERE Route_From='.$_ POST['Desde'].'') or die(mysql_error ());[/PHP]

[rohitsavalia]

Just replace your $query2 line with this :

$query2=mysql_q uery("SELECT * FROM route WHERE Route_From='".$ _POST["Desde"]."'") or die(mysql_error ());

[wish]

Hello,
I had faced same problem..if user select the item in <select> then the below got one textarea to display the description of item..what should i write the code..

from <select> i retrieve all the item from db and then i want based on the user select the item then show the description of the item that what user was selected..

Can some one guide me...

[ak1dnar]

wish,
After getting the result set row values print it in side the teaxarea.

[PHP]<textarea name="descripti on" cols="5" rows="5"><? echo $the_descriptio n ?></textarea>[/PHP]

[wish]

Thank you my friend...
One problem is settle but another problem prompt out..
because based on the above example, the <form name="change" method="post"> the action property is no use then when i click button to submit..it can not work..if i add action property in the form tag then when i select the item then it direct go to another page...then what should i write the code...

Can you guide me again..

[ak1dnar]

Thank you my friend...
One problem is settle but another problem prompt out..
because based on the above example, the <form name="change" method="post"> the action property is no use then when i click button to submit..it can not work..if i add action property in the form tag then when i select the item then it direct go to another page...then what should i write the code...

Can you guide me again..

Its better to post a new thread for this question,Its not a good idea to continue with this tread here after.because this is not related to the thread of yo123.