Drop Down box to pull info from mysql

tazdog

I'm trying to have\create a drop down box that when someone selects the right option from the list and hit GO, that it will query mysql and and display the results on the same page.

Is this possible.?

thanks
Scott

btw - I'm fairly new to php\mysql but learning fast

vt_dave

i'm assuming you are creating the select box from html and not pulled database values, so i'll continue from the user selecting a value.

first you need to create your html form.
<form name="myform" action="/processform.php" method="post">

you can also specify method="get" and depending on which you choose you will have to pull the selected value from that array $POST or $GET

When you create your select box in the form you should specify a name for it as well.
<select name="userchoice"><option value="first">First Value</option></select>

Now in processform.php you can just pull the selected value from the appropriate array (GET or POST)

$user_value = $_POST['userchoice'];

Now you can insert that into a SQL statement as you need:

$q = "Select * from table where choice=".$user_value;

Hope that helps
-Dave

tazdog

Thanks Dave,

This will get me started.. 🙂

tazdog

I'm getting closer but not close enough, lol

Here is my code I have:

$user_value = $_POST['myform'];

			$dbcxn = mysql_connect("$host","$user","$pswd");
			if (!$dbcxn) die ("Could not connect MySQL");
			mysql_select_db($dbname,$dbcxn) or die ("Could not open database");

			$query = "SELECT * FROM destinations WHERE 'Destination_ID'='.$user_value'";
			$result = mysql_query($query) or die ("Query Failed");

And then below that code I have formatting for the results display.

So now when I try to execute this from the form, I goto the new page and I show the table row header but no results.. Do I have something out of place..?

Also when i run this query in sql {phpmyadmin}

SELECT * FROM destinations WHERE 'Destination_ID' = 'Domestic'

It will return results with no error..
But I have not been able to execute this on a standard php page..

thanks
Scott

TheBeginner

Try with that instead in the php code:

$query = "SELECT * FROM destinations WHERE Destination_ID='$user_value'";