Drop down menus

steve55

I have a mysql database with three fields in it and I would like to create three drop down menus populated with data from each field in the database so that users can search the database.

The user can then select any combination of the three dropdowns and then hit a submit button and then a results page based on the users selections would be displayed...

I am new to php and can't find any resources on how to do this, even though such an search option seems to be on the home page of any database driven site..

Can anyone help me out...
Thank you.

DigitalExpl0it

id do this

connect to the database

use the Distenct option in your SQL statement

the echo the contents to the drop down menu.

// Database Connection String ------Start------
$connect = mysql_connect("your_host", "your_Database_username", "your_database_password") or die ("Couldnt connect");
$db = mysql_select_db("your_database_name", $connect) or die ("Couldnt open Database");
$sql = "select DISTENCT from your_table where your_filed='your_search_string'";
$sql_result = mysql_query($account_sql, $connect) or die ("SQL Statement Error");
$count = mysql_num_rows($sql_result);
// Database Connection String -------End-------

//List
<select name="select">
<option Selected> </option>
// Get Record From Database -------Start-------
while ($row = mysql_fetch_array($sql_result)){
$field1 = $row["your_field_here"];
//Time for loop

echo "<option>$field1</option>";
}
</select>
// Get Record From Database -------End-------
// Close Datbase Connection
mysql_close($connect);

Thora_Fan

A useful solution above, but this is how I would do it. Just a matter of personal coding preferance.

$db = mysql_connect("localhost", "username", "password");
mysql_select_db("databasename", $db);
$SQL = "SELECT fieldOne, fieldTwo, fieldThree FROM yourTable";
$myResult = mysql_query($SQL, $db);
while(list($fieldOne, $fieldTwo, $fieldThree) = mysql_fetch_array($myResult))
{
	$menuOne .= "<option value=\"$fieldOne\">$fieldOne</option>\r";
	$menuTwo .= "<option value=\"$fieldTwo\">$fieldTwo</option>\r";
	$menuThree .= "<option value=\"$fieldThree\">$fieldThree</option>\r";
}
mysql_close($db);

echo "<form action=\"WHERE YOU WANT THIS TO GO\" method=\"GET\">\r";
echo "<select name=\"fieldOne\">$menuOne</select>\r";
echo "<select name=\"fieldTwo\">$menuTwo</select>\r";
echo "<select name=\"fieldThree\">$menuThree</select>\r";
echo "<input type=\"Submit\" name=\"Submit\" value=\"Submit\">\r";
echo "</form>\r";

Make sure, obviously, that you change the database username/password, the field names in the SQL query, the form action, and the names in the <select> tags.

steve55

The last solution works (thank you), but how do I pass a value to the results page..
For example my results page would need to start with something like this:-

if ($search) // perform search only if a string was entered.

and then select the relevant information from the database..

As I am passing 3 values to the results page I am unsure what value I need to put in my 'if' statement.............

RedDragon

enclose the html by the form tag:

<form method="post" action="file.php?sub=do_this">
//html...
</form>

then use
if($submit=="Submit") {
if($fieldone) {
if($fieldone=="Delete") {
//...
}
}
if($fieldtwo) {
// do something with it
}
}

steve55

still a bit confused (sorry)..