Tutorial: Flash Animation for Beginners
Content Aware Scaling in Photoshop CS4Populating a Combo-box in PHP Dynamically from MySQL
For this article, let’s pretend you have the following database table. There are two columns (I’m separating the fields with commas for readability), Item and Price:
Cheese Pizza, 1.00
Pepporoni Pizza, 1.50
Sausage Pizza, 1.50
Cheese Calzone, 1.50
Ham Calzone, 2.00
What if you want to populate a combo-box with your possible options? We can write php that generates a drop-down box that has five item choices in it, based on our table above. Here is the sample code:
<?php
// Connect to the database
mysql_connect("localhost", "user", "password") or die(mysql_error());
mysql_select_db("name") or die(mysql_error());
// Has the form been submitted?
if (isset($_POST['item'])) {
// The form has been submitted, query results
$queryitem = "SELECT * FROM table WHERE item = '".$_POST['item']."';";
// Successful query?
if($result = mysql_query($queryitem)) {
// More than 0 results returned?
if($success = mysql_num_rows($result) > 0) {
// For each result returned, display it
while ($row = mysql_fetch_array($result)) echo $row[serial];
}
// Otherwise, no results, tell user
else { echo "No results found."; }
}
// Error connecting? Tell user
else { echo "Failed to connect to database."; }
}
// The form has NOT been submitted, so show the form instead of results
else {
// Create the form, post to the same file
echo "<form method='post' action='example.php'>";
// Form a query to populate the combo-box
$queryitem = "SELECT DISTINCT item FROM table;";
// Successful query?
if($result = mysql_query($queryitem)) {
// If there are results returned, prepare combo-box
if($success = mysql_num_rows($result) > 0) {
// Start combo-box
echo "<select name='item'>n";
echo "<option>-- Select Item --</option>n";
// For each item in the results...
while ($row = mysql_fetch_array($result))
// Add a new option to the combo-box
echo "<option value='$row[item]'>$row[item]</option>n";
// End the combo-box
echo "</select>n";
}
// No results found in the database
else { echo "No results found."; }
}
// Error in the database
else { echo "Failed to connect to database."; }
// Add a submit button to the form
echo "<input type='submit' value='Submit' /></form>";
}
?>
March 31st, 2009 | Tags: code, combo-box, drop-down box, mysql, PHP
Posted in: iD Tech Bloggers
having problem with this it says failed to connect to database even though i have listed the database names i have database users and 2 tables Item and Price. then why its saying failed to connect to database user password is also correnct. help me out with this example.
One way to test if you are properly connecting to your database is to try the connect lines and leave everything else out:
mysql_connect(“localhost”, “user”, “password”) or die(mysql_error());
mysql_select_db(“name”) or die(mysql_error());
If you get an error message from just those two lines, then you know the error is related to the connection information you are supplying.
Hi!
Nice basic work BUT, what if you would like to have a link added to combo-box results? How would it then look like? I mean for each item in the combo-box.
thank you very much! works like a charm!
This is an easy to use code to select records from MySQL database table and display in dropdown combo box using PHP.
$cn=mysql_connect($db_host,$db_user,$db_password) or die(mysql_error());
mysql_select_db($db_name,$cn) or die(mysql_error());
$sql = “SELECT field_name FROM table_name”;
$rs = mysql_query($sql) or die(mysql_error());
echo “”;
while($row = mysql_fetch_array($rs)){
echo “”.$row["field_name"].”";
}mysql_free_result($rs);
echo “”;
Source:
http://phphelp.co/2012/05/10/how-to-fill-a-dropdown-combo-box-in-php-from-mysql-database-table/
OR
http://addr.pk/a8cf