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How to Populate HTML Drop Down List From Mysql Table Using Php Function?

by Justine Dangelo (2020-02-29)

id="mod_47435238">Populating HTML drop down list instead of filling it with static values has many advantages such as-

<u>List grows/ shrinks automatically and dynamically as the new values are added/ removed.</u>

<u>List can be edited without affecting the code or <a href="">카지노</a> user interface.</u>

<b>Data in the list can be easily sorted.</b>
therefore creating a dynamic list from database is the most used technique in web development.

<u><b>The working of dynamic drop down list is illustrated in following figure-</b></u>

The working of application is very simple. The php object retrieves the value from MySql table and fills the html drop down list.

For creating dynamic drop down list, first we need to have a database table that actually holds our data. In this example I have created a very simple table called City with only two fields i.e. city id and city name. The city id is the primary key. The code for <a href="">카지노사ì´íŠ¸</a> table is following:

<u><strong>CREATE TABLE IF NOT EXISTS `city` (</strong></u>
`city_id` int(11) NOT NULL AUTO_INCREMENT,
`city_name` varchar(150) DEFAULT NULL,

<b>PRIMARY KEY (`city_id`)</b>
) ENGINE=InnoDB DEFAULT CHARSET=latin1 AUTO_INCREMENT=1; After the table is created, you need to input some records. You may input your own data or use following data that I am using in this example for testing purpose:

<b>INSERT INTO `city` (`city_name`) VALUES</b>
( 'Bhopal'),
( 'Delhi'),
( 'Kolkata'),
( 'Mumbai'),
( 'Pune'),
( 'Indore'),
( 'Jhansi'),
('Chennai'); Please note that I am not using city id in insert statement because it is auto generated primary key which the MySql itself takes care.

After our data is ready, <a href="">카지노사ì´íŠ¸ì£¼ì†Œ</a> we need to write code to extract data. For this I have created a PHP class with a method to fetch data from table. The code is given below:

<u>class MyClass</u>

<b>private $host = "localhost";</b>
<b> private $user = "root";</b>
<strong><u> private $password = "";</u></strong>
<b> private $database = "test";</b>

<u> private $con;</u>

<strong>function __construct()</strong>
<b> $this->con = $this->connectDB();</b>

<b>function connectDB()</b>
<strong> $con = mysqli_connect($this->host,$this->user,$this->password,$this->database);</strong>
<b> return $con;</b>

<u><b> function getData($query)</b></u>
<u><i> $result = mysqli_query($this->conn, $query);</i></u>
<b> $resultset[] = $row;</b>

<u><b> return $resultset;</b></u>

In the PHP class from above code I have defined two functions. The connectDB() functions connects to the database and returns a connection object. The second function is the getData() function which accepts one parameter and returns a result set. The parameter to this function is the select statement. The getData() function actually returns the array of data. This function is therefore an example of function that returns multiple values.

<u><strong>Copy the above code in the file MyClass.php or any name that you wold like.</strong></u>

The next step is to create a list box and populate it. The following code illustrates it. Create a PHP file and copy and <A HREF=''>카지노사ì´íŠ¸ì¿ í°</A> paste following code in it.

<u><strong> include "MyClass.php";</strong></u>

<u>!DOCTYPE html></u>

<u>$obj = new MyClass();</u>
<u> $row = $obj->getData("select city_name from city");</u>
?php foreach($row as $row) ?>
option>?php echo $row['city_name'] ?>/option>

?php ?>
/html> As in above code, first include <a href="">php file</a> that contain php class. After <a href="">including</a> file, create the object of class and call the function by passing select statement as parameter to fetch the record. Catch the data returned by function getData into variable row and loop through row to retrieve each record and add it to drop down list.

The drop down list displays the names of city. However when you use this drop down list in an input form that store data to another sql table, you will never want to store the names of city in the table because then you will create a de-normalized table structure. So instead of storing names of city you would want to store city id of associated city.

<b>Consider the following table structure:</b>

In this case the primary key of your city table would become foreign key of another table and the details are extracted using the primary key and foreign key. To associate city id which each city names use the following code.

<b> include "MyClass.php";</b>

<u><i>!DOCTYPE html></i></u>

<u>$obj = new MyClass();</u>
<strong><u> $row = $obj->getData("select city_id, city_name from city");</u></strong>
?php foreach($row as $row) ?>
option value='?php echo $row['city_id'] ?>'>?php echo $row['city_name'] ?>/option>

?php ?>
/html> In the above code I have only changed the select statement to add city id field and added value attribute in drop down list and bind it to city id field. So whenever user select a city name, the corresponding city id will be selected and when you submit the form the data of the value attribute will be submitted.

<b>Is this article helpful?</b>


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