CODEWITHSUNDEEP
phpjquerymysql
GLRotterdam
GLRotterdam

Reputation: 143

How to show drop down selected values

I am trying to create a dropdown menu which has values from the database and when selected it should show all the information that are connected to the selected value.

In MySQL I have three tables, person, address and resume. I created two drop-down menu's(see code below) which are showing the values from address > address_state and address > address_city. What needs to happen is when I selected a state it will show only the persons living in the selected state. And when selecting city it has to show all the persons living in the same city.

My drop-down menu select state only:(got the same code city) 

$servername = "localhost";
$username = "root";
$password = "usbw";
$dbname = "persons";

// CREATE A CONNECTION WITH THE DATABASE
// CONNECTIE MAKEN MET DATABASE
$conn = new mysqli($servername, $username, $password, $dbname);
if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
} 

//$sql="SELECT name,id FROM student"; 

$sql="SELECT DISTINCT address_state FROM address ORDER BY address_state asc"; 

/* You can add order by clause to the sql statement if the names are to be displayed in alphabetical order */

echo "<select name=address value=''>Student Name</option>"; // list box select command

foreach ($conn->query($sql) as $row){//Array or records stored in $row

echo "<option value=>$row[address_state]</option>"; 

/* Option values are added by looping through the array */ 

}

echo "</select>";// Closing of list box
?>

I know you cannot provide any code because I don't have any code to show which actually "selects" the persons. But I have searched on Google for tutorials and I did not find one which did the thing I needed. So I hope you guys can help me or can provide any tutorials.

Upvotes: 0

Views: 97

Answers (2)

Tomasz Ferfecki
Tomasz Ferfecki

Reputation: 1263

First replace:

echo "<select name=address value=''>Student Name</option>";

To:

echo '<select name="address">';

Next replace this:

echo "<option value=>$row[address_state]</option>";

With: 

echo '<option value="'.$row[address_state].'">'.$row[address_state].'</option>';

Upvotes: 1

Amit Rajput
Amit Rajput

Reputation: 2059

Replace this line:

echo "<option value=>$row[address_state]</option>";

With 

echo "<option value='".$row['address_state']."'>".$row['address_state']."</option>";

Upvotes: 4

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