Populating DropDown List with PHP from MYSQL Database

Question

I'm trying to bring data from MYSQL Database called ebms_db. The table is events and the fields are Event_ID and Event_Name.

The code I'm using currently to show the Event_Name only in the dropdown list is:

    <select name="mySelect"> 
        <?php

            include 'db.php';

            $sql = "SELECT Event_Name FROM events";
            $result = mysql_query($sql);

            echo "<select name='Event_Name'>";
            while ($r = mysql_fetch_array($result)) {
                echo '<option value="'.$row["Event_Name"].'">'.$row["Event_Name"].'</option>';
            }
            echo "</select>";
            ?>

        <input type = "submit" name="Search" value="Search">
    </select>

Db.Php looks like this...

    $servername = "localhost";
    $username = "test";
    $password = "test";
    $dbname = "ebms_db";

    $conn = new mysqli($servername, "test", "test", $dbname);
    if ($conn->connect_error) {
        die("Connection failed: " . $conn->connect_error);
        echo "Error";
    }

What am I doing wrong?

The output shows a combo box with

 - '.$row["Events_Name"].'

inside it.

you use while ($r = mysql_fetch_array($result)) {... so instead of $row["Event_Name"] use $r["Event_Name"] — Goikiu
– Goikiu, Commented Dec 13, 2014 at 0:02
possible duplicate of Populate a PHP Dropdown List from MySQL Database — Nerdroid
– Nerdroid, Commented Dec 13, 2014 at 0:07
I am using $r["Event_Name"] ... I still get the same result. I changed $row to $r ... — IHS
– IHS, Commented Dec 13, 2014 at 1:21

iatboy · Accepted Answer · 2014-12-13 00:04:06Z

1

while ($row = mysql_fetch_array($result)) { 
       ^^^^ here
    echo '<option value="'.$row["Event_Name"].'">'.$row["Event_Name"].'</option>';
}

answered Dec 13, 2014 at 0:04

iatboy

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1 Comment

IHS Over a year ago

I changed it to $row in the while loop and in the echo statement. The result is still the same. I get a dropbox with $row["Event_Name"] this written inside it.

Rao Biz · Accepted Answer · 2015-02-20 05:44:34Z

1

   <?php

        include 'db.php';
       // you just fetching here (Event_Name) take all the values from the database or 
        $sql = "SELECT * FROM events";
        $result = mysql_query($sql);

        echo "<select name='Event_Name'>";
        while ($row = mysql_fetch_array($result)) {
            echo '<option value="'.$row["Event_Name"].'">'.$row["Event_Name"].'</option>';
        }
        echo "</select>";
        ?>

    <input type = "submit" name="Search" value="Search">

Why are you taking two selects? I just removed one select just beginning above the php tags.

answered Feb 20, 2015 at 5:44

Rao Biz

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Comments

user4343646 · Accepted Answer · 2014-12-13 00:13:08Z

0

MYSQL is deprecated, you should really use something else, if and/or when they remove it your website will be defunct.

answered Dec 13, 2014 at 0:13

user4343646

1 Comment

IHS Over a year ago

I started coding in PHP and HTML two days ago. So I'm not sure what else I should use. I've gone through many of the solutions on this website but none have helped me.

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