我无法在php mysql中插入所选下拉菜单的值

Following is select option to retrieve the data from mysql db in php. On selecting the dropdown, I need to insert option value to mysql db.

<select device_id='device_id'>;

<?php
require_once('sqlcon.php');

$drop=mysql_query("select * from device");

while($row = mysql_fetch_array($drop)){
    echo "<option value='" .$row['device_id']. "'>" . $row['device_id']. "</option>";   
}

?>
</select>

I am using the following in add.php . Error is showing only in this line.

$device_id=$_POST['device_id'];

Also check that the device_id is set or not.

isset() function in PHP determines whether a variable is set and is not NULL. It returns a Boolean value, that is, if the variable is set it will return true and if the variable value is null it will return false.

if(isset($_POST['device_id'])){ $device_id = $_POST['device_id']; }

your error says it all,your select box does not have a "name" attribute, you need to give it one..

<select name='device_id'>

undefined index: device_id in C:\...\ in add.php on line 55

It tells what is wrong, $_POST['device_id'] (Index = device_id) is not setted. Make sure you have <select name='device_id'>...</select>

name='device_id' = $_POST['device_id']

Use attribute onchange="document.forms[0].submit()" to submit the form on selected option.

Add hidden field in form.

<input type="hidden" name="action" value="submit" />

Get device_id

if(isset($_POST["action"])) {    
    echo $_POST['device_id'];
}