Implement data using query - php

Question

I want to create a php page that contains a html drop down list of people's names (as the option
text) and then their age (as the value). Below is my form for you to see (almost what I mean) which I hard coded:
<form>
<select name="nameOption">
<option value="">Select your name:</option>
<option value="45">Mary Smith</option>
<option value="16">Lily Roe</option>
<option value="32">Elliot Perkins</option>
</select>
<p><input type="submit" name="submit" value="Submit"/>
<input type="reset" value="Reset" />
</form>
What I want to do (or been trying to do) is to create the drop down list by running a SQL query to obtain the data (the people's name and age) from my database (unlike what I written above) and then when I click on one of the options, only their value or age should appear. So basically, I need to implement the data from the database into a drop down list
Now it's here where I am stuck. I am familiar with writing SQL statements for tables but I seem to get puzzled when I try to create a SQL statement for a drop down list in a php tag.
How would I write it? Like:
$sql = "SELECT name, age FROM person WHERE name = ". $person. ";
or
$nameOption = $_POST['nameOption'];
print_r ($nameOption);
with selecting a database:
$conn = mysql_connect("localhost", " ", " ");
mysql_select_db(" ", $conn)
I know it may seem like a dull answer but I need help. How would I implement SQL query to a drop down list? I would love your help.

Answer 1

As you have to enclose string in quotes, change your query to
$sql = "SELECT name, age FROM person WHERE name = '$person'";
and for showing dropdown dynamically you can do like
$query=mysql_query($sql);
echo '<select name="nameOption">
<option value="">Select your name:</option>';
while($result=mysql_fetch_array($query))
{
echo '<option value="'.$result['age'].'">'.$result['name'].'</option>';
}
echo '</select>';

Answer 2

You should do like that
<select name="nameOption">
<?php
$query = mysql_query($sql);
while( $row = mysql_fetch_array($query) ) {
echo '<option value="'.$row['age'].'">"'.$row['name'].'"</option>';
}
?>
</select>

Answer 3

You need to get the full list of people from the database first, then iterate through that outputting each option tag for each row:
$cnx = new Mysqli('localhost', 'username', 'password', 'database_name');
$people = $cnx->query("SELECT name, age FROM person");
echo '<select name="nameOption">';
while($person = $people->fetch_assoc())
{
echo '<option value="' . $person['age'] . '">' . $person['name'] . '</option>';
}
echo '</select>';