populate dropdown list with database value - php

Question

I want to populate a drop down list with data from a specific field in the database. Here is my sample code
<?php
$con = mysql_connect("localhost","root","");
if (!$con)
{
die('Could not connect: ' . mysql_error());
}
mysql_select_db("disertation ", $con);
$results = mysql_query("SELECT name FROM user_parent;");
?>
<select name="name">
<option value="name">Select one</option>
<?php
while($row=mysql_fetch_array($results))
{ echo '<option value=" ' . $row['name'] . ' ">' . $row['name'] . '</option>'; }
?>
</select>
It's currently displaying nothing from db, any help?

Answer 1

Try this, some white space in your code mysql_select_db("disertation", $con);
mysql_select_db("disertation", $con);
$results = mysql_query("SELECT name FROM user_parent") or die (mysql_error());

Answer 2

Make your mysql_fetch_array call read:
mysql_fetch_array($results, MYSQL_ASSOC)
Without MYSQL_ASSOC you can't refer to column names in $row.
Also, consider using MYSQLI or PDO. MYSQL is considerably outdated.

Answer 3

I would recommend you to use mysqli instead of mysql
<?php
$con=mysqli_connect("localhost","root","","disertation ");
// Check connection
if (mysqli_connect_errno()) {
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$resource= mysqli_query($con,"SELECT * FROM user_parent");
echo "<select class="name"><option value="name">Select one</option>";
while($result = mysqli_fetch_array($resource)){
echo '<option value="'.$result["name"].'">'.$result["name"].'</option>';
}
echo "</select>";
mysqli_close($con);
?>

Answer 4

To answer your question, directly, you should be first checking if there are any errors (mysql_error()) and then checking there are some results (mysql_num_rows) - these make for easier debugging of your code, since it will tell you what is wrong.
Try this;
<?php
// Connect with user
$con = mysql_connect("localhost","root","");
if (!$con)
{
die('Could not connect: ' . mysql_error());
}
// Select database
mysql_select_db("disertation", $con);
// Run query
$results = mysql_query("SELECT `name` FROM `user_parent`;") or die (mysql_error());
// Check for no results
if (mysql_num_rows($results) == 0)
{
echo 'There are no options for you to select.';
}
else
{
// If results, loop them.
// If the names are user input, make sure they're displayed in non-raw form
echo '<select name="name">
<option value="name">Select one</option>';
while($row = mysql_fetch_assoc($results))
{
$name = htmlentities($row['name'], ENT_QUOTES, "UTF-8");
echo '<option value=" ' . $name . ' ">' . $name . '</option>';
}
echo '</select>';
}
Will edit with a mysqli_ solution, if that is an option for you, since mysql_ is deprecated and will be dropped from PHP support, sooner or later.
MySQLi solution;
<?php
// Connect to database;
$mysqli = new mysqli("localhost", "my_user", "my_password", "data_base");
if (mysqli_connect_errno())
{
die("Connect failed: " . mysqli_connect_error());
}
$result = $mysqli->query("SELECT `name` FROM `user_parent`");
if ($result->num_rows > 0)
{
echo '<select name="name">
<option value="name">Select one</option>';
while($row = $result->fetch_assoc)
{
$name = htmlentities($row['name'], ENT_QUOTES, "UTF-8");
echo '<option value=" ' . $name . ' ">' . $name . '</option>';
}
echo '</select>';
$result->close();
}
else
{
echo 'There are no options for you to select.';
}