How to show selected user input in my HTML form? - php

Question

I have a PHP form which saves user data to a MySQL database.
I want to show the user which information is held about them and display it in a form in order for them to update or edit the values.
I have a problem in getting the user's saved data from the database in a PHP loop and show that to user in order for them to update or edit it.
Below is the piece of code:
<?php
$id = $_GET['id'];
$conn = mysqli_connect('localhost', 'phpmyadmin', 'Test#2000', 'user');
$sql1 = "SELECT * FROM usr WHERE id='$id'";
$result = mysqli_query($conn, $sql1);
$row = mysqli_fetch_assoc($result);
?>
<fieldset><label>Birthday</label>
<select name="birthday">
<?php
for ($i = 1300; $i <= 1397; $i++) {
echo "<option >$i</option>";
}
?>
</select>
<fieldset>
<button name="btn" type="submit" id="contact-submit">Submit</button>
</fieldset>
</form>
</div>
</body>
</html>
I want to show into the form's Select input the birthday value that user selected originally, in order to edit or update by user.

Answer 1

The <select> children elements <option> does support a selected tag to indicate that it was the selected value, so by adding the selected tag like so <option value='1' selected> you can have that as the selected value.
You'll also want to probably add the $i value into your option element to ensure that the values are being submitted properly.
Mozilla documentation:
https://developer.mozilla.org/en-US/docs/Web/HTML/Element/select

Answer 2

I have edited your code in this way.
<?php
$id = $_GET['id'];
$conn = mysqli_connect('localhost', 'phpmyadmin', 'Test#2000', 'user');
$sql1 = "SELECT * FROM usr WHERE id='$id'";
$result = mysqli_query($conn, $sql1);
$row = mysqli_fetch_assoc($result);
?>
<fieldset><label>Birthday</label>
<select name="birthday">
<?php for ($i = 1300; $i <= 1397; $i++) {
echo "<option" . (($i == $row['birthdayYear']) ? 'selected="true"' : '') . ">$i</option>";
}
?>
</select>
<fieldset>
<button name="btn" type="submit" id="contact-submit">Submit</button>
</fieldset>
</form>
</div>
</body>
</html>
Hope this helps, thanks.

Answer 3

A.) WITH PDO MODULE
It is best practice today to use prepared statements to avoid SQL injection. This is done through the PDO object.
Set for select the autocomplete="off" attribute, because Firefox apparently has a bug with the selected="selected" that needs this to be set.
See if it is the user's birthday, we can use intval to compare.
<?php
$id = $_GET['id'];
$host = 'localhost';
$username = 'phpmyadmin';
$password = 'Test#2000';
$db_name = 'user';
$conn = new PDO('mysql:host:=' . $host . '; dbname=' . $db_name, $username, $password);
$sql1 = "SELECT * FROM usr WHERE id = :id;";
$stmt = $conn->prepare($sql1);
$stmt->bindParam(':id', $id);
$stmt->execute();
$row = $stmt->fetch(PDO::FETCH_ASSOC);
?>
<fieldset><label>Birthday</label>
<select name="birthday" autocomplete="off">
<?php for ($i = 1300; $i <= 1397; $i++) {
if($i === intval($row['birthdayYear'])){
echo "<option selected='selected'>$i</option>";
} else {
echo "<option>$i</option>";
}
}
?>
</select>
<fieldset>
<button name="btn" type="submit" id="contact-submit">Submit</button>
</fieldset>
</form>
</div>
</body>
</html>
B.) Please at least try option A.), but if it doesn't work:
<?php
$id = $_GET['id'];
$conn = mysqli_connect('localhost', 'phpmyadmin', 'Test#2000', 'user');
$sql1 = "SELECT * FROM usr WHERE id='$id'";
$result = mysqli_query($conn, $sql1);
$row = mysqli_fetch_assoc($result);
?>
<fieldset><label>Birthday</label>
<select name="birthday" autocomplete="off">
<?php for ($i = 1300; $i <= 1397; $i++) {
if($i === intval($row['birthdayYear'])){
echo "<option selected='selected'>$i</option>";
} else {
echo "<option>$i</option>";
}
}
?>
</select>
<fieldset>
<button name="btn" type="submit" id="contact-submit">Submit</button>
</fieldset>
</form>
</div>
</body>
</html>