confused in echo mysql result inside echo - php

Question

after sign in and redirect to next page, i check the user with $_SESSION.
Then if username valid, i want to show dropdown list which the list is get from database. I am confuse how to echo.
<?php
session_start(); //Start the session
if(!isset($_SESSION['pic']))
{
header('Location:index.php?p=signin');
}
else {
echo '<div class="form-group">
<label for="symcat" class="control-label">Symptom Category</label>
<select id="symcat" name="symcat" class="selectlist form-control">
<option value=""></option>
$sql='SELECT category FROM sym_category';
if ($sql) {
$res=mysqli_query($dbc,$sql) or die(_ERROR26.': '.mysqli_connect_error());
}
while ($dat = mysqli_fetch_array($res, MYSQLI_NUM)) {
echo '\t<option value="'.$dat[0].'">'.$dat[0].'</option>\n';
}
mysqli_free_result($res);
</select>
</div> <!-- /form-group -->';
}
?>

Answer 1

This would probably be better:
<?php
session_start(); //Start the session
if(!isset($_SESSION['pic']))
{
header('Location:index.php?p=signin');
}
else {
$sql = 'SELECT category FROM sym_category';
if ($sql) {
$res = mysqli_query($dbc,$sql) or die(_ERROR26.': '.mysqli_connect_error());
}
?>
<div class="form-group">
<label for="symcat" class="control-label">Symptom Category</label>
<select id="symcat" name="symcat" class="selectlist form-control">
<option value=""></option>
<?php if (isset($res)): ?>
<?php while ($dat = mysqli_fetch_array($res, MYSQLI_NUM)): ?>
<option value="<?php echo $dat[0] ?>"><?php echo $dat[0] ?></option>
<?php endwhile ?>
<?php mysqli_free_result($res); ?>
<?php endif ?>
</select>
</div> <!-- /form-group -->
<?php } ?>
Honestly I still don't think this is a very nice way. If possible, please use view files.
I think it's reasonable to say that you shouldn't use an echo within another echo. Echo is used to print something to your screen, so it shouldn't be necessary to use an echo within that same echo.

Answer 2

Check out this sample code. Maybe it'll help you.
// Create connection
$conn = mysqli_connect($servername, $username, $password, $dbname);
// Check connection
if (!$conn) {
die("Connection failed: " . mysqli_connect_error());
}
// Fetch the option list details from DB
$optionList = array();
$sql = "SELECT category FROM sym_category";
$result = mysqli_query($conn, $sql);
if (mysqli_num_rows($result) > 0) {
// output data of each row
while($row = mysqli_fetch_assoc($result)) {
$optionList[] = $row['category']
}
} else {
echo "0 results";
}
// Create the dropdown list
echo "<div class='form-group'>";
echo '<label for="symcat" class="control-label">Symptom Category</label>';
echo '<select id="symcat" name="symcat" class="selectlist form-control">';
echo '<option value=""></option>';
foreach ($optionList as $key => $value)
{
echo "<option value='$value'>$value</option>";
}
echo '</select>';
echo '</label>';
echo "</div>"

Answer 3

As Akintunde & SheperOfFire said the nice and tidy way is like :
<?php
session_start(); //Start the session
if(!isset($_SESSION['pic']))
{
header('Location:index.php?p=signin');
}
else {
header('Location:index.php?p=forminputcar');
}
?>
Because, inside of forminputcar has php tag and echo also. Or you can also put above script at the top of forminputcar page. So, if session match we stay on current page.
<?php
session_start();
if(isset($_SESSION['pic']))
{
header('charset=UTF-8');
}
else {
header('Location:index.php?p=signin');
}
?>