echo is showing output in page source instead of on page - php

Question

I made a table in php and wanted to show the Id's in the dropdown select menu by making a separate file for php. So the code in main file is:
<?php include "functions.php";?>
<form action="login_update.php" method="post">
<div class="form-group">
<label for="username">username</label>
<input type="text" name="username" class="form-control">
</div>
<div class="form-group">
<label for="password">password</label>
<input type="password" name="password" class="form-control">
</div>
<div class="form-group">
<select name="id" id="">
<?php
showAllData();
echo "<br>"."askfkldfjl;adfafladfdf";
?>
</select>
</div>
<input class="btn btn-primary" type="submit" name="submit" value="update">
</form>
The code of functions.php is :
<?php
function showAllData(){
$connection = mysqli_connect('localhost','root','****','loginapp');
if($connection){
echo "We are connected. a=".$a."<br>";
}else{
die("Database connection failed");
}
$query = "SELECT * FROM users";
$result = mysqli_query($connection,$query);
if($result){
echo("<br>"." <b><h6>We are successful</h6></b>");
}
else {
die("Query FAILED" . mysqli_error());
}
while($row = mysqli_fetch_assoc($result)) {
$id = $row["id"];
echo"<option value='$id'>$id</option>";
}
}
?>
The expected output was :
But the output is:
So the top two lines in the above screenshot are not printing.
These lines are shown in the INSPECT ELEMENT in chrome.
I forgot to mention the echo command:
echo "<br>"."askfkldfjl;adfafladfdf";
below show all data is also not working.

Answer 1

You made a mistake.
Actually you wrote a code in selectbox and you dont add option so thats why it is not show in html
So write a code like below code so its show in select box as option.
<div class="form-group">
<select name="id" id="">
<option> <?php
showAllData();
echo "<br>"."askfkldfjl;adfafladfdf";
?></option>
</select>
</div>
And If you want to show option from showAllData(); function, the you have return the html.
For this update your showAllData(); function with below code:
function showAllData(){
$options="";
$connection = mysqli_connect('localhost','root','****','loginapp');
if($connection){
echo "We are connected. a=".$a."<br>";
}else{
die("Database connection failed");
}
$query = "SELECT * FROM users";
$result = mysqli_query($connection,$query);
if($result){
echo("<br>"." <b><h6>We are successful</h6></b>");
}
else {
die("Query FAILED" . mysqli_error());
}
while($row = mysqli_fetch_assoc($result)) {
$id = $row["id"];
$options.="<option value='$id'>$id</option>";
}
return $options;
}

Answer 2

Move the PHP function showAllData() before the HTML <select> element.
Because the <select> element awaits for an <option> element, but all another text will not be visible on page.
E.g.:
<div class="form-group">
<?php showAllData(); ?>
</div>
<?php
function showAllData(){
$connection = mysqli_connect('localhost','root','****','loginapp');
if($connection){
echo "We are connected. a=".$a."<br>";
}else{
die("Database connection failed");
}
$query = "SELECT * FROM users";
$result = mysqli_query($connection,$query);
if($result){
echo("<br>"." <b><h6>We are successful</h6></b>");
}
else {
die("Query FAILED" . mysqli_error());
}
echo '<select name="id" id="">';
while($row = mysqli_fetch_assoc($result)) {
$id = $row["id"];
echo"<option value='$id'>$id</option>";
}
echo "</select>";
}
?>

Answer 3

Your code is mixed up.
You can use below code. Create an array which gives you values which you needs to show in select.
<?php
function showAllData(){
$idArr = array('msg'=>'','data'=>'','status'=>0);
$connection = mysqli_connect('localhost','root','****','loginapp');
if($connection){
$idArr['msg'] = "We are connected";
$idArr['status'] = 1;
}else{
$idArr['msg'] = "Database connection failed";
$idArr['status'] = 0;
}
if($idArr['status'] == 1){
$query = "SELECT * FROM users";
$result = mysqli_query($connection,$query);
if($result){
$idArr['msg'] = "We are successful";
$idArr['status'] = 1;
}else {
$idArr['msg'] = "Query FAILED" . mysqli_error();
$idArr['status'] = 0;
}
if($idArr['status'] == 1){
while($row = mysqli_fetch_assoc($result)) {
$idArr['data'][] = $row["section_id"];
}
}
}
return $idArr;
}
$idArr = showAllData();
?>
<?php
if(!empty($idArr['data'])){
echo "We are connected<br>";
echo("<br>"." <b><h6>We are successful</h6></b>");
?>
<form action="login_update.php" method="post">
<div class="form-group">
<label for="username">username</label>
<input type="text" name="username" class="form-control">
</div>
<div class="form-group">
<label for="password">password</label>
<input type="password" name="password" class="form-control">
</div>
<div class="form-group">
<select name="id" id="">
<option value="0">--Select--</option>
<?php
foreach ($idArr['data'] as $key => $value) {
echo"<option value='$value'>$value</option>";
}
?>
</select>
</div>
<input class="btn btn-primary" type="submit" name="submit" value="update">
</form>
<?php }else{
echo $idArr['msg'];
}
?>

Answer 4

Why you are putting <br> inside the select tag? select tag only accept the options tag under it, so please remove br tag and all extra strings inside the select tag. and you should echo the message above the select tag if you want to show your users.
Thanks