Getting items from a database not working as expected - php

Question

I'm learning PHP and I've hit a wall.
When the user submits to the form, it adds to the database.
I am also trying to display all items in the database on the same page as the form.
However,this only works if the form has just been submitted. If the form has not been submitted (but there is still content in the database), nothing is shown.
How can I always show what is in the current database?
<form action="" method="post">
<input type="text" name="todo_content" id="">
<input type="submit" value="Submit">
</form>
<?php
if ( isset($_POST['todo_content'])) {
$latest_content = $_POST['todo_content'];
} else {
die();
}
$todo_db = mysqli_connect('localhost', 'root', 'root');
mysqli_select_db($todo_db, 'todo_list');
mysqli_query(
$todo_db,
"INSERT INTO todo_items (item_content) VALUES ('$latest_content')"
);
$all_todos = mysqli_query( $todo_db, "SELECT item_content FROM todo_items" );
$all_todos_result = mysqli_fetch_array($all_todos);
var_dump($all_todos_result); // these show nothing
var_dump($all_todos); // these show nothing
?>

Answer 1

The die() is your problem:
if ( isset($_POST['todo_content'])) {
echo 'is set';
$latest_content = $_POST['todo_content'];
} else {
die();
}
If $_POST is not set, you will never reach the part where you start printing your items. And since the form is not submitted, $_POST is empty.
EDIT
You could do it like this:
<form action="" method="post">
<input type="text" name="todo_content" id="">
<input type="submit" value="Submit">
</form>
<?php
$todo_db = mysqli_connect('localhost', 'root', 'root');
mysqli_select_db($todo_db, 'todo_list');
if ( isset($_POST['todo_content'])) {
$latest_content = $_POST['todo_content'];
mysqli_query($todo_db, "INSERT INTO todo_items (item_content) VALUES ('$latest_content')");
}
$all_todos = mysqli_query( $todo_db, "SELECT item_content FROM todo_items" );
$all_todos_result = mysqli_fetch_array($all_todos);
var_dump($all_todos_result); // these show nothing
var_dump($all_todos); // these show nothing
?>

Answer 2

You need to move your insert query inside your if condition and you will need not to die if the form is not submitted but print the form and your query results
$todo_db = mysqli_connect('localhost', 'root', 'root');
mysqli_select_db($todo_db, 'todo_list');
if ( isset($_POST['todo_content'])) {
echo 'is set';
$latest_content = $_POST['todo_content'];
mysqli_query($todo_db, "INSERT INTO todo_items (item_content) VALUES ('$latest_content')");
} else {
?>
<form action="" method="post">
<input type="text" name="todo_content" id="">
<input type="submit" value="Submit">
</form>
<?php
$all_todos = mysqli_query( $todo_db, "SELECT item_content FROM todo_items" );
$all_todos_result = mysqli_fetch_array($all_todos);
var_dump($all_todos_result);
var_dump($all_todos);
}
As side note i'd say you are at high risk of mysql injection. You should use prepaed statments and not inserting $_POST data inside your database directly