PHP If statement returning early(amateur) - php

Question

I'm currently struggling with a page that allows a user to complete one of two options. They can either update an existing item in the SQL database or they can delete it. When the customer deletes an option everything runs perfectly well, however whenever a customer updated an item it displays the Query failed statement from the delete function before applying the update.
It seems obvious to me that the problem must be in my IF statement and that the DeleteButton function isn't exiting if the $deleteno variable isn't set. Any help would be appreciated. Excuse the horribly messy code PHP isn't a language I am familiar with. (I have not included the connect information for privacy reasons)
function DeleteButton(){
#mysqli_select_db($con , $sql_db);
//Checks if connection is successful
if(!$con){
echo"<p>Database connection failure</p>";
} else {
if(isset($_POST["deleteID"])) {
$deleteno = $_POST["deleteID"];
}
if(!isset($deleteno)) {
$sql = "delete from orders where orderID = $deleteno;";
$result = #mysqli_query($con,$sql);
if((!$result)) {
echo "<p>Query failed please enter a valid ID </p>";
} else {
echo "<p>Order $deleteno succesfully deleted</p>";
unset($deleteno);
}
}
}
}
That is the code for the delete button and the following code is for the UpdateButton minus the connection information (which works fine).
if(isset($_POST["updateID"])) {
$updateno = $_POST["updateID"];
}
if(isset($_POST["updatestatus"])) {
if($_POST["updatestatus"] == "Fulfilled") {
$updatestatus = "Fulfilled";
} elseif ($_POST["updatestatus"] == "Paid") {
$updatestatus = "Paid";
}
}
if(isset($updateno) && isset($updatestatus)) {
$sql ="update orders set orderstatus='$updatestatus' where orderID=$updateno;";
$result = #mysqli_query($con,$sql);
if(!$result) {
echo "<p>Query failed please enter a valid ID</p>";
} else {
echo "<p>Order: $updateno succesfully updated!</p>";
}
}
Once again these are incomplete functions as I have omitted the connection sections.

Answer 1

if(!isset($deleteno)) {
$sql = "delete from orders where orderID = $deleteno;";
Are you sure you want to execute that block if $deleteno is NOT set?
P.S. You shouldn't rely on $_POST['deleteId'] being a number. Please read about SQL injections, how to avoid them and also about using prepared statements.

Answer 2

I've update your code, but you need to write cleaner code ( spaces, indents, etc ) this won't only help you to learn but to find your errors easily.
<?php
function DeleteButton()
{
#mysqli_select_db($con , $sql_db);
/*
Checks if connection is successful
*/
if(!$con){
echo"<p>Database connection failure</p>";
} else {
/*
Check if $_POST["deleteID"] exists, is not empty and it is numeric.
*/
if(isset($_POST["deleteID"]) && ! empty($_POST["deleteID"]) && ctype_digit(empty($_POST["deleteID"]))
$deleteno = $_POST["deleteID"];
$sql = "delete from orders where orderID='$deleteno'";
$result = #mysqli_query($con,$sql);
if(!$result){
echo "<p>Query failed please enter a valid ID </p>"
} else {
echo "<p>Order $deleteno succesfully deleted</p>";
unset($deleteno);
}
} else {
echo "<p>Please enter a valid ID </p>" ;
}
}
}
/*
Part 2:
===========================================================================
Check if $_POST["updateID"] exists, is not empty and it is numeric.
Check if $_POST["updatestatus"] exists, is not empty and equal to Paid or Fullfilled
*/
if( isset($_POST["updateID"]) &&
! empty($_POST["updateID"]) &&
ctype_digit(empty($_POST["updateID"]) &&
isset($_POST["updatestatus"]) &&
! empty($_POST["updatestatus"]) &&
( $_POST["updatestatus"] == "Fulfilled" || $_POST["updatestatus"] == "Paid" ) )
{
$updateno = $_POST["updateID"];
$updatestatus = $_POST["updatestatus"];
$sql ="update orders set orderstatus='$updatestatus' where orderID=$updateno;";
$result = #mysqli_query($con,$sql);
if(!$result){
echo "<p>Query failed please enter a valid ID</p>";
} else {
echo "<p>Order: $updateno succesfully updated!</p>";
}
}

Answer 3

There is an error in MySQL Syntax
$sql = "delete from orders where orderID = $deleteno;";
$deleteno after orderID must be inside single quotes.
change it to this $sql = "delete from orders where orderID = '$deleteno';";