if (isset($_POST["Name"]==null $_POST["Email"]==null, $_POST["Password"]==null, $_POST["Phone"]==null)) what is wrong with this in php

Question

i have a database and want to check if the elements are set or not before they pass to database because i dont want to pass null values for entities mentioned but while using this statement i find out values are passing without being check for availability.

i tried to pass those elements to the database but check for nulls first and they were not checked for nulls and sometimes got this error as well:

Fatal error: Cannot use isset() on the result of an expression (you can use "null !== expression" instead) in C:\xampp\htdocs....on line 3

Answer 1

The isset function works differently, you need to put a variable into it, not a condition.

For example:

$arr = [
  'a' => 'Hello',
];

$bool = isset($arr['a']) // return true
$bool = isset($arr['b']) // return false

this page detailed

---

Let's try writing the following to fix the error in your code.

if (
  !isset($_POST["Name"]) ||
  !isset($_POST["Email"]) ||
  !isset($_POST["Password"]) ||
  !isset($_POST["Phone"])
) {
  // some fields is null, not valid.
}

or

if (!isset($_POST["Name"], $_POST["Email"], $_POST["Password"], $_POST["Phone"])) {
  // some fields is null, not valid.
}

Answer 2

If you want to prevent null values from being inserted into the DB:

CREATE TABLE User(
    ID int NOT NULL,
    LastName varchar(255) NOT NULL,
    FirstName varchar(255) NOT NULL
);

also, use the required attribute in your html document:

<form action="/action_page.php"> 
    <input type="text" id="username" required>  
</form>