如何在将数据插入表中时调试错误？

Question

PHP code didn't insert.

$conn = mysqli_conncect("localhost","root","","auth");

I tried everything but not found the mistake to insert the query into the table.

Answer 1

Remove the single quotation in your else statement. PHP will interpret it as a string instead of a POST variable.

That is:

$email = $_POST['email'];
$pass = $_POST['password'];