php - Why is my nested IF condition not working? -

i want check if record exists , if go page a, otherwise go page b. goes page whether there record or not.

<?php $ini = parse_ini_file("../phpconfig.ini");  $conn = mysqli_connect($ini['hostaddress'], $ini['username'], $ini['password'], $ini['databasename']);  if (mysqli_connect_errno()) {     echo "failed connect mysql: " . mysqli_connect_error(); }  $options = ['cost' => 10,];  $number = mysqli_real_escape_string($conn, $_post['number']); $password = password_hash((mysqli_real_escape_string($conn, $_post['password'])), password_bcrypt, $options);  $sql = "insert usertemp (employee_number, password) values ('$number', '$password')";  if (mysqli_query($conn, $sql)) {      $sql2 = "select exists(select 1 employee number = '$number')";     $row2 = mysqli_query($conn, $sql2);          if (mysqli_num_rows($row2) > 0) {             //has record in employee table             header("location: display_createaccount_a.php");         } else {             //no record in employee table             header("location: display_createaccount_b.php");         }  } else {     echo "error: " . $sql . "<br>" . mysqli_error($conn); }  mysqli_close($conn); ?> 

your select exists(...) going return 1 row true or false.

in case, read first index of result , use conditional value.

or better yet, use limit instead , keep app logic is:

$sql2 = "select * employee number = '$number' limit 1"; 

also, recommend checking out php documents on prepared statements. safer way of querying mysql.