in line 45 mysqli give error

Question

Why is error occur in line 45 mysqli. and i strictly want to use only mysqli? I tried connection without "new" by creating object it does not works. but when i skip new in connection database line it still does not works fine. please refer me best solution for this Insert, View, edit and delete with mysqli check statement? use

 <?php
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "myDB";

// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);

// Check connection
if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
}

if(isset($_POST['submit'])){

         $firstname     = $_POST['firstname'];
         $lastname      = $_POST['lastname'];
         $email          = $_POST['email'];



// prepare and bind
$prep_stmt = $conn->prepare("INSERT INTO MyGuests (firstname, lastname, email) VALUES (?, ?, ?)");

if($stmt = $conn->prepare($prep_stmt)){

$stmt->bind_param("sss", $firstname, $lastname, $email);
$stmt->execute();
            echo "<script>alert('data inserted')</script>"; 

        }
}

?>

Possible duplicate of [mysqli\_fetch\_array()/mysqli\_fetch\_assoc()/mysqli\_fetch\_row() expects parameter 1 to be resource or mysqli\_result, boolean given](http://stackoverflow.com/questions/2973202/mysqli-fetch-array-mysqli-fetch-assoc-mysqli-fetch-row-expects-parameter-1) — S.I., Nov 07 '16 at 07:09
correct this line `if($stmt = $conn->prepare($prep_stmt)){` pass query instead of `$prep_stmt` — amit, Nov 07 '16 at 07:10
Possible duplicate of [mysqli::query() expects parameter 1 to be string, object given](http://stackoverflow.com/questions/30675593/mysqliquery-expects-parameter-1-to-be-string-object-given) — devpro, Nov 07 '16 at 07:11
you add two time $conn->prepare add only $prep_stmt = $conn->prepare("INSERT INTO MyGuests (firstname, lastname, email) VALUES (?, ?, ?)"); . — pawan sen, Nov 07 '16 at 07:12

score 0 · Answer 1 · answered Nov 07 '16 at 07:25

I have Solved my code now.

<?php
//$servername = "localhost";
//$username = "root";
//$password = "";
//$dbname = "myDB";

// Create connection
//$conn = new mysqli($servername, $username, $password, $dbname);

$conn = mysqli_connect("localhost", "root", "", "myDB");




// Check connection
if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
}

if(isset($_POST['submit'])){

         $firstname     = $_POST['firstname'];
         $lastname      = $_POST['lastname'];
         $email          = $_POST['email'];



// prepare and bind
$prep_stmt ="INSERT INTO MyGuests (firstname, lastname, email) VALUES (?, ?, ?)";

if($stmt = $conn->prepare($prep_stmt)){

$stmt->bind_param("sss", $firstname, $lastname, $email);
$stmt->execute();
            echo "<script>alert('data inserted')</script>"; 

        }
}

?>

in line 45 mysqli give error

1 Answers1