0

im adding some values to my table in mysql to test it but nothing happens. it doesnt even give me an error.

<?php
//get values from index.php

$lecturerid = $_POST['studentid'];
$password = $_POST['Pass'];



// Create connection
$conn = mysqli_connect("localhost", "root", "", "coursework");

// Check connection
if (!$conn) {
    die("Connection failed: " . mysqli_connect_error());
}
echo "Connected successfully";


$sql = "INSERT INTO studentlogin (studentid, password)
VALUES ('John', 'Doe')";




?>

can anyone tell me why.

4
  • 2
    You're not executing the query. mysqli::prepare() (php.net/mysqli.prepare) / mysqli::query() (php.net/mysqli.query) Commented Apr 22, 2017 at 17:44
  • how to execute ? im new to this Commented Apr 22, 2017 at 17:45
  • Use mysqli::prepare() if you're using variables in the query, mysqli::query() can be used for static queries. Commented Apr 22, 2017 at 17:46
  • Can you mark one of the answer below as 'solved' since you wrote 'thanks it works'. Commented Apr 22, 2017 at 18:07

3 Answers 3

1

You must add mysqli_query()

//get values from index.php

$lecturerid = $_POST['studentid'];
$password = $_POST['Pass'];



// Create connection
$conn = mysqli_connect("localhost", "root", "", "coursework");

// Check connection
if (!$conn) {
    die("Connection failed: " . mysqli_connect_error());
}
echo "Connected successfully";


$sql = "INSERT INTO studentlogin (studentid, password)
VALUES ('John', 'Doe')";

mysqli_query($conn, $sql);


?>
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1 Comment

It's ok. Welcome
1

Quite simple,

<?php
    //get values from index.php

    $lecturerid = $_POST['studentid'];
    $password = $_POST['Pass'];

   // Create connection
   $conn = mysqli_connect("localhost", "root", "", "coursework");

   // Check connection
   if (!$conn) {
          die("Connection failed: " . mysqli_connect_error());
   }
   echo "Connected successfully";

   //Build query
   $sql = "INSERT INTO studentlogin (studentid, password) VALUES ('John', 'Doe')";

   //execute query
   mysqli_query($conn,$sql) or die(mysqli_error($conn));

?>

1 Comment

Pleasure is mine @Java_NewBie :)
0

You need to use the below methods to insert it

$firstname = "John";
$lastname = "Doe";
$stmt = $conn->prepare("INSERT INTO studentlogin (studentid, password) 
VALUES (:studentLogin, :password)");
$stmt->bindParam(':firstname', $firstname);
$stmt->bindParam(':lastname', $lastname);
$stmt->execute();

1 Comment

isn't this PDO syntax? the OP is using mysqli. Good call, bad number :D

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