SQL error (ERROR: Could not able to execute INSERT INTO) [duplicate] - php

Question

This question already has an answer here:
What to do with mysqli problems? Errors like mysqli_fetch_array(): Argument #1 must be of type mysqli_result and such
(1 answer)
Closed 3 years ago.
I have a database that I have made on phpmyAdmin that consists of three columns: id, name and number.
I have added 3 rows of data to the database through phpmyadmin.
I now wish to add data to this database through my php file. This is the code that I use to add in the data and display the data on the browser:
<?php
$servername = "localhost";
$username = "root";
$password = "root";
$dbname = "myfirstsite";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
} else {
echo "wooo connected";
}
$sql = "INSERT INTO hi (id, name, number)
VALUES ('99', 'Doe', '999999')";
if(mysqli_query($link, $sql)){
echo "Records inserted successfully.";
} else{
echo "ERROR: Could not able to execute $sql. " . mysqli_error($link);
}
//displaying data
$sql = "SELECT id, name, number FROM hi";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
// output data of each row
while($row = $result->fetch_assoc()) {
echo "id: " . $row["id"]. " - Name: " . $row["name"]. " " . $row["number"]. "<br>";
}
} else {
echo "0 results";
}
$conn->close();
?>
The thing is that I don't understand why the new data isn't placed into the database but the current data is displayed onscreen.

Answer 1

In your code you are making reference to $link which doesnt exists, it should be $conn
Chaneg this:
if(mysqli_query($link, $sql)){
echo "Records inserted successfully.";
} else{
echo "ERROR: Could not able to execute $sql. " . mysqli_error($link);
}
To:
if(mysqli_query($conn, $sql)){
echo "Records inserted successfully.";
} else{
echo "ERROR: Could not able to execute $sql. " . mysqli_error($link);
}

Answer 2

you need to fix this one
$conn = new mysqli($servername, $username, $password, $dbname);
to
$conn = new mysqli_connect($servername, $username, $password, $dbname);
and also this need to fix
if(mysqli_query($conn, $sql)){
echo "Records inserted successfully";
}else{
echo "ERROR: Could not able to execute $sql. " . mysqli_error($conn);
}