Directly echo the resulting data show an error msg [duplicate] - php

Question

This question already has answers here:
Object of class mysqli_result could not be converted to string
(5 answers)
Closed 1 year ago.
When I am echo the $result variable. It display an error msg like this:
Catchable fatal error: Object of class mysqli_result could not be converted to string in D:\Inspiration\server\Table2_yahooData_db.php
on line 13
<?php
// connect to the dataBase
$conn = mysqli_connect('localhost', 'root', '', 'yahooData');
//checking Connection.
if (!$conn) {
die('Error connection failed: ' . mysqli_connect_error($conn));
}
// Creating select query Using sql commands.
$sql = "SELECT * FROM User;" ;
$result = mysqli_query($conn, $sql);
echo $result;
mysqli_close($conn);
?>

Answer 1

$result is an object.
You cannot print it with string's echo function.
You can rather print it with print_r($result);
echo is for scalar variables (which have single value or are single dimensions) like number, string. For multi-dimensional variables e.g. array, objects we need to use print_r() which prints the whole tree.

Answer 2

When using a SELECT query the mysqli_query() function returns a resource, not a string. You need to use mysqli_fetch_assoc() to put the results into an array. Here is an object-oriented example:
$mysqli = new mysqli("localhost", "my_user", "my_password", "world");
/* check connection */
if ($mysqli->connect_errno) {
printf("Connect failed: %s\n", $mysqli->connect_error);
exit();
}
$query = "SELECT name FROM User";
if ($result = $mysqli->query($query)) {
/* fetch associative array */
while ($row = $result->fetch_assoc()) {
echo $row['name'];
}
/* free result set */
$result->free();
}