Json total result mysql - php

Question

I am having difficulty displaying the total number of the result of the mysql query using json.
I tried this rule but it only returns like this:
{"Total1": "10"}
I need to return it with the []:
[{"Total1": "10"}]
Follows the php code:
<?php
include 'DatabaseConfig.php';
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT revenda, quantidade FROM app_venda";
$result = $conn->query($sql);
$total1 = mysql_num_rows($result);
if ($result->num_rows > 0) {
// output data of each row
$row[total1] = $total1;
$json = json_encode($row);
} else {
echo "0 results";
}
echo $json;
$conn->close();
?>

Answer 1

Then you need to add another array around the data
$total1 = 10;
echo json_encode( array(array('Total1' => $total1)) );
Result
[{"Total1":10}]
Or using your code with a correction in the array definition. Notice $row['total1'] = 10; uses quotes around the array name to stop this error
Notice: Use of undefined constant total1 - assumed 'total1'
$row['total1'] = 10;
echo json_encode( array( $row ) );
Can I suggest in furutre you test code with these lines added to the top of all your scripts
ini_set('display_errors', 1);
ini_set('log_errors',1);
error_reporting(E_ALL);
mysqli_report(MYSQLI_REPORT_ERROR | MYSQLI_REPORT_STRICT);
This will force any mysqli_ errors to
generate an Exception that you can see on the browser and other errors will also be visible on your browser.

Answer 2

$total1 = mysql_num_rows($result);
$row = array();
if ($result->num_rows > 0) {
// output data of each row
$row[][total1] = $total1;
$json = json_encode($row);
} else {
echo "0 results";
}
echo $json;