php - 一次从数据库中回显两个表

Question

这是我用于回显一张表的代码。任何人都可以告诉如何在同一个数据库中回显两个表？

<?php
include 'dbconfig.php';

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

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

$sql = "SELECT * FROM table1";
$result = $conn->query($sql);

if ($result->num_rows >0) {
    // output data of each row
    while($row[] = $result->fetch_assoc()) {

        $tem = $row;

        $json = json_encode($tem);
    }
} else {
    echo "0 results";
}
echo $json;
$conn->close();

Answer 1

你只发出一行，对吗？问题是您为每一行一次又一次地覆盖 $json 。这应该效果更好:

<?php
// as before

$sql = "SELECT * FROM table1";
$result = $conn->query($sql);

$rows = array();

while($row[] = $result->fetch_assoc()) {
    $rows[] = $row;
}

echo json_encode($rows);