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();
最佳答案
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);
关于php - 一次从数据库中回显两个表,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/47837067/