php 代码没有从数据库返回任何结果
$conn = new mysqli($host, $username, $password, $dbname);
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT * FROM pelicula";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
while($row = $result->fetch_assoc()) {
array_push($result_array, $row);
}
}
echo json_encode($result_array);
$conn->close();
最佳答案
$conn = new mysqli($host, $username, $password, $dbname);
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT * FROM pelicula";
$result = $conn->query($sql);
$result_array=[];
if ($result->num_rows > 0) {
while($row = $result->fetch_assoc()) {
array_push($result_array, $row);
}
}
echo json_encode($result_array);
$conn->close();`enter code here`
在将 $result_array 与 array_push 函数一起使用之前，必须先创建并实例化它!!!
关于php - php 中的数据库查询失败，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/57365900/