Php 转换OOP JSON
我有两个文件:apix.php和crud.php。我正在尝试将从php获取的数据转换为jason打印。问题是,该函数没有在代码的下一节中为我解析结果 CRUD.php:Php 转换OOP JSON,php,Php,我有两个文件:apix.php和crud.php。我正在尝试将从php获取的数据转换为jason打印。问题是,该函数没有在代码的下一节中为我解析结果 CRUD.php: class crud { private $db; function __construct($DB_con) { $this->db = $DB_con; } public function dataview_new($query) {
class crud
{
private $db;
function __construct($DB_con)
{
$this->db = $DB_con;
}
public function dataview_new($query)
{
$stmt = $this->db->prepare($query);
$stmt->execute();
$users = array();
while($row = $stmt->fetch(PDO::FETCH_ASSOC)){
array_push($users, $row);
}
return $users;
}
}
APIX.php文件:
require_once 'crud.php';
$crud = new crud($DB_con);
$res = array('error' => false);
$action = 'read';
if(isset($_GET['action'])){
$action = $_GET['action'];
}
if($action == 'read'){
$query = "SELECT * FROM test";
$crud->dataview_new($query);
$res['users'] = $users;
}
header("Content-type: application/json");
echo json_encode($res);
die();
我遗漏了一些明显的东西,因为我在IF语句中的crud之外测试了代码,它运行良好:
if($action == 'read'){ // alternative code to the previous IF statement
$result = $DB_con->prepare("SELECT * FROM `test`");
$result->execute();
$users = array();
while($row = $result->fetch(PDO::FETCH_ASSOC)){
array_push($users, $row);
}
$res['users'] = $users;
}
您没有将
$crud->dataview\u new($query)
的返回值分配给变量
您需要使用作业:
if($action == 'read'){
$query = "SELECT * FROM test";
$users = $crud->dataview_new($query); // <--assignment
$res['users'] = $users;
}
if($action=='read'){
$query=“从测试中选择*”;
$users=$crud->dataview\u new($query);//您没有将$crud->dataview\u new($query)
的返回值分配给变量
您需要使用作业:
if($action == 'read'){
$query = "SELECT * FROM test";
$users = $crud->dataview_new($query); // <--assignment
$res['users'] = $users;
}
if($action=='read'){
$query=“从测试中选择*”;
$users=$crud->dataview\u new($query)//