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php - PHP 值 Mysqli stmt 没有结果

转载 作者:搜寻专家 更新时间:2023-10-30 20:25:06 25 4
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我的 sqli 请求没有结果..就像空数据..我很确定我的数据库中有很多数据

这是我的代码,如果我的代码有误请纠正我

<?php
// include db handler
class DB_Functions {

private $conn;

// constructor
function __construct() {
require_once 'include/DB_Connect.php';
// connecting to database
$db = new Db_Connect();
$this->conn = $db->connect();
}

// destructor
function __destruct() {

}

function getSliderList(){

$stmt = $this->conn->prepare("SELECT cPID, image FROM sliderImage");
$stmt->execute();
if ($stmt->num_rows > 0 ) {
$result = $stmt->get_result()->fetch_assoc();
$response[] = $result;
$stmt->close();
echo json_encode($response);
return true;
} else {
// user not found
return false;
}
}
}

$x = new DB_Functions();
$user = $x->getSliderList();
$response = Array();
if($user){
$user;
return false;
} else {
$response['error'] = "Sorry an error occured. Our Problem, not you.";
return true;
}

?>

我的数据库请求连接

<?php
class DB_Connect {
private $conn;

// Connecting to database
public function connect() {
require_once 'include/Config.php';

// Connecting to mysql database
$this->conn = new mysqli(DB_HOST, DB_USER, DB_PASSWORD, DB_DATABASE);

// return database handler
return $this->conn;
}
}

?>

config.php文件

<?php

/**
* Database config variables
*/
define("DB_HOST", "localhost");
define("DB_USER", "bxxx");
define("DB_PASSWORD", "xxxx");
define("DB_DATABASE", "xxxx");
?>

我想将结果放入数组中...并使用 json_encode 发送此数据以在我的应用中使用...

最佳答案

我认为您在这里不需要准备好的语句,因为您没有参数化任何数据。改为尝试简单的查询

$stmt = $this->conn->query("SELECT cPID, image FROM sliderImage");
$response = array();
while($result = $stmt->fetch_assoc()) {
$response[] = $result;
}
echo json_encode($response);
return true;

关于php - PHP 值 Mysqli stmt 没有结果,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/38680392/

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