如何在 PHP 中使用 for 循环来返回 mysql 数据 - How can I use a for loop in PHP to return mysql data

我正在制作一个应用程序，该应用程序将在列表视图中按某个商店列出员工。我在DB_Functions.php文件中的当前函数是：

public function getEmployeeList($name) {
    $stmt = $this->con->prepare("SELECT employee_name FROM employees WHERE name = ?");
    $stmt->bind_param('s', $name);
    if ($stmt->execute()) {
        $employee_list = $stmt->get_result()->fetch_assoc();
        $stmt->close();
        if (empty($employee_list)) {
            return NULL;
        } else {
            return $employee_list;
        }
    }
}

在我的员工.php文件中，我有以下代码：

<?php
require_once 'include/DB_Functions.php';
$db = new DB_Functions();
$response = array('error' => FALSE);
if (isset($_POST['name'])) {
    $name = $_POST['name'];
    $employee_list = $db->getEmployeeList($name);
    if ($employee_list != false) {
        $response['error'] = FALSE;
        //EMPLOYEE LIST OBJECT HERE
    } else {
        $response['error'] = TRUE;
        $response['error_msg'] = 'No employees have been added to this profile.';
        echo json_encode($response);
    }
} else {
    $response['error'] = TRUE;
    $response['error_msg'] = 'You have not logged in to your store''s account, please log in first.';
    echo json_encode($response);
}
?>

我想在上面的注释空间中有一个employee_list对象。像这样：

$response['employee_list']['0'] = $employee_list['0'];
$response['employee_list']['1'] = $employee_list['1'];
$response['employee_list']['2'] = $employee_list['2'];

等等...

之后，JSONObject返回到安卓应用程序，内容将在列表视图中列出。我需要一个 for 循环（我认为），因为员工编号永远不会知道，因为每个商店都可以根据需要添加和删除员工。有人可以指出我正确的方向，并建议我是否在其余代码中使用正确的方法。谢谢。

首先，在DB_Functions.php中，您应该返回mysqli_result对象。因此，您的DB_Functions应该是这样的：

public function getEmployeeList($name) {
    $stmt = $this->con->prepare("SELECT employee_name FROM employees WHERE name = ?");
    $stmt->bind_param('s', $name);
    if ($stmt->execute()) {
        // we get the mysqli_result object without calling the fetch_assoc() on it
        $result = $stmt->get_result();
        $stmt->close();
        // if the count is less than 1, no result found, hence return null
        if ($result->num_rows < 1) {
            return null;
        } else {
            // we return the mysqli_result object without calling the fetch_assoc() on it
            return $result;
        }
    }
}

在你的员工.php中，你想要的是这样的东西：

<?php
require_once 'include/DB_Functions.php';
$db = new DB_Functions();
$response = array('error' => FALSE);
if (isset($_POST['name'])) {
    $name = $_POST['name'];
    $result = $db->getEmployeeList($name);
    // do an early check for if result returns null or is not set
    if (is_null($result) || !$result) {
        $response['error'] = TRUE;
        $response['error_msg'] = 'No employees have been added to this profile.';
    } else {
        $response['error'] = FALSE;
        //EMPLOYEE LIST OBJECT HERE
        // since $result->fetch_assoc() returns one row at a time, you want to loop through each row and get the appropriate data
        while ($row = $result->fetch_assoc()) {
            // inject the current into the employee_list array
            $response['employee_list'][] = $row;
        }
    }
} else {
    $response['error'] = TRUE;
    $response['error_msg'] = 'You have not logged in to your store''s account, please log in first.';
}
// echo response gets called no matter what
echo json_encode($response);

希望对你有帮助