Question

我正在制作一个应用程序，它会在列表视图中按特定商店列出员工。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employees.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格中有一个employee_list对象。类似的东西：

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

等等...等等......

之后，JSONObject返回到Android应用程序，内容将列在listview中。我需要一个for循环（我认为），因为员工编号永远不会被知道，因为每个商店都可以按照自己的意愿添加和删除员工。有人可以指出我正确的方向，并建议我是否使用正确的方法，其余的代码。感谢。

Answer 1

首先，在DB_Functions.php中，您应该返回mysqli_result对象。因此你的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;
        }
    }
}

在您的employees.php中，您想要的是这样的：

<?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);

希望有所帮助

如何在PHP中使用for循环来返回mysql数据

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