PHP get_curent_userの例

コード例 #1

ファイル: index.php プロジェクト: experimental-platform/office-cleaning-app

function done($conn, $done_type)
{
    switch ($done_type) {
        case "muell":
            $work_id = 1;
            break;
        case "pfand":
            $work_id = 2;
            break;
        case "ein":
            $work_id = 3;
            break;
        case "aus":
            $work_id = 4;
            break;
        case "kaffee":
            $work_id = 5;
            break;
    }
    $user_id = get_curent_user($conn, $work_id);
    $user_info = get_user_info($conn, $user_id);
    $count = $user_info['count_' . $done_type];
    $count++;
    $sql = "UPDATE user SET count_" . $done_type . "=" . $count . " WHERE id=" . $user_id;
    $result = $conn->query($sql);
    $desc = $done_type . '_done';
    $timedate = date("Y-m-d H:i:s");
    $sql = "INSERT INTO work_log (Description, Date, User_ID) VALUES ('" . $desc . "','" . $timedate . "','" . $user_id . "')";
    $result = $conn->query($sql);
}

コード例 #2

ファイル: abwesend.php プロジェクト: experimental-platform/office-cleaning-app

function done($conn, $done_type)
{
    switch ($done_type) {
        case "muell":
            $work_id = 1;
            break;
        case "pfand":
            $work_id = 2;
            break;
        case "ein":
            $work_id = 3;
            break;
        case "aus":
            $work_id = 4;
            break;
        case "kaffee":
            $work_id = 5;
            break;
    }
    $user_id = get_curent_user($conn, $work_id);
    $user_info = get_user_info($conn, $user_id);
    $count = $user_info['count_' . $done_type];
    $count++;
    $sql = "UPDATE user SET count_" . $done_type . "=" . $count . " WHERE id=" . $user_id;
    $result = $conn->query($sql);
}