复选框以过滤数据 php jQuery AJAX

问：

我有一个名为“活动”的复选框，当员工进入系统时，它总是被选中，它将显示活动工作人员的数据。现在，如果我取消选中该复选框，它应该显示非活动工人的数据，而在我的情况下，它不起作用。请检查我的代码，如果您为我提供一些解决方案，它将对我有所帮助

这是我的复选框代码

<div class="col-lg-3">
    <label for="activeWorker">
        <input type="checkbox" id="activeWorker" name="activeWorker" value="active" checked> Active Workers
    </label>
</div>

这是我根据所选日期范围重新加载数据的函数代码

function reload() {
    var dateFrom = $("#dateFrom").val();
    var dateTo = $("#dateTo").val();
    var coid = $("#selectCompanyId").val();
    var activeStatus = $("#activeWorker").val();

    var dataURL;

    alert("You have clicked the reload button. Reloading the data for the date range:" + dateFrom + " to " + dateTo + " For Company : " + coid);

    if (activeTab == "#tab-1") { // Pending
        dataURL = "data/data_workerSurvey.php?mod=pending&dateFrom=" + dateFrom + "&dateTo=" + dateTo + "&coid";
        $('#tab1_tblSurveyPending').DataTable().ajax.url(dataURL).load();
    } else if (activeTab == "#tab-2") { // Completed
        dataURL = "data/data_workerSurvey.php?mod=completed&dateFrom=" + dateFrom + "&dateTo=" + dateTo + "&coid=" + coid;
        $('#tab2_tblSurveyCompleted').DataTable().ajax.url(dataURL).load();
    } else { // All
        dataURL = "data/data_workerSurvey.php?mod=all&dateFrom=" + dateFrom + "&dateTo=" + dateTo + "&coid=" + coid;
        $('#tab3_tblSurveyAll').DataTable().ajax.url(dataURL).load();
    }
}

这是我的数据库代码

case "pending":
    $dateFrom = $_GET['dateFrom']; // Every time will have value
    $dateTo = $_GET['dateTo']; // Every time will have value
    $coid = $_GET['coid']; // This is not certain, depends on user selection
    $activeStatus = TRUE; // TRUE - Showing active worker; FALSE - showing inactive worker

    if ($activeStatus) {
        $whereClause = " AND a.biodata_status_id IN ('E', 'P') ";
    } else {
        $whereClause = " AND a.biodata_status_id = 'N' ";
    }

    if ($coid !== "" && $coid !== null) {
        $sql = "SELECT a.worker_id, a.worker_name, a.batchno, a.passport_no, a.company, a.arrival_date, a.plks, a.co_id, IF(DATEDIFF(a.plks, a.arrival_date) < 365 * 2, DATE_ADD(a.plks, INTERVAL 1 YEAR), a.plks) AS contract_end_date
            FROM rbs_worker a
            LEFT JOIN tbl_contract_extension_survey b ON (b.workerid = a.id AND b.status = 1 AND b.contract_end_date >= a.plks)
            LEFT JOIN rbs_payroll_company c ON c.co_id = a.co_id
            WHERE a.worker_id IS NOT NULL AND a.arrival_date IS NOT NULL AND a.plks IS NOT NULL 
                AND a.status = 'A' 
                AND a.co_id = %s
                AND IF(DATEDIFF(a.plks, a.arrival_date) < 365 * 2, DATE_ADD(a.plks, INTERVAL 1 YEAR), a.plks) >= %s 
                AND IF(DATEDIFF(a.plks, a.arrival_date) < 365 * 2, DATE_ADD(a.plks, INTERVAL 1 YEAR), a.plks) <= %s AND b.id IS NULL" . $whereClause;
        $rs = DB::query($sql, $coid, $dateFrom, $dateTo);
    } else {
        $sql = "SELECT a.worker_id, a.worker_name, a.batchno, a.passport_no, a company, a.arrival_date, a.plks, a.co_id, IF(DATEDIFF(a.plks, a.arrival_date) < 365 * 2, DATE_ADD(a.plks, INTERVAL 1 YEAR), a.plks) AS contract_end_date 
            FROM rbs_worker a
            LEFT JOIN tbl_contract_extension_survey b ON (b.workerid = a.id AND b.status = 1 AND b.contract_end_date >= a.plks)
            LEFT JOIN rbs_payroll_company c ON c.co_id = a.co_id
            WHERE a.worker_id IS NOT NULL AND a.arrival_date IS NOT NULL AND a.plks IS NOT NULL 
                AND a.status = 'A' 
                AND IF(DATEDIFF(a.plks, a.arrival_date) < 365 * 2, DATE_ADD(a.plks, INTERVAL 1 YEAR), a.plks) >= %s 
                AND IF(DATEDIFF(a.plks, a.arrival_date) < 365 * 2, DATE_ADD(a.plks, INTERVAL 1 YEAR), a.plks) <= %s AND b.id IS NULL" . $whereClause;
        $rs = DB::query($sql, $dateFrom, $dateTo);
    }

我希望如果员工取消选中该框，他们应该能够查看非活动工作人员数据。