Javascript 如何防止php（int）将null转换为0

我正在使用JQuery Ajax将POST变量发送到php，其中一些变量尚未定义，因此我声明它们并将它们

null

如下所示：

if ($(this).is('.filters')) {
    alert("works");
}
else {
    var type = status = bhk = null;
}
$.ajax({
    method: 'POST',
    url: '/search',
    data: {
        'do': getUrlParameter('do'),
        'use': use,
        'type': type, //null for now
        'status': status, //null for now
        'bhk': bhk, //null for now
        'city': $('select[name="city"]').val(),
        'zone': $('select[name="zone"]').val(),
        'page': page
    }
}).done(function(data) {
    if ($('#search').is(':visible'))
        $('#search').hide();

    var new_content = $(data).find('#search-filters, #scroll-to-list');
    $( '#results' ).html( new_content );

    $('html, body').animate({
        scrollTop: $('#scroll-to-list').offset().top
    }, 1000);
});

在php页面中，我声明并过滤它们：

$use            = isset( $_POST['use'] ) ? (int) $_POST['use'] : '';        // int AJAX
$filter_type    = isset( $_POST['type'] ) ? (int) $_POST['type'] : '';      // int AJAX
$filter_status  = isset( $_POST['status'] ) ? (int) $_POST['status'] : '';  // int AJAX
$filter_bhk     = isset( $_POST['bhk'] ) ? (int) $_POST['bhk'] : '';        // int AJAX
$filter_city    = isset( $_POST['city'] ) ? (int) $_POST['city'] : 0;         // int AJAX
$filter_zone    = isset( $_POST['zone'] ) ? (int) $_POST['zone'] : 0;         // int AJAX
$page_number    = isset( $_POST['page'] ) ? (int) $_POST['page'] : '';      // int AJAX

我正在使用这些变量为MySQLi Select创建附加条件，但如果它们不是null，而是0，则会出现错误：

if ($filter_type != NULL || $filter_type != '') {
    $filter_type = 'AND t2.type = ' . $filter_type;
} else $filter_type = NULL;
if ($filter_status != NULL || $filter_status != '') {
    $filter_status = 'AND t2.status = ' . $filter_status;
} else $filter_status = NULL;
if ($filter_bhk != NULL || $filter_bhk != '') {
    $filter_bhk  = 'AND t2.bhk = ' . $filter_bhk;
} else $filter_bhk = NULL;

if ($filter_city != 0) {
    $filter_city = 'AND t2.city = ' . $filter_city;
    $filter_zone = NULL;

    if ($filter_zone != 0) {
        $filter_zone = 'AND t2.zone = ' . $filter_zone;
    }
} else $filter_city = $filter_zone = NULL;

if ($stmt = $mysqli->prepare(' SELECT t1.id, t2.*
                               FROM ' . $table . ' t1
                               INNER JOIN property t2 ON t2.id = t1.id
                               WHERE t2.use = ?
                               ' . $filter_type
                                 . $filter_status
                                 . $filter_bhk
                                 . $filter_city
                                 . $filter_zone . '
                               LIMIT ?, ?')) {

---

除此之外，有没有更干净或优化的方法？我会很感激的，谢谢

将您的

NULL

从jquery右侧更改为

”

，也可以在ifs

$filter\u type！=''中更改？只需添加
| |$filter\u type！=“
我在第5行发现mysql错误接近
'000限制？，？”