Question

这里我想做的变量不是空的，我想在数据中传递一个变量，在这种情况下，区域是空的，所以我想传递city和listing类型的参数，但我不知道怎么办？

var city = "Karnadaka";
var area = "";
var listing type = "RENT";

$.ajax({
    type: 'GET',
    url: "http://www.domain.com/api/get/searchProperties",
    data: {
        area: area,
        city: city,
        listingType: listing_type
    },
    success: function(data) {
        console.log(data);
    }
});

Answer 1

您可以使用delete删除对象

中的对象

var city = "Karnadaka";
var area = "";
var listing_type = "RENT";

var data={
        area: area,
        city: city,
        listingType: listing_type
    }

for (k in data){
    if(data[k]=="") delete data[k];
}

$.ajax({
    type: 'GET',
    url: "http://www.domain.com/api/get/searchProperties",
    data: data,
    success: function(data) {
        console.log(data);
    }
});

Answer 2

试试这个：

var city = "Karnadaka";
var area = "";
var listing_type = "RENT";

var data = {};

if(city != '')
    data['city'] = city;

if(area != '')
    data['area'] = area;

if(listing_type != '')
    data['listing_type'] = listing_type;

$.ajax({
    type: 'GET',
    url: "http://www.domain.com/api/get/searchProperties",
    data: data,
    success: function(response) {
        console.log(response);
    }
});

仅将参数传递给AJAX数据中的非空变量

2 个答案: