Question

尝试使用一系列过滤器来搜索商店位置的javascript对象，我需要使用过滤器数组中的两个过滤器获得完全匹配，过滤器数组现在有2个项目，但会有更多。下面的代码有效，但它找到了每个商店的匹配项。由于.json文件中的一个商店没有ATM机，因此它不应出现在控制台中。任何帮助将不胜感激！

上传.json：

[
{
    "storenumber": "5210",
    "services": [
        {
            "img": "/locator/img/icons/atm.svg",
            "alt": "ATM Machine"
        },
        {
            "img": "/locator/img/icons/carwash.svg",
            "alt": "Car Wash"
        }
    ]
},
{
    "storenumber": "1066",
    "services": [
        {
            "img": "/locator/img/icons/carwash.svg",
            "alt": "Car Wash"
        }

    ]
}
]

的.js：

var jsonData,
    filteredJsonData = [],
    filters = ["Car Wash","ATM Machine"];

   $.getJSON("/locator/js/locations.json")
    .done(function(data) {
        jsonData = data;

        $.each(jsonData,function(locationsIndex,locationsItem){
            var services = locationsItem.services;

            $.each(services,function(servicesIndex,servicesItem){
                if (servicesItem.alt && ($.inArray(servicesItem.alt,filters) != -1)) {
                    // This is finding matches in the filters array, but i need to match all items in the array
                    console.log("Match Found: " + servicesItem.alt + " at Index " + locationsIndex)
                    filteredJsonData.push(locationsItem)
                }
            })
        })  
        console.log(filteredJsonData)
    })

Answer 1

您可以使用#include <math.h> #include <stdio.h> #include <string.h> #include <stdlib.h> #include <assert.h> #include <limits.h> #include <stdbool.h> int main() { int n; scanf("%d",&n); int arr[n]; int i; int p = 0; int arr_i; for (arr_i = 0; arr_i < n; arr_i++) { scanf("%d",&arr[arr_i]); } for (i = 0; i < n; i++) { int arr_index=i; if (arr[arr_index] > 0) { p++; } } printf("%d",p); return 0; }，filter和some的组合：

Set

const jsonData = [{"storenumber": "5210","services": [{ "img": "/locator/img/icons/atm.svg","alt": "ATM Machine"},{"img": "/locator/img/icons/carwash.svg","alt": "Car Wash"}]}, {"storenumber": "1066","services": [{"img": "/locator/img/icons/carwash.svg","alt": "Car Wash"}]}],
    filters = ["Car Wash","ATM Machine"];

const locations = jsonData.filter( location => {
    const filterSet = new Set(filters);
    return location.services.some( service => {
        filterSet.delete(service.alt);
        return !filterSet.size;
    });
});
console.log(locations);

注意：我为了这个演示片段而删除了Ajax调用。

适用于旧版浏览器

一个可以在旧浏览器（ES3，主要是IE）中使用的版本，使用jQuery：

.as-console-wrapper { max-height: 100% !important; top: 0; }

var jsonData = [{"storenumber": "5210","services": [{ "img": "/locator/img/icons/atm.svg","alt": "ATM Machine"},{"img": "/locator/img/icons/carwash.svg","alt": "Car Wash"}]}, {"storenumber": "1066","services": [{"img": "/locator/img/icons/carwash.svg","alt": "Car Wash"}]}],
    filters = ["Car Wash","ATM Machine"];

var locations = $.grep(jsonData, function (location) {
    var filterSet = {};
    for (var i = 0; i < filters.length; i++) {
        filterSet[filters[i]] = 1;
    }
    var filtersLeft = filters.length;
    $.each(location.services, function (i, service) {
        if (filterSet[service.alt]) {
            filtersLeft--;
            filterSet[service.alt] = 0; 
        };
        return filtersLeft > 0;
    });
    return filtersLeft === 0;
});
console.log(locations);

.as-console-wrapper { max-height: 100% !important; top: 0; }

使用Jquery Each在Javascript对象和数组中查找匹配项

1 个答案:

适用于旧版浏览器