如何通过名称、角色、分支三个多条件过滤对象数组

实现思路 #

• 三个过滤条件为且的关系，只有满足所有非空过滤条件的对象才会被保留
• 过滤参数为空时（name为空字符串、role/branch数组为空）自动跳过对应条件的校验
• 数组类型的过滤参数（role、branch）只要和目标对象的对应数组存在至少一个共同元素，即判定符合该条件

代码实现（JavaScript） #

function filterUserList(userList, filterName, filterRole, filterBranch) {
    // 所有过滤条件为空时直接返回原列表
    const isAllFilterEmpty = !filterName && filterRole.length === 0 && filterBranch.length === 0;
    if (isAllFilterEmpty) return userList;

    return userList.filter(item => {
        // 名称匹配，可根据需求把模糊匹配的includes改为全匹配的===
        const nameMatch = !filterName ? true : item.userName.includes(filterName);
        // 角色匹配，可根据需求把存在交集的some改为全部包含的every
        const roleMatch = filterRole.length === 0 ? true : filterRole.some(role => item.roleArray.includes(role));
        // 分支机构匹配，可根据需求把存在交集的some改为全部包含的every
        const branchMatch = filterBranch.length === 0 ? true : filterBranch.some(branch => item.branchArray.includes(branch));

        return nameMatch && roleMatch && branchMatch;
    })
}

// 测试调用
const list = [{
    "userName": "name1",
    "roleArray": [
        "role1",
        "role2"
    ],
    "branchArray": [
        "branch1",
        "branch2",
        "branch3"
    ]
},
{
    "userName": "name2",
    "roleArray": [
        "role2",
        "role3"
    ],
    "branchArray": [
        "branch3",
        "branch4",
        "branch5"
    ]
},
{
    "userName": "name3",
    "roleArray": [
        "role1",
        "role3"
    ],
    "branchArray": [
        "branch1",
        "branch2",
        "branch4"
    ]
}];
const filter_name = "";
const filter_role = ["role1","role3"];
const filter_branch = ["branch1","branch5"];

console.log(filterUserList(list, filter_name, filter_role, filter_branch))

内容的提问来源于stack exchange，提问作者Tuan Do Huu