请问大家，这两组数据，我如何做出筛选

Question

tab2:[
    {
        name: 111',
        numbers: 222
    },
 {
        name: 222,
        numbers: 666
    },
 {
        name: 333,
        numbers: 888
    }
]

tab3:{
        "fid": 111,
        "pigBatchEntry": [
            {
            "fid": 111,
            "parentId": 111,
            "pigBatchId": 333
           }
        ],
"fid": 111,
        "pigBatchEntry": [
            {
            "fid": 111,
            "parentId": 111,
            "pigBatchId": 666
           }
        ],
"fid": 111,
        "pigBatchEntry": [
            {
            "fid": 111,
            "parentId": 111,
            "pigBatchId": 777
           }
        ]
}

tab2 和 tab3 两组数据，请问我如何通过tab2的numbers属性值，和tab3的pigBatchId属性值，做比较，如果tab2的numbers == tab3的pigBatchId属性值，那就删除tab3的这组数据，请问如何用js实现

Answer 1

题主要的代码如下，有帮助麻烦点个采纳【本回答右上角】，谢谢~~

    var obj = {
        tab2: [
            {
                name: 111,
                numbers: 222
            },
            {
                name: 222,
                numbers: 666
            },
            {
                name: 333,
                numbers: 888
            }
        ],
        tab3: {
            "fid": 111,
            "pigBatchEntry": [
                {
                    "fid": 111,
                    "parentId": 111,
                    "pigBatchId": 333
                },
                {
                    "fid": 111,
                    "parentId": 111,
                    "pigBatchId": 666
                },
                {
                    "fid": 111,
                    "parentId": 111,
                    "pigBatchId": 777
                }
            ]
        }
    };


    for (var v of obj.tab2) {
        var index = obj.tab3.pigBatchEntry.findIndex(item => item.pigBatchId == v.numbers);
        if (index != -1) obj.tab3.pigBatchEntry.splice(index, 1);//找到numbers==pigBatchId的项下标执行删除

    }
    alert(JSON.stringify(obj.tab3))