Question

此代码适用于一条记录，但我正在尝试使用大约500条记录的数组...

var aaa = {
    aaa_id: 123,
    abbr_name: "j_doe",
    name: "john doe"
}
var bbb = {
    bbb_id: 789,
}


for (var attrname in bbb) {

    aaa[attrname] = bbb[attrname];

}

console.log(aaa.aaa_id)

输出：

Object {aaa_id: 123, abbr_name: "j_doe", name: "john doe", bbb_id: 789}

以下是我的JSON对其他记录的看法：

var aaaRecords [   {
        "first_name": "Hasheem",
        "last_name": "Thabeet",
        "aaa_player_id": "4562"
    },
    ...
]

var bbbRecords [{
        "first_name": "Hasheem",
        "last_name": "Thabeet",
        "aaa_player_id": "4562"
    },
    ....
]

关于如何使这个功能适用于这些数组的任何想法？非常感谢您的帮助！

Answer 1

已经过测试：

<html>
<head>
    <script src="//ajax.googleapis.com/ajax/libs/jquery/1.11.1/jquery.min.js"></script>
    <script language="javascript">

        var aaaRecords = [{
            "first_name": "Hasheem",
            "last_name": "Thabeet",
            "aaa_player_id": "456"
        },{
            "first_name": "Mariah",
            "last_name": "Carey",
            "aaa_player_id": "721"
        }];

        var bbbRecords = [{
            "first_name": "Hasheem",
            "last_name": "Thabeet",
            "bbb_player_id": "489"
        },{
            "first_name": "Mariah",
            "last_name": "Carey",
            "bbb_player_id": "198"
        }];

        $(document).ready(function() {
            mergeRecords(function() {
                displayRecords();
            });
        });

        function mergeRecords(callback) {
            for (var i = 0; i < aaaRecords.length; i++) {
                aFName = aaaRecords[i].first_name;
                aLName = aaaRecords[i].last_name;
                for (var j = 0; j < bbbRecords.length; j++) {
                    bFName = bbbRecords[j].first_name;
                    bLName = bbbRecords[j].last_name;
                    bPlayerID = bbbRecords[j].bbb_player_id;
                    if (aFName == bFName && aLName == bLName) {
                        aaaRecords[i].bbb_player_id = bPlayerID;
                    }
                }
            } callback();
        }

        function displayRecords() {
            for (var i = 0; i < aaaRecords.length; i++) {
                $('#results').append('<tr><td>' + 
                    aaaRecords[i].first_name + '</td><td>' +
                    aaaRecords[i].last_name + '</td><td>' +
                    aaaRecords[i].aaa_player_id + '</td><td>' +
                    aaaRecords[i].bbb_player_id + '</td></tr>');
            }
        }

    </script>
</head>

<body>
    <table id="results" border="1" cellpadding="2" cellspacing="2" width="400">
        <tr>
            <th>First Name:</th>
            <th>Last Name:</th>
            <th>Player ID (a):</th>
            <th>Player ID (b):</th>
        </tr>
        <!-- Display results here -->
    </table>
</body>
</html>

结果输出：

enter image description here

如何在保留两个值的同时将两个对象组合在不同的数组中？

1 个答案: