javascript - 请问php中如何将查询出来的结果数组转化成自己想要的格式，并在前台利用js输出到html中

Question

考试类型的表jx_exam_type，可后台添加内容

考试成绩的表jx_result，可后台添加内容

期中考试成绩表中的exam_id对应考试类型表中的id，也就是添加的成绩是属于期中还是期末

然后使用php查询

$sql="SELECT re.type, re.score, re.exam_id, et.title, DATE_FORMAT(et.addtime, '%Y-%m-%d') AS etime FROM jx_result AS re LEFT JOIN jx_exam_type AS et ON re.exam_id = et.id WHERE re.uid = '$uid' ORDER BY et.addtime DESC";

$result=$db->query($sql);

while($row=$result->fetch_assoc()){
    $arr[]=$row;
}

echo json_encode($arr);

输出的格式如下

[
    {
        "type": "语文",
        "score": "91",
        "exam_id": "2",
        "title": "三年级期末考试",
        "etime": "2017-06-02"
    },
    {
        "type": "英语",
        "score": "89",
        "exam_id": "2",
        "title": "三年级期末考试",
        "etime": "2017-06-02"
    },
    {
        "type": "数学",
        "score": "60",
        "exam_id": "2",
        "title": "三年级期末考试",
        "etime": "2017-06-02"
    },
    {
        "type": "数学",
        "score": "91",
        "exam_id": "1",
        "title": "三年级期中考试",
        "etime": "2017-05-25"
    },
    {
        "type": "语文",
        "score": "85",
        "exam_id": "1",
        "title": "三年级期中考试",
        "etime": "2017-05-25"
    },
    {
        "type": "英语",
        "score": "87",
        "exam_id": "1",
        "title": "三年级期中考试",
        "etime": "2017-05-25"
    }
]

请问我如何才能将以上输出的json格式变为以下这种

{
    "title": "三年级期中考试",
    "etime": "2017-05-25",
    "exam_id": [
        {
            "type": "数学",
            "score": "91",
            "exam_id": "1"
        },
        {
            "type": "语文",
            "score": "85",
            "exam_id": "1"
        },
        {
            "type": "英语",
            "score": "87",
            "exam_id": "1"
        }
    ],
    "title": "三年级期末考试",
    "etime": "2017-06-02",
    "exam_id": [
        {
            "type": "语文",
            "score": "91",
            "exam_id": "2"
        },
        {
            "type": "英语",
            "score": "89",
            "exam_id": "2"
        },
        {
            "type": "数学",
            "score": "60",
            "exam_id": "2"
        }
    ]
}

变为以上这种格式后输出到前台，通过JS来输出到html上面
（可能我写的想要的格式有问题，不过大概意思就是将原来的数据根据exam_id来归类一下再输出）

目前正在学习中，很多地方不是很懂，求指教~~谢谢

Answer 1

这是我理解的架构,这样前端应该可以遍历到

while ($row = $result->fetch_assoc()) {
    $scorearr = array("type" => $row["type"], "score" => $row["type"], "exam_id" => $row["exam_id"]);
    if (isset($arr[$row["exam_id"]])) {
        $arr[$row["exam_id"]]["exam_id"][] = $scorearr;
    } else {
        $arr[$row["exam_id"]]["title"] = $row["title"];
        $arr[$row["exam_id"]]["etime"] = $row["etime"];
        $arr[$row["exam_id"]]["exam_id"][0] = $scorearr;
    }
}

$result = array();
foreach($arr as $val) {
    $result[] = $val;
}
echo $json_encode($result)

Answer 2

应该是少包了一层，可以遍历一下你的数组，然后就可以得到你想要的了，然后在json转码就可以了

Answer 3

你取数据和处理数据的逻辑有问题，不过你这种的话也可以写

<?php
$str = '[
    {
        "type": "语文",
        "score": "91",
        "exam_id": "2",
        "title": "三年级期末考试",
        "etime": "2017-06-02"
    },
    {
        "type": "英语",
        "score": "89",
        "exam_id": "2",
        "title": "三年级期末考试",
        "etime": "2017-06-02"
    },
    {
        "type": "数学",
        "score": "60",
        "exam_id": "2",
        "title": "三年级期末考试",
        "etime": "2017-06-02"
    },
    {
        "type": "数学",
        "score": "91",
        "exam_id": "1",
        "title": "三年级期中考试",
        "etime": "2017-05-25"
    },
    {
        "type": "语文",
        "score": "85",
        "exam_id": "1",
        "title": "三年级期中考试",
        "etime": "2017-05-25"
    },
    {
        "type": "英语",
        "score": "87",
        "exam_id": "1",
        "title": "三年级期中考试",
        "etime": "2017-05-25"
    }
]';

$arr = json_decode($str,true);

$result = [];
$examId = [];
foreach($arr as $key=>$val){

    if(!isset($result[$val['exam_id']])){
        $result[$val['exam_id']]['title'] = $val['title'];
        $result[$val['exam_id']]['etime'] = $val['etime'];
    }

    $result[$val['exam_id']]['exam_id'][] = array("type"=>$val['type'],"score"=>$val['score'],"exam_id"=>$val['exam_id']);

}


echo json_encode(array_values($result));

不过你这种取数据的方式不建议，如果数据复杂点 量大点，处理起来就会麻烦