如何从 NodeJS 中的 MySQL 查询中生成嵌套的 JSON 结果？答案

【问题标题】：How to make nested JSON result from MySQL query in NodeJS?如何从 NodeJS 中的 MySQL 查询中生成嵌套的 JSON 结果？
【发布时间】：2020-09-07 17:06:26
【问题描述】：

我有 2 个 MySQL 表，我们称它为 kj，它是 jj 的父表作为子表。目前，我在我的 NodeJS 代码中编写了 SQL 查询以返回此输出：

[{
    "id_kj": 1,
    "title_kj": "title1",
    "description_kj": "description1",
    "image_kj": "image1",
    "id_jj": 66,
    "title_jj": "title1",
    "description_jj": "description1",
    "image_jj": "image1"
  },
  {
    "id_kj": 1,
    "title_kj": "title1",
    "description_kj": "description1",
    "image_kj": "image1",
    "id_jj": 67,
    "title_jj": "title1",
    "description_jj": "description1",
    "image_jj": "image1"
  },
  {
    "id_kj": 1,
    "title_kj": "title1",
    "description_kj": "description1",
    "image_kj": "image1",
    "id_jj": 68,
    "title_jj": "title1",
    "description_jj": "description1",
    "image_jj": "image1"
  },
  {
    "id_kj": 2,
    "title_kj": "title2",
    "description_kj": "description2",
    "image_kj": "image2",
    "id_jj": 71,
    "title_jj": "title1",
    "description_jj": "description2",
    "image_jj": "image2"
  },
  {
    "id_kj": 2,
    "title_kj": "title2",
    "description_kj": "description2",
    "image_kj": "image2",
    "id_jj": 69,
    "title_jj": "title1",
    "description_jj": "description2",
    "image_jj": "image2"
  },
  {
    "id_kj": 2,
    "title_kj": "title2",
    "description_kj": "description2",
    "image_kj": "image2",
    "id_jj": 70,
    "title_jj": "title1",
    "description_jj": "description2",
    "image_jj": "image2"
  },
  {
    "id_kj": 3,
    "title_kj": "title3",
    "description_kj": "description3",
    "image_kj": "image3",
    "id_jj": 72,
    "title_jj": "title3",
    "description_jj": "description3",
    "image_jj": "image3"
  },
  {
    "id_kj": 3,
    "title_kj": "title3",
    "description_kj": "description3",
    "image_kj": "image3",
    "id_jj": 73,
    "title_jj": "title3",
    "description_jj": "description3",
    "image_jj": "image3"
  },
  {
    "id_kj": 3,
    "title_kj": "title3",
    "description_kj": "description3",
    "image_kj": "image3",
    "id_jj": 74,
    "title_jj": "title3",
    "description_jj": "description3",
    "image_jj": "image3"
  }
]

如何使结果嵌套？我需要id_jj, title_jj, description_jj, image_jj 成为每个id_kj 键的孩子。这是我需要的结果：

[{
    "id_kj": 1,
    "title_kj": "title1",
    "description_kj": "description1",
    "image_kj": "image1",
    "children": [{
        "id_jj": 66,
        "title_jj": "title1",
        "description_jj": "description1",
        "image_jj": "image1"
      },
      {
        "id_jj": 67,
        "title_jj": "title1",
        "description_jj": "description1",
        "image_jj": "image1"
      },
      {
        "id_jj": 68,
        "title_jj": "title1",
        "description_jj": "description1",
        "image_jj": "image1"
      }
    ]
  },
  {
    "id_kj": 2,
    "title_kj": "title2",
    "description_kj": "description2",
    "image_kj": "image2",
    "children": [{
        "id_jj": 71,
        "title_jj": "title1",
        "description_jj": "description2",
        "image_jj": "image2"
      },
      {
        "id_jj": 69,
        "title_jj": "title1",
        "description_jj": "description2",
        "image_jj": "image2"
      },
      {
        "id_jj": 70,
        "title_jj": "title1",
        "description_jj": "description2",
        "image_jj": "image2"
      }
    ]
  },
  {
    "id_kj": 3,
    "title_kj": "title3",
    "description_kj": "description3",
    "image_kj": "image3",
    "children": [{
        "id_jj": 72,
        "title_jj": "title3",
        "description_jj": "description3",
        "image_jj": "image3"
      },
      {
        "id_jj": 73,
        "title_jj": "title3",
        "description_jj": "description3",
        "image_jj": "image3"
      },
      {
        "id_jj": 74,
        "title_jj": "title3",
        "description_jj": "description3",
        "image_jj": "image3"
      }
    ]
  }
]

这是我的代码：

const getData = async function() {
  return new Promise(function(resolve) {
    const query = "SELECT a.`id` AS 'id_kj', a.`title` AS 'title_kj', a.`description` AS 'description_kj', a.`image` AS 'image_kj',\n" +
      "b.`id` AS 'id_jj', b.`title` AS 'title_jj', b.`description` AS 'description_jj', b.`image` AS 'image_jj' FROM `kj` AS a LEFT JOIN `jj` AS b\n" +
      "ON a.`id` = b.`id_kj` GROUP BY a.`id`, b.`id` ORDER BY  a.`title` ASC, b.`title` DESC";
    con.query(query, function(error, results) {
      if (results.length > 0) {
        resolve(results);
      } else {
        resolve(null);
      }
    });
  });
};

module.exports = {
  getNestedData: async function(baseResponse) {
    const data = await getPelatihan();
    return data;
  },
};

我还创建了我的表的 SQL Fiddle 方案，这是小提琴链接： db-fiddle

任何帮助将不胜感激

谢谢

【问题讨论】：

标签： javascript mysql node.js

【解决方案1】：

一个应该非常有效且逻辑上相对简单的选项是在内存中执行连接。

我们通过连接字段 title_kj 创建 result_kj 表的映射，然后遍历 result_jj 表以填充 result_kj 子级。

let result_kj = [{ "id_kj": 1, "title_kj": "title2", "description_kj": "description1", "image_kj": "image1", "id_jj": 66, "title_jj": "title2", "description_jj": "description1", "image_jj": "image1" }, { "id_kj": 1, "title_kj": "title2", "description_kj": "description1", "image_kj": "image1", "id_jj": 67, "title_jj": "title2", "description_jj": "description1", "image_jj": "image1" }, { "id_kj": 1, "title_kj": "title1", "description_kj": "description1", "image_kj": "image1", "id_jj": 68, "title_jj": "title1", "description_jj": "description1", "image_jj": "image1" }, { "id_kj": 2, "title_kj": "title2", "description_kj": "description2", "image_kj": "image2", "id_jj": 71, "title_jj": "title1", "description_jj": "description2", "image_jj": "image2" }, { "id_kj": 2, "title_kj": "title2", "description_kj": "description2", "image_kj": "image2", "id_jj": 69, "title_jj": "title1", "description_jj": "description2", "image_jj": "image2" }, { "id_kj": 2, "title_kj": "title2", "description_kj": "description2", "image_kj": "image2", "id_jj": 70, "title_jj": "title1", "description_jj": "description2", "image_jj": "image2" }, { "id_kj": 3, "title_kj": "title3", "description_kj": "description3", "image_kj": "image3", "id_jj": 72, "title_jj": "title3", "description_jj": "description3", "image_jj": "image3" }, { "id_kj": 3, "title_kj": "title3", "description_kj": "description3", "image_kj": "image3", "id_jj": 73, "title_jj": "title3", "description_jj": "description3", "image_jj": "image3" }, { "id_kj": 3, "title_kj": "title3", "description_kj": "description3", "image_kj": "image3", "id_jj": 74, "title_jj": "title3", "description_jj": "description3", "image_jj": "image3" } ]; 
let result_jj = [{ "id_jj": 66, "title_jj": "title1", "description_jj": "description1", "image_jj": "image1" }, { "id_jj": 67, "title_jj": "title1", "description_jj": "description1", "image_jj": "image1" }, { "id_jj": 68, "title_jj": "title1", "description_jj": "description1", "image_jj": "image1" }, { "id_jj": 71, "title_jj": "title2", "description_jj": "description2", "image_jj": "image2" }, { "id_jj": 69, "title_jj": "title2", "description_jj": "description2", "image_jj": "image2" }, { "id_jj": 70, "title_jj": "title2", "description_jj": "description2", "image_jj": "image2" }, { "id_jj": 72, "title_jj": "title3", "description_jj": "description3", "image_jj": "image3" }, { "id_jj": 73, "title_jj": "title3", "description_jj": "description3", "image_jj": "image3" }, { "id_jj": 74, "title_jj": "title3", "description_jj": "description3", "image_jj": "image3" } ]; 
   
// Sort by the relevant key
const sortKey = "title_kj";
result_kj.sort((a, b) => {
    if (a[sortKey] < b[sortKey]) {
        return -1;
    }
    if (a[sortKey] > b[sortKey]) {
        return 1;
    }
    return 0;
})

let kjMap = result_kj.reduce((map, row) => {
    key = row["title_kj"]; 
    map[key] = row;
    return map;
}, {})

let resultMap = result_jj.reduce((map, row) => { 
    let key = row["title_jj"];
    if (map[key]) { 
        if (!map[key].children) map[key].children = [];
        map[key].children.push(row);
    }
    return map;
}, kjMap)

let result = Object.values(resultMap);
console.log("Result:", result);

【讨论】：

您好，感谢您的回答。如果我错了，请纠正我，为了生成 result_kj 和 result_jj 我需要 2 个查询？
嗨@MANsyori，你绝对做到了！但是，这 2 个查询可能比 1 个复杂的连接查询占用更少的资源。在任何情况下都不会有很大的不同。
哇，它有效！谢谢。如果你不介意，我还有一个问题。如何安排父结果按字母顺序排序？ jj 结果映射已按原样排序。在父查询中使用 ORDER BY title ASC 不起作用。
如果我们愿意，我们可以在内存中进行排序，尽管它也应该在 SQL 领域中工作。为了完整起见，我将添加一个在内存中排序的示例！
parent没有排序的原因是我把parent中的“key”改成id_kj，把children的“key”改成id_kj，默认排序是基于@ 987654328@。在这种情况下，我最终添加了一个新功能并按照您所说的在内存中排序，现在一切都完成了！再次感谢。干杯。