有什么方法可以“连接”两个 javascript 数组的内容，就像在 SQL 中进行连接一样

我有两个数组：问题和用户配置文件

• The userProfiles: [] 数组包含{ id, name } objects
• The questions: [] 数组包含{ id, text, createdBy } objects

The createdBy问题中的整数始终是 id 值之一userProfiles.

有没有一种方法可以“连接”数组，就像使用数据库时连接两个 SQL 表一样。

我需要的最终结果是一个包含

{ id, text, name }

对应的SQL是：

SELECT u.id, q.text, u.name 
FROM userProfiles u 
JOIN questions q ON q.createdBy=u.id

我认为你想要的是内部联接，这很简单，可以用 JavaScript 实现：

const innerJoin = (xs, ys, sel) =>
    xs.reduce((zs, x) =>
    ys.reduce((zs, y) =>        // cartesian product - all combinations
    zs.concat(sel(x, y) || []), // filter out the rows and columns you want
    zs), []);

出于演示目的，我们将使用以下数据集（谢谢@AshokDamani）：

const userProfiles = [
    {id: 1, name: "Ashok"},
    {id: 2, name: "Amit"},
    {id: 3, name: "Rajeev"},
];

const questions = [
    {id: 1, text: "text1", createdBy: 2},
    {id: 2, text: "text2", createdBy: 2},
    {id: 3, text: "text3", createdBy: 1},
    {id: 4, text: "text4", createdBy: 2},
    {id: 5, text: "text5", createdBy: 3},
    {id: 6, text: "text6", createdBy: 3},
];

您将这样使用它：

const result = innerJoin(userProfiles, questions,
    ({id: uid, name}, {id, text, createdBy}) =>
        createdBy === uid && {id, text, name});

用 SQL 术语来说，这类似于：

SELECT questions.id, questions.text, userProfiles.name
FROM userProfiles INNER JOIN questions
ON questions.createdBy = userProfiles.id;

把它们放在一起：

const innerJoin = (xs, ys, sel) =>
    xs.reduce((zs, x) =>
    ys.reduce((zs, y) =>        // cartesian product - all combinations
    zs.concat(sel(x, y) || []), // filter out the rows and columns you want
    zs), []);

const userProfiles = [
    {id: 1, name: "Ashok"},
    {id: 2, name: "Amit"},
    {id: 3, name: "Rajeev"},
];

const questions = [
    {id: 1, text: "text1", createdBy: 2},
    {id: 2, text: "text2", createdBy: 2},
    {id: 3, text: "text3", createdBy: 1},
    {id: 4, text: "text4", createdBy: 2},
    {id: 5, text: "text5", createdBy: 3},
    {id: 6, text: "text6", createdBy: 3},
];

const result = innerJoin(userProfiles, questions,
    ({id: uid, name}, {id, text, createdBy}) =>
        createdBy === uid && {id, text, name});

console.log("Open your browser console to see the output.");

console.table(result);

Edit:然而这并不是最好的解决方案。由于上述解决方案循环通过笛卡尔积它需要O(m × n)是时候跑步了。通过一点点修改我们就可以让它运行O(m + n)时间-@pebbl首先找到它:

const equijoin = (xs, ys, primary, foreign, sel) => {
    const ix = xs.reduce((ix, row) => // loop through m items
        ix.set(row[primary], row),    // populate index for primary table
    new Map);                         // create an index for primary table

    return ys.map(row =>              // loop through n items
        sel(ix.get(row[foreign]),     // get corresponding row from primary
        row));                        // select only the columns you need
};

现在您可以按如下方式使用它：

const result = equijoin(userProfiles, questions, "id", "createdBy",
    ({name}, {id, text}) => ({id, text, name}));

把它们放在一起：

const equijoin = (xs, ys, primary, foreign, sel) => {
    const ix = xs.reduce((ix, row) => ix.set(row[primary], row), new Map);
    return ys.map(row => sel(ix.get(row[foreign]), row));
};

const userProfiles = [
    {id: 1, name: "Ashok"},
    {id: 2, name: "Amit"},
    {id: 3, name: "Rajeev"},
];

const questions = [
    {id: 1, text: "text1", createdBy: 2},
    {id: 2, text: "text2", createdBy: 2},
    {id: 3, text: "text3", createdBy: 1},
    {id: 4, text: "text4", createdBy: 2},
    {id: 5, text: "text5", createdBy: 3},
    {id: 6, text: "text6", createdBy: 3},
];

const result = equijoin(userProfiles, questions, "id", "createdBy",
    ({name}, {id, text}) => ({id, text, name}));

console.log("Open your browser console to see the output.");

console.table(result);