Node.js 根据其他表的行建议结果

我在postgresql数据库中有两个表。表1=空缺，表2=搜索查询。我想根据用户之前的搜索查询向他们推荐职位空缺（只有最后10个搜索查询是重要的）

当用户搜索空缺时，他必须输入一些关键字，并且他可以选择一个位置（单个）、一个公司（单个）、多个部门（多个）、多个工作域（多个）和就业类型（多个）

行业和工作领域是关键，如“建筑”、“销售”、“技术”。就业类型、公司和地点是UUID，在其他表中有更多的规定（因为新公司可以加入，新地点可以加入，但不可能添加新部门或工作域）

我的搜索查询表如下所示：

CREATE TABLE search_queries(
  search_query_id uuid DEFAULT gen_random_uuid() NOT NULL,
  user_id uuid,
  keywords text[] NOT NULL,
  company_id uuid,
  location_id uuid,
  sectors text[],
  job_domains text[],
  employment_types uuid[]
);

--- EXAMPLE ROW
search_query_id = 9dde0eae-dfda-4e3d-9e87-55076111eee2;
user_id = 1d52fd2f-a830-4bf2-8249-e9cdf3681c3a;
keywords = {keyword1,keyword2};
location_id = NULL;
company_id = NULL;
sectors = {construction,sales};
job_domains = NULL;
employment_types = {0a1a702e-2f45-4c4d-b870-a41f83149835};

const keywordsArray = ['keyword1', 'keyword2', ...];
const locationIdArray = ['uuid1', ...];
const sectorsArray = ['construction', 'sales', ...];

// 1) join all keywords together
const keywordsQuery = `to_tosvector(title || ' ' || keywords) @@ to_tsquery('${keywordsArray.join(' | ')}')`;
// i'm using the | operator to search for 'OR' instead of 'AND' because not all keywords must be present in a vacancy to match.

// 2) create a postgresql array of all location ID's and check if a location ID exists in a vacancy.
// I only create the dynamic query part if the locationIdArray has values. If not, I just skip it.
const locationIdQuery = locationIdArray.length > 0 ? `OR location_id = ANY(ARRAY[${locationIdArray.join(',')}]::uuid[])` : '';

// 3) create a postgresql array of all sectors and check if a vacancy sector exists in the array
const sectorsQuery = sectorsArray.length > 0 ? `OR sector = ANY(ARRAY[${sectors.join(',')}])`: '';

// ... and so on for the other properties

const query = `
SELECT
    title,
    company,
    timestamp,
    (tsrank(to_tosvector(title || ' ' || keywords),to_tsquery('${keywordsArray.join(' | ')}'))::float AS relevance
FROM
    vacancies
WHERE
    ${keywordsQuery}
    ${locationIdQuery}
    ${sectorsQuery}
ORDER BY
    relevance DESC,
    timestamp DESC
`

在示例行中，用户搜索了“施工”和“销售”部门中的工作，工作类型为“临时”（UUID表示工作类型为临时），关键字为关键字1和关键字2

目前，当我想推荐作业时，我会做以下工作：获取用户的最后10个搜索查询，并将所有不同字段合并到数组中。我用最后10行中的所有关键字创建了一个关键字数组（我删除了重复项），我用所有位置ID创建了一个位置数组（我删除了重复项和空值）。。。等等然后我会得到这样的结果：

CREATE TABLE search_queries(
  search_query_id uuid DEFAULT gen_random_uuid() NOT NULL,
  user_id uuid,
  keywords text[] NOT NULL,
  company_id uuid,
  location_id uuid,
  sectors text[],
  job_domains text[],
  employment_types uuid[]
);

--- EXAMPLE ROW
search_query_id = 9dde0eae-dfda-4e3d-9e87-55076111eee2;
user_id = 1d52fd2f-a830-4bf2-8249-e9cdf3681c3a;
keywords = {keyword1,keyword2};
location_id = NULL;
company_id = NULL;
sectors = {construction,sales};
job_domains = NULL;
employment_types = {0a1a702e-2f45-4c4d-b870-a41f83149835};

const keywordsArray = ['keyword1', 'keyword2', ...];
const locationIdArray = ['uuid1', ...];
const sectorsArray = ['construction', 'sales', ...];

// 1) join all keywords together
const keywordsQuery = `to_tosvector(title || ' ' || keywords) @@ to_tsquery('${keywordsArray.join(' | ')}')`;
// i'm using the | operator to search for 'OR' instead of 'AND' because not all keywords must be present in a vacancy to match.

// 2) create a postgresql array of all location ID's and check if a location ID exists in a vacancy.
// I only create the dynamic query part if the locationIdArray has values. If not, I just skip it.
const locationIdQuery = locationIdArray.length > 0 ? `OR location_id = ANY(ARRAY[${locationIdArray.join(',')}]::uuid[])` : '';

// 3) create a postgresql array of all sectors and check if a vacancy sector exists in the array
const sectorsQuery = sectorsArray.length > 0 ? `OR sector = ANY(ARRAY[${sectors.join(',')}])`: '';

// ... and so on for the other properties

const query = `
SELECT
    title,
    company,
    timestamp,
    (tsrank(to_tosvector(title || ' ' || keywords),to_tsquery('${keywordsArray.join(' | ')}'))::float AS relevance
FROM
    vacancies
WHERE
    ${keywordsQuery}
    ${locationIdQuery}
    ${sectorsQuery}
ORDER BY
    relevance DESC,
    timestamp DESC
`

然后，我创建一个动态查询。我现在是这样做的：

CREATE TABLE search_queries(
  search_query_id uuid DEFAULT gen_random_uuid() NOT NULL,
  user_id uuid,
  keywords text[] NOT NULL,
  company_id uuid,
  location_id uuid,
  sectors text[],
  job_domains text[],
  employment_types uuid[]
);

--- EXAMPLE ROW
search_query_id = 9dde0eae-dfda-4e3d-9e87-55076111eee2;
user_id = 1d52fd2f-a830-4bf2-8249-e9cdf3681c3a;
keywords = {keyword1,keyword2};
location_id = NULL;
company_id = NULL;
sectors = {construction,sales};
job_domains = NULL;
employment_types = {0a1a702e-2f45-4c4d-b870-a41f83149835};

const keywordsArray = ['keyword1', 'keyword2', ...];
const locationIdArray = ['uuid1', ...];
const sectorsArray = ['construction', 'sales', ...];

// 1) join all keywords together
const keywordsQuery = `to_tosvector(title || ' ' || keywords) @@ to_tsquery('${keywordsArray.join(' | ')}')`;
// i'm using the | operator to search for 'OR' instead of 'AND' because not all keywords must be present in a vacancy to match.

// 2) create a postgresql array of all location ID's and check if a location ID exists in a vacancy.
// I only create the dynamic query part if the locationIdArray has values. If not, I just skip it.
const locationIdQuery = locationIdArray.length > 0 ? `OR location_id = ANY(ARRAY[${locationIdArray.join(',')}]::uuid[])` : '';

// 3) create a postgresql array of all sectors and check if a vacancy sector exists in the array
const sectorsQuery = sectorsArray.length > 0 ? `OR sector = ANY(ARRAY[${sectors.join(',')}])`: '';

// ... and so on for the other properties

const query = `
SELECT
    title,
    company,
    timestamp,
    (tsrank(to_tosvector(title || ' ' || keywords),to_tsquery('${keywordsArray.join(' | ')}'))::float AS relevance
FROM
    vacancies
WHERE
    ${keywordsQuery}
    ${locationIdQuery}
    ${sectorsQuery}
ORDER BY
    relevance DESC,
    timestamp DESC
`

我最终得到了这样一个查询：

CREATE TABLE search_queries(
  search_query_id uuid DEFAULT gen_random_uuid() NOT NULL,
  user_id uuid,
  keywords text[] NOT NULL,
  company_id uuid,
  location_id uuid,
  sectors text[],
  job_domains text[],
  employment_types uuid[]
);

--- EXAMPLE ROW
search_query_id = 9dde0eae-dfda-4e3d-9e87-55076111eee2;
user_id = 1d52fd2f-a830-4bf2-8249-e9cdf3681c3a;
keywords = {keyword1,keyword2};
location_id = NULL;
company_id = NULL;
sectors = {construction,sales};
job_domains = NULL;
employment_types = {0a1a702e-2f45-4c4d-b870-a41f83149835};

const keywordsArray = ['keyword1', 'keyword2', ...];
const locationIdArray = ['uuid1', ...];
const sectorsArray = ['construction', 'sales', ...];

// 1) join all keywords together
const keywordsQuery = `to_tosvector(title || ' ' || keywords) @@ to_tsquery('${keywordsArray.join(' | ')}')`;
// i'm using the | operator to search for 'OR' instead of 'AND' because not all keywords must be present in a vacancy to match.

// 2) create a postgresql array of all location ID's and check if a location ID exists in a vacancy.
// I only create the dynamic query part if the locationIdArray has values. If not, I just skip it.
const locationIdQuery = locationIdArray.length > 0 ? `OR location_id = ANY(ARRAY[${locationIdArray.join(',')}]::uuid[])` : '';

// 3) create a postgresql array of all sectors and check if a vacancy sector exists in the array
const sectorsQuery = sectorsArray.length > 0 ? `OR sector = ANY(ARRAY[${sectors.join(',')}])`: '';

// ... and so on for the other properties

const query = `
SELECT
    title,
    company,
    timestamp,
    (tsrank(to_tosvector(title || ' ' || keywords),to_tsquery('${keywordsArray.join(' | ')}'))::float AS relevance
FROM
    vacancies
WHERE
    ${keywordsQuery}
    ${locationIdQuery}
    ${sectorsQuery}
ORDER BY
    relevance DESC,
    timestamp DESC
`

此查询检查关键字存在的所有空缺或位置ID位于数组或中的所有空缺 其中扇区位于数组中，结果根据相关性（和发布日期）排序

这是可行的，但我对这种方法有一些问题。

首先，它根据相关性对结果进行排序，但相关性仅基于关键词。我想订购“比赛”的结果。同时匹配关键字和位置ID的空缺在结果集中应高于仅匹配关键字的空缺

其次，我不知道这是否是正确的解决方案。目前，它是有效的，我将结果缓存在redis数据库中，因此这种“计算”不是每次用户刷新页面时都进行的，而是每小时进行一次

有人能帮我解决这个问题吗？其他人如何根据另一个表中的行“建议”内容？这是正确的方式，还是我在监督什么