需要使用"/projects?technologies=AI&industries=Security“处理查询字符串nodeJS

Question

这是我第一次创建查询字符串参数，所以请帮助我。

• 我想过滤基于数据的json字段“技术、行业和成熟度”--每个领域(技术、行业和成熟度)都有一系列元素。我想过滤每个字段的元素。在这里输入图像描述
• 我编写了后端代码：

​

const { pick } = require('lodash');

const toArray = props =>
  (props && props.split ? props.split(',') : props || []).map(item => item.trim());
 console.log("this is in filter props: ", toArray)

module.exports = (entities, query) => {

  const fields = toArray(query.pick);
  const industries = toArray(query.industries);
  const technologies = toArray(query.technologies);
  const maturity = toArray(query.maturity);


  return entities.reduce((accumulator, entity) => {
    const hasTecnology = technologies.length

      ? technologies.every(technology => entity.technologies && entity.technologies.includes(technology))
      : true;
    
    const hasIndustury = industries.length
      ? industries.every(industry => entity.industries && entity.industries.includes(industry))
      : true;
    
      const hasMaturity = maturity.length
      ? maturity.every(maturty => entity.maturity && entity.maturity.includes(maturty))
      : true;
    const condition = hasTecnology && hasIndustury && hasMaturity;

    return condition
      ? [...accumulator, fields.length ? pick(entity, fields) : entity]
      : accumulator;
  }, []);
};

​

• 我的控制器功能：

​

  async find(ctx) {
    const { query } = ctx.request;
    const email = ctx.state.user.email; 
    const userProjects = strapi.cache.users[email].displayCatalog;

    let projects = strapi.cache.projects.filter(project => userProjects.includes(project.slug));

    if (query.pick || query.industries || query.technologies || query.maturity) {
      projects = filter(projects, query);
    }
    return buildResponse(projects.map(project => hideSensitive(project, userProjects, email)));
  },

​

• 此代码仅对单个参数"/projects?technologies=AI“有效，但我正在尝试获取此类型的filter usign。

注:如果我们选择“技术”字段的“前安全”元素或行业字段的“前人工智能”元素，则数据将同时显示“技术安全”和“行业AI.”。

请帮帮我

Answer 1

您可以查询所有字段。

把这个放在controller.js文件中。

        const queryObj = { ...req.query };
        let queryStr = JSON.stringify(queryObj);
     
        queryStr = queryStr.replace(/\b(gt|gte|lt|lte|eq|ne)\b/g, match => `$${match}`);

        console.log(queryStr);

       const doc1 = await Model.find(JSON.parse(queryStr));

在上述代码中

api/projects?technologies[eq]="AI"&industries[eq]="Security"

Answer 2

• 谢谢你的建议，我修改了一些代码，得到了解决方案。

​

const toArray = props =>
  (props && props.split ? props.split(',') : props || []).map(item => item.trim());

const hasField = (entity, fields) => {
  if (!fields.length) {
    return false;
  }
  for (let i = 0; i < fields.length; i + 1) {
    if (entity && entity.includes(fields[i])) {
      return true;
    }
  }
  return false;
};

module.exports = (entities, query) => {
  const industries = toArray(query.industries);
  const technologies = toArray(query.technologies);
  const maturity = toArray(query.maturity);

  return entities.filter(entity => {
    const hasTecnology = hasField(entity.industries, industries);
    const hasIndustury = hasField(entity.technologies, technologies);
    const hasMaturity = hasField(entity.maturity, maturity);

    const condition = hasTecnology || hasIndustury || hasMaturity;

 
    if (condition) {
      return entity;
    }
    return null;
  }, []);
};

​

• 我的控制器也会一样