Bootstraping Strapi角色权限

时间:2019-06-27 16:47:20

标签: javascript node.js graphql strapi

我是一位首次从事前端工作,stradi和javascript的开发人员。我希望有人可以怜悯我,并提供一个示例,说明如何通过bootstrap.js脚本设置公共角色权限。 node.js v10.16.0
Strapi v3.0.0-next.11
Graphql 14.3.1
MongoDB:3.6
全部在Windows 10上

在Strapi UI中,它是公共角色的角色和权限 Public Role 我想将这些框设置为CHECKED Toggle Checkboxes

另一个开发人员已使用bootstrap.js文件将项目添加到我们创建的服务(菜单)中。我什至不知道如何返回有关角色权限的最基本信息。 我的函数称为test(),我在其中搜索了示例,发现的最好的是在stackoverflow上: Strapi Plugin Route Default Permission

strapi.plugins['users-permissions'].models.role.find 

但我不知道如何使用它:

WORKING
function add_widgets_from_sheet(sheet_name, model_object){
  console.log(`adding ${sheet_name}`)
  let xlsxSheet = Sheets[sheet_name]
  const widgets = XLSX.utils.sheet_to_json(xlsxSheet)

  widgets.forEach(function (widget) {
    //See if the object is already in the db before adding it
    model_object.count(widget)
      .then(result => {
        if (result == 0) {
          console.log('Adding '+sheet_name+': ' + JSON.stringify(widget))
          return model_object.add(widget)
        }
      })
  })
}

NOT WORKING
function test(){
  console.log(`Testing ${strapi.plugins['users-permissions'].models.role.find}`)
}

module.exports = next => {

  console.log('Starting Strapi bootstrap')
  add_widgets_from_sheet('Menus', strapi.services.menu) //adding menus
  test() // Returning nothing
  console.log('Ending Strapi bootstrap')
  next()
}

我想将这些复选框切换为TRUE,CHECKED或其他名称。这样我们就不必在每次转储数据库时都通过UI手动进行操作。

我从例子中学习得最好……希望您能提供帮助。谢谢!

4 个答案:

答案 0 :(得分:1)

我觉得这更简单。

// In your bootstrap.js file
'use strict';
module.exports = async () => {

    const authenticated = await strapi.query('role', 'users-permissions').findOne({ type: 'authenticated' });
    authenticated.permissions.forEach(permission => {

        if (permission.type === 'application'){ // Whatever permissions you want to change
            let newPermission = permission;
            newPermission.enabled = true; // Editing permission as needed

            strapi.query('permission', 'users-permissions').update( { id: newPermission.id }, newPermission ); // Updating Strapi with the permission
        }
    });
    return;
};

答案 1 :(得分:1)

在上述两个答案的基础上,您似乎可以摆脱一个循环,并且可以为publicauthenticated用户设置权限。

这是针对trapi 3.2.4编写的,而我正在使用NodeJS 12,因此可以使用诸如散布运算符...之类的东西。

  const permOrm = strapi.query('permission', 'users-permissions')
  const perms = await permOrm.find({ type: 'application' })
  for (const curr of perms) {
    if (curr.role.type === 'authenticated') {
      strapi.log.info(
        `Allowing authenticated to call ${curr.controller}.${curr.action}`,
      )
      permOrm.update({ id: curr.id }, { ...curr, enabled: true })
      continue
    }
    // permission is for public
    const isReadEndpoint = ['find', 'findone', 'count'].includes(curr.action)
    if (isReadEndpoint) {
      strapi.log.info(
        `Allowing public to call ${curr.controller}.${curr.action}`,
      )
      permOrm.update({ id: curr.id }, { ...curr, enabled: true })
      continue
    }
    // TODO add custom logic for any non-standard actions here
    strapi.log.info(
      `Disallowing public from calling ${curr.controller}.${curr.action}`,
    )
    permOrm.update({ id: curr.id }, { ...curr, enabled: false })
  }

答案 2 :(得分:0)

因此,以下代码来自名为PROJECT / STRAPI / config / functions / bootstrap.js的文件 这会利用我们保存在excel电子表格中的信息自动创建内容类型和内容。但是,为了使用这些内容类型,必须激活一些角色和权限,以便Web ui可以访问它们。 基本上,我们不想手动进入Strapi UI来创建您的用户,创建内容类型,创建内容或更新权限。我们希望脚本能够完成所有这些工作。

'use strict'

我们的环境变量

require('dotenv').config({ path:'../.env' })

保存我们的数据的Excel电子表格(随附)

const XLSX = require('xlsx')
const BOOTSTRAP_DATA = XLSX.readFile(process.env.BOOTSTRAP_DATA).Sheets

从.env中提取的变量

const ADMIN_USERNAME = process.env.ADMIN_USERNAME
const ADMIN_PASSWORD = process.env.ADMIN_PASSWORD
const ADMIN_EMAIL    = process.env.ADMIN_EMAIL

在XLSX中阅读

async function bootstrap_resource(resource_type, resource_service) {
  strapi.log.info(`Bootstrapping ${resource_type}`)

  const resources = XLSX.utils.sheet_to_json(BOOTSTRAP_DATA[resource_type])

  for (let resource of resources) {

    if (await resource_service.count(resource) === 0) {
      strapi.log.warn(`Bootstrapping ${resource_type}: ${JSON.stringify(resource)}`)

      await resource_service.create(resource)
    }
  }
}

为trapi创建初始用户

async function bootstrap_admin() {
  strapi.log.info(`Bootstrapping Admin`)

  const admin_orm = strapi.admin.queries('administrator', 'admin')

  const admins = await admin_orm.find({username: ADMIN_USERNAME})

  if ( admins.length === 0) {
    const blocked  = false
    const username = ADMIN_USERNAME
    const password = await strapi.admin.services.auth.hashPassword(ADMIN_PASSWORD)
    const email    = ADMIN_EMAIL
    const user     = { blocked, username, password, email }

    const data = await admin_orm.create(user)

    strapi.log.warn(`Bootstrapped Admin User: ${JSON.stringify(user)}`)
  }
}

以下是get_roles()-get_permissions()必需,而get_permissions()是enable_permissions()必需。在这里,我们打开这些内容类型,以便Web ui可以看到它。

async function get_roles() {
  const role_orm = strapi.plugins['users-permissions'].queries('role', 'users-permissions')

  const role_list = await role_orm.find({}, [])

  const roles = {}

  for (let role of role_list) {
    roles[ role._id ] = role
    roles[ role.name ] = role
  }

  return roles
}

async function get_permissions( selected_role, selected_type, selected_controller ) {
  const roles          = await get_roles()
  const permission_orm = strapi.plugins['users-permissions'].queries('permission', 'users-permissions')

  let permission_list  = await permission_orm.find({_limit: 999}, [])

  if ( selected_role       ) permission_list = permission_list.filter( ({ role       }) => `${role}`       === `${roles[selected_role]._id}` )
  if ( selected_type       ) permission_list = permission_list.filter( ({ type       }) => `${type}`       === `${selected_type}`            )
  if ( selected_controller ) permission_list = permission_list.filter( ({ controller }) => `${controller}` === `${selected_controller}`      )

  return permission_list
}

async function enable_permissions(role, type, controller) {
  strapi.log.info(`Setting '${controller}' permissions for '${role}'`)

  const permission_orm = strapi.plugins['users-permissions'].queries('permission', 'users-permissions')

  const permissions = await get_permissions(role, type, controller)

  for (let { _id } of permissions) {
    permission_orm.update({ _id }, { enabled: true })
  }
}

最后,我们运行程序

module.exports = async next => {

  await bootstrap_admin()

  await bootstrap_resource( 'Clients', strapi.services.client )
  await bootstrap_resource( 'Menus',   strapi.services.menu   )

  enable_permissions('Public', 'application', 'client'     )
  enable_permissions('Public', 'application', 'github'     )
  enable_permissions('Public', 'application', 'menu'       )
  enable_permissions('Public', 'application', 'confluence' )

  next()
}

拿出我的评论,您便拥有了完整的bootstrap.js文件。下图显示了demo.xlsx工作簿的3个选项卡,这些选项卡用于填充所有内容。 clients tab menus tab users tab

最后,显示结果。使用Nuxt的菜单(内容),权限集和公共网站。 list of menus permissions for the public users Nuxt Generate Web Page leveraging strapi

答案 3 :(得分:0)

对于trapi 3.3.x以及可能的将来版本
效率略高(一次查询角色,使用查询过滤权限)

'use strict';

module.exports = async () => {
  const publicRole = await getRoleByName('Public')
  await grantPermissions(publicRole, 'application', 'images', ['upload', 'remove'])  // upload, remove in 'images' controller
  await grantPermissions(publicRole, 'application', 'project') // any action in 'project' controller                      
};

async function getRoleByName(name) {
  return strapi.query('role', 'users-permissions').findOne({ name }, [])
}

async function getPermissions(role, permissionType, controller, actions = null) {
  const permissionQuery = strapi.query('permission', 'users-permissions')
  const permissionRequest = {
    _limit: 1000,
    role: role.id,
    type: permissionType,
    controller: controller
  }

  if (actions) {
    permissionRequest.action_in = Array.isArray(actions) ? actions : [actions]
  }

  return permissionQuery.find(permissionRequest, [])
}

async function grantPermissions(role, permissionType, controller, actions) {
  if (actions && !Array.isArray(actions)) {
    actions = [ actions ]
  }
  strapi.log.info(`Setting '${controller}' [${actions ? actions.join(', ') : '*'}] permissions for '${role.name}'`)

  const permissionQuery = strapi.query('permission', 'users-permissions')
  const permissions = await getPermissions(role, permissionType, controller, actions)
  if (permissions.length === 0) {
    throw new Error(`Error enabling permissions: ${role.name}, ${permissionType}, ${controller}, ${actions}`)
  }

  for (const { id } of permissions) {
    await permissionQuery.update({ id }, { enabled: true })
  }
}