Question

我的Firebase数据以这种方式组织：

+ myappname
  + customers
    + -JV2NQv3GmoM81zdUfTe
      + name: "Mary"
      + age: "24"
      + ...
    + -JV2N9NnItCfz5vB04RS
      + name: "John"
      + age: "32"
      + ...
    + ...
  + ...

如何通过名称检索客户？
该名称保证是独一无二的这是我的客户服务，目前：

app.factory('Customer', function ($firebase, FIREBASE_URL) {
  var ref = new Firebase(FIREBASE_URL + 'customers');
  var customers = $firebase(ref);

  var Customer = {
    all: customers,
    create: function (customer) {
      return customers.$add(customer).then(function (ref) {
        var customerId = ref.name();
        return customerId;
      });
    },
    set: function(customerId, customer) {
      return customers.$child(customerId).$set(customer);
    },
    find: function (customerId) {
      return customers.$child(customerId);
    },
    findByName: function (customerName) { // TODO...
    },
    delete: function (customerId) {
      var customer = Customer.find(customerId);
      customer.deleted = true;
      customer.$on('loaded', function () {
        customers.$child(customerId).$set(customer);
      });
    }
  };
  return Customer;
});

我应该在每个findByName（）调用上扫描所有客户吗？或者我应该建立类似“二级指数”的东西？请一些建议，我刚开始......： - （

Answer 1

感谢Kato指示和Frank van Puffelen的建议，我终于解决了自己的问题我确实在我的Firebase中添加了一个“index”，“customersByName”（记住“磁盘空间便宜，用户的时间不是”Firebase座右铭...... :-)。我没有按照提到的答案中的指示，因为我认为这个解决方案更常用：它可以扩展多个“索引”...... 我想在这里发布，希望它对其他人有用。
我想看看评论：这个解决方案有可能存在缺陷吗？对于某些用例，总体而言，这是一个明智的解决方案吗？

app.factory('Customer', function ($firebase, FIREBASE_URL) {
  var ref = new Firebase(FIREBASE_URL + 'customers');
  var customers = $firebase(ref);
  var refByName = new Firebase(FIREBASE_URL + 'customersByName');
  var customersByName = $firebase(refByName);

  var Customer = {
    all: customers,
    create: function (customer) {
      return customers.$add(customer).then(function (ref) {
        var customerId = ref.name();
        customersByName.$child(customer.name).$set(customerId);
        return customerId;
      });
    },
    set: function(customerId, customer) {
      var oldname = customers.$child(customerId).name;
      if (customer.name !== oldname) {
        customersByName.$remove(oldname);
      }
      customersByName.$child(customer.name).$set(customerId);
      return customers.$child(customerId).$set(customer);
    },
    find: function (customerId) {
      return customers.$child(customerId);
    },
    findByName: function (customerName) {
      return customersByName.$child(customerName);
    },
    delete: function (customerId) {
      var customer = Customer.find(customerId);
      customer.deleted = true;
      customer.$on('loaded', function () {
        customersByName.$remove(customer.name);
        customers.$child(customerId).$set(customer);
      });
    }
  };

  return Customer;
});

Angularfire：如何通过其中一个属性访问项目？

1 个答案: