如何有条件地向我的对象框查询添加条件？

Question

我有这个方法可以找到我所有的

Clientes

:

static Future<List<Cliente>> findAll({
  int limit = 50,
  String? cnpjCpf,
  String? fantasia,
}) async {
  List<Cliente> clientes = [];

  final boxCliente = objectBox.store.box<Cliente>();

  Query<Cliente> query = boxCliente
      .query(Cliente_.cnpjCpf
          .contains(cnpjCpf ?? "")
          .or(Cliente_.fantasia.contains(fantasia ?? "")))
      .order(Cliente_.fantasia)
      .build();

  query.limit = limit;

  clientes = await query.findAsync();

  return clientes;
}

现在，我不需要像上面的示例那样添加所有条件，而是需要有条件地添加我的条件，即，如果

cnpjCpj != null

则将其添加到

QueryBuilder

。

我在想，也许是这样的：

static Future<List<Cliente>> findAll({
  int limit = 50,
  String? cnpjCpf,
  String? fantasia,
}) async {
  List<Cliente> clientes = [];

  final boxCliente = objectBox.store.box<Cliente>();

  final queryBuilder = boxCliente.query();

  if (cnpjCpf != null && cnpjCpf.isNotEmpty) {
    queryBuilder.equal(Cliente_.cnpjCpf, cnpjCpf);
  }

  if (fantasia != null && fantasia.isNotEmpty) {
    queryBuilder.contains(Cliente_.fantasia, fantasia);
  }

  queryBuilder.order(Cliente_.fantasia);

  Query<Cliente> query = queryBuilder.build();

  query.limit = limit;

  clientes = await query.findAsync();

  return clientes;
}

有人知道该怎么做吗？ QueryBuilder 对象没有 equals 或 contains 。

Answer 1

首先组合条件，然后将它们传递给

.query(conditions)

方法。所以类似：

var Condition<Cliente>? conditions = null;

if (cnpjCpf != null && cnpjCpf.isNotEmpty) {
  conditions = Cliente_.cnpjCpf.equal(cnpjCpf);
}

if (fantasia != null && fantasia.isNotEmpty) {
  if (conditions == null) {
    conditions = Cliente_.fantasia.equal(fantasia);
  } else {
    conditions = conditions.or(Cliente_.fantasia.equal(fantasia));
  }
}

if (conditions != null) {
  final query = boxCliente.query(conditions).build();
}