Question

我需要从下面代码的第二个函数返回一个布尔值。但是，返回class Main_func: def calculate(self,data): #do some things with the DF return df.copy(deep=True) <- one of many attempts to fix the problem, but didnt work cpus = 24 actors = [] for _ in range(cpus): actors.append(Main_func.remote()) from ray.util import ActorPool pool = ActorPool(actors) import os arr = os.listdir("/some/files") def to_ray(): try: filename = arr.pop(0) pf = ParquetFile("/some/files/" + filename) df = pf.to_pandas() pool.submit(lambda a,v:a.calculate.remote(v),df.copy(deep=True) except Exception as e: print(e) for _ in range(cpus): to_ray() while(True): res = pool.get_next_unordered() write('./temp/' + random_filename, res,compression='GZIP') del res to_ray()给我isValidOrg，而console.logging变量给了我所需的答案。在我的情况下，是否有可能返回0或1？

Promise { <pending> }

const GetOrganization = userId => {
    return User.aggregate([
            { $match: { _id: mongoose.Types.ObjectId(userId) } },
            {
              $project: {
                _id: 0,
                organization: 1
              }
            }
        ])
}

Answer 1

这个用例是完全错误的。

const isValidOrg = await GetOrganization(_id).then(res => {
        return xOrg !== res[0].organization ? 0 : 1
})

为什么要链接。然后，只需将其分配给isValidOrg并在下一行对其进行评估。

NodeJS返回布尔值，而不是未决承诺

1 个答案: