Question

实际上我想找到旁边没有来源的容器，所以我的卡车可以更灵活地在基地周围传递能量而不是有一个巨大的存储空间。（我的控制器很远）

// find all containers in room
var containers = creep.room.find(FIND_STRUCTURES, {
    filter: (s) => s.structureType == STRUCTURE_CONTAINER
});
// find all sources
var sources = creep.room.find(FIND_SOURCES);
// if there is no source next to the container
if (!(sources.length < 0)) {
}

这完成没有错误但我无法设法获取容器ID 旁边没有任何消息来源。

我相信我必须在过滤器中包含最有可能遍历每个容器的逻辑吗？

我不了解循环/迭代，但还没有让这个工作。我尝试了很多东西但是现在我感到很无助试图找出这个难题。

Answer 1

我认为你可以更新你的第一个过滤器以检查附近的来源，如：

    let containers = creep.room.find(FIND_STRUCTURES, {
        filter: (s) => s.structureType === STRUCTURE_CONTAINER && s.pos.findInRange(FIND_SOURCES, 2).length === 0
    });

我使用2作为范围，但如果您的容器始终与您的源相邻（或非常远），您可以将其更改为1.

如果将结果存储在内存中，则可以节省CPU。如果您想知道如何做，我可以提供更多信息。

编辑：对于缓存，您可以像这样在Room对象上创建一个属性（可能需要对此进行一些调试 - 我写了但没有测试它）：

Object.defineProperty(Room.prototype, 'nonSourceContainers', {
    get: function () {
        if (this === Room.prototype || this === undefined) return undefined;

        const room = this;

        // If any containers have been constructed that don't have isNearSource defined, fix that first.
        let newContainers = room.find(FIND_STRUCTURES, {
            filter: (s) => s.structureType === STRUCTURE_CONTAINER && s.memory.isNearSource === undefined
        });

        // Found some new containers?
        if (newContainers.length) {
            for (let i = 0; i < newContainers.length; ++i) {
                newContainers[i].memory.isNearSource = newContainers[i].pos.findInRange(FIND_SOURCES, 2).length > 0;
            }

            // Set the list of non-source container ids in the room's memory. This lasts forever.
            room.memory.nonSourceContainerIds = room.find(FIND_STRUCTURES, {filter: {isNearSource: false}});

            // Reset the cached set of containers.
            room._nonSourceContainers = undefined;
        }

        // If there is no cached list, create it. Caching will mean the following runs once per tick or less.
        if (!room._nonSourceContainers) {
            if (!room.memory.nonSourceContainerIds) {
                room.memory.nonSourceContainerIds = [];
            }

            // Get the containers by their ids.
            room._nonSourceContainers = room.memory.nonSourceContainerIds.map(id => Game.getObjectById(id));
        }

        return room._nonSourceContainers;
    },
    enumerable: false,
    configurable: true
});

Screeps - 按范围1来过滤容器（容器采矿）

1 个答案: