Screeps

Question

实际上我想找到旁边没有资源的容器，这样我的卡车就可以更灵活地在基地周围传输能量而不是拥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 旁边没有来源。

我相信我必须在过滤器中包含逻辑，很可能循环遍历每个容器？

我对循环/迭代的理解还不够，无法使它正常工作。我尝试了很多事情，但现在我感到无助试图找出这个谜题。

Answer 1

我认为您可以像这样更新您的第一个过滤器来检查附近的来源：

    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，但如果你的容器总是与你的源相邻（或很远），你可以将其更改为 1。

如果将结果存储在内存中，您将节省 CPU。如果你想知道如何做到这一点，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）：

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 1 范围内的来源过滤容器（容器挖掘）

Screeps - filter containers by sources in range of 1 (container mining)

javascript