表格结构：
Collection
name (String)
Image
collection (Pointer<Collection>)
url (String)
position (Number)
Image类具有列collection，它是指向Collection类的指针。
position用于对Images内的Collection进行排序。
假设我有~3000张图片和3个收藏品。
构建一个返回数组的查询的最佳方法是什么，该数组包含至少一个关联Image且仅显示 的所有集合每个集合的5张图片，按position
每个集合的相关图像都需要包含在响应中，并且可能看起来像这样：
results: [{
collection: {
name: 'foo'
},
images: [{
position: 0,
url: 'test.jpg'
},
{
position: 1,
url: 'test.gif'
}]
}, {
...
}]
目前我所能想到的只是做两个查询，一个是获取所有Collections而另一个是预先获得所有Images然后过滤它们，这看起来相当倒退，再加上事实Parse仅限于最大1000的结果集。我是否需要重新考虑我的桌面结构？
答案 0 :(得分：1)
使用该数据结构，您将拥有多个（嵌套）查询，但在CloudCode中完成时，如果使用适当的限制和承诺，它仍应足够快。
在这里尝试卷曲：
curl -X POST -H "X-Parse-Application-Id: 1ClVjWioWtW4uz8Nt7zYiXHkoNYQao6Z1rdXd8Gt" -H "X-Parse-REST-API-Key: VlYbFT8uhPpsYeRW5ezIn1A8bWa5N9OW9riqs4ji" -H "Content-Type: application/json" -d '{}' https://api.parse.com/1/functions/collectionSummary
Cloud Code：
var _ = require('underscore');
Parse.Cloud.define("collectionSummary", function(request, response) {
var collections = [];
var collectionQuery = new Parse.Query("Collection");
collectionQuery.find({
success: function(results) {
var promises = [];
_.each(results, function(collection) {
var name = collection.get("name");
var imageQuery = new Parse.Query("Image");
imageQuery.equalTo("collection", collection);
imageQuery.ascending("position");
imageQuery.limit(5);
promises.push(imageQuery.find({
success: function(images) {
if (images.length > 0) {
collections.push({ "collection":collection, "images":images });
}
},
error: function() {
response.error("Could not read images.");
}
}));
});
Parse.Promise.when(promises).then(function() {
response.success(collections);
});
},
error: function() {
response.error("Could not read collections.");
}
});
});
输出：
{
"result": [
{
"collection": {
"__type": "Object",
"className": "Collection",
"createdAt": "2015-05-11T21:52:28.406Z",
"name": "Apple",
"objectId": "cpTsJBNc9q",
"updatedAt": "2015-05-11T21:52:28.406Z"
},
"images": [
{
"__type": "Object",
"className": "Image",
"collection": {
"__type": "Pointer",
"className": "Collection",
"objectId": "cpTsJBNc9q"
},
"createdAt": "2015-05-11T22:12:35.152Z",
"objectId": "0AmtlLrlHT",
"position": 0,
"updatedAt": "2015-05-11T22:12:49.463Z",
"url": "test0"
},
{
"__type": "Object",
"className": "Image",
"collection": {
"__type": "Pointer",
"className": "Collection",
"objectId": "cpTsJBNc9q"
},
"createdAt": "2015-05-11T22:12:58.185Z",
"objectId": "WbomPr3hUK",
"position": 1,
"updatedAt": "2015-05-11T22:13:05.523Z",
"url": "test1"
},
{
"__type": "Object",
"className": "Image",
"collection": {
"__type": "Pointer",
"className": "Collection",
"objectId": "cpTsJBNc9q"
},
"createdAt": "2015-05-11T22:19:02.836Z",
"objectId": "ASLInM7Hu5",
"position": 15,
"updatedAt": "2015-05-11T22:19:08.990Z",
"url": "test8"
},
{
"__type": "Object",
"className": "Image",
"collection": {
"__type": "Pointer",
"className": "Collection",
"objectId": "cpTsJBNc9q"
},
"createdAt": "2015-05-11T22:19:14.719Z",
"objectId": "VkCF98Ts0N",
"position": 20,
"updatedAt": "2015-05-11T22:19:20.699Z",
"url": "test9"
},
{
"__type": "Object",
"className": "Image",
"collection": {
"__type": "Pointer",
"className": "Collection",
"objectId": "cpTsJBNc9q"
},
"createdAt": "2015-05-11T22:19:27.032Z",
"objectId": "I2mEG20bfp",
"position": 40,
"updatedAt": "2015-05-11T22:19:37.554Z",
"url": "test10"
}
]
},
{
"collection": {
"__type": "Object",
"className": "Collection",
"createdAt": "2015-05-11T21:52:35.297Z",
"name": "Banana",
"objectId": "zxPfnWlm1T",
"updatedAt": "2015-05-11T21:52:35.297Z"
},
"images": [
{
"__type": "Object",
"className": "Image",
"collection": {
"__type": "Pointer",
"className": "Collection",
"objectId": "zxPfnWlm1T"
},
"createdAt": "2015-05-11T22:16:05.924Z",
"objectId": "vK9O7b5vR6",
"position": 0,
"updatedAt": "2015-05-11T22:16:12.607Z",
"url": "test2"
},
{
"__type": "Object",
"className": "Image",
"collection": {
"__type": "Pointer",
"className": "Collection",
"objectId": "zxPfnWlm1T"
},
"createdAt": "2015-05-11T22:16:26.676Z",
"objectId": "oMEMgwauEi",
"position": 1,
"updatedAt": "2015-05-11T22:16:30.750Z",
"url": "test3"
},
{
"__type": "Object",
"className": "Image",
"collection": {
"__type": "Pointer",
"className": "Collection",
"objectId": "zxPfnWlm1T"
},
"createdAt": "2015-05-11T22:16:35.378Z",
"objectId": "0sAbNK1LbN",
"position": 2,
"updatedAt": "2015-05-11T22:16:39.475Z",
"url": "test4"
},
{
"__type": "Object",
"className": "Image",
"collection": {
"__type": "Pointer",
"className": "Collection",
"objectId": "zxPfnWlm1T"
},
"createdAt": "2015-05-11T22:16:42.984Z",
"objectId": "QVZJmHQJ6E",
"position": 3,
"updatedAt": "2015-05-11T22:16:52.405Z",
"url": "test5"
},
{
"__type": "Object",
"className": "Image",
"collection": {
"__type": "Pointer",
"className": "Collection",
"objectId": "zxPfnWlm1T"
},
"createdAt": "2015-05-11T22:16:56.827Z",
"objectId": "NGS39LCAJL",
"position": 4,
"updatedAt": "2015-05-11T22:17:01.401Z",
"url": "test6"
}
]
}
]
}
答案 1 :(得分：1)
根据您的约束，您可以在一次通话中执行此操作。
如果您有少量Collection个对象并且您正在使用相对位置（每个Collection都有一个图像position = 1,2等），那么无需过度思考问题。这些约束意味着“位置”<1的项目相对较少。 6（或5，如果0是你的第一个指数位置）。
只需在Image类上调用并将其限制为位置小于或等于5的对象。使用.include()返回附加到每个Collection的对象{1}}对象作为返回的一部分。然后以您认为合适的任何格式在本地对Image个对象进行排序。
请记住，Parse限制的对象数量可以返回到100，并且您可以将其增加到最多1000个。如果Image为.include()每个Collection，它表示每个返回的Image计算2个对象，因此最多可以返回500个Image个对象。听起来这远远超出了你的预期需求。
Voilá - 一个电话，你的所有物品。
Image<强>更新 值得注意的是，这种方法优于“嵌套查询”方法，无论是少量集合还是大规模集合（1000或10,000集合）