Question

我必须在我的数据浏览器中解析类，＆＃39;详细信息＆＃39;和＆＃39;内容＆＃39;。＆＃39;细节＆＃39; class有以下 - ＆gt; ＆＃39; objectId＆＃39;，＆＃39; uuid＆＃39;并且＆＃39;接近＆＃39; ＆＃39;内容＆＃39;课程有＆＃39; objectId＆＃39;，＆＃39; descId＆＃39;和＆＃39;提供＆＃39;。

我使用Javascript SDK创建了一个Web UI，因此当用户输入uuid时，邻近度和优惠，uuid和proximity将存储在＆＃39;详细信息中。 class，成功后我获得新创建的对象的objectId。然后我将该objectId存储在＆＃39;内容＆＃39; descId下的类和用户输入的商品。

我的问题是我有一个需要填充的html表，所以我需要从两个类中提取数据。来自＆＃39;细节＆＃39;的uuid和接近度和来自内容的提议＆＃39;所以我需要在一个查询中执行此操作。这是我存储＆＃39;详细信息的原因。 objectId在＆＃39;内容＆＃39; class作为一种外键。

我被困在这条十字路口，并尝试过包括等等，但我只是在尝试，我不知道自己需要做什么。如果有人可以提供帮助，也许可以给我看一个样本，我非常感激

这是我的js保存代码：

 //Creating Beacon Content Parse Object
    var iBeaconContent = Parse.Object.extend("Content");
    var beaconContent = new iBeaconContent();

    //Creating Object to save to iBeacon Description Table
    var iBeaconDescription = Parse.Object.extend("Details");
    var beaconDescription = new iBeaconDescription();

    beaconDescription.set("uuid", tdUuid.children("input[type=text]").val().toString());
    beaconDescription.set("proximity", parseInt(tdProximity.children($('prox')).val().toString()));

    beaconDescription.save(null, {
        success: function(beaconDescriptionObject) {
            var query = new Parse.Query("Details");
            query.equalTo("uuid", tdUuid.children("input[type=text]").val().toString());
            query.find({
                success: function(results) {

                    objectId = results[0].id;

                    beaconContent.set("descId", objectId);
                    beaconContent.set("offer", tdOffer.children("input[type=text]").val().toString());
                    beaconContent.save(null, {
                        success: function(object) {
                            document.location.reload(true);
                        }, error: function(beaconContent, error) {
                        }
                    });
                }
            });
        },
        error: function(error) {
        }
    });

新JAVASCRIPT

var BeaconDetail = Parse.Object.extend("Details");
var BeaconContent = Parse.Object.extend("Content");

var innerQuery = new Parse.Query(BeaconDetail);
innerQuery.exists("descId");
var query = Parse.Query(BeaconDetail);

query.matchesQuery("objectId", innerQuery);
    query.find({
        success:function(beaconContent){
            alert("Success----lenght:  " + beaconContent.length);

        }
    })

Answer 1

听起来像是需要使用复合查询或关系查询。这是一些链接

https://www.parse.com/docs/js_guide#queries-relational

https://www.parse.com/docs/js_guide#queries-compound

https://parse.com/questions/compound-relational-queries

来自两个类的查询示例如下

看到代码也很好，这有助于给出更相对的答案。

CODE

var lotsOfWins = new Parse.Query("Player");
lotsOfWins.greaterThan("wins", 150);

var fewWins = new Parse.Query("Player");
fewWins.lessThan("wins", 5);

var mainQuery = Parse.Query.or(lotsOfWins, fewWins);
mainQuery.find({
  success: function(results) {
     // results contains a list of players that either have won a lot of games or won only a few games.
  },
  error: function(error) {
    // There was an error.
  }
});

Answer 2

如果我理解正确，您的Content类包含指向Details属性中descId类的指针，并且您希望能够根据某些Details进行查询}字段并返回两个对象？

注意：我必须指出descId是一个名字很差的财产，只会引起混淆。如果它是指针，只需给它命名为desc，不要使用Id后缀。

无论如何，如果那是你想要的：

var query = new Parse.Query("Content");
var uuid = tdUuid.children("input[type=text]").val().toString();
var proximity = parseInt(tdProximity.children($('prox')).val().toString());
// consider logging those two variable to confirm they're what you want

// query properties of a pointer
query.equalTo("descId.uuid", uuid);
query.equalTo("descId.proximity", proximity);

// include the pointer in the output
query.include("descId");

query.find({
    success: function(beaconContent) {
        alert("Success -- length: " + beaconContent.length);
        // sample output of first result:
        var content = beaconContent[0];
        var detail = content.get("descId");
        console.log("uuid", detail.get("uuid"));
        console.log("proximity", detail.get("proximity"));
        console.log("offer", content.get("offer"));
    }
});

根据一个表中的列从两个解析类中获取数据

2 个答案: