Question

我正在尝试首先加载用户的最新帖子，并以降序排列它们，但在将endAt().limitToLast()与orderByChild()一起使用时遇到了问题。

我可以使用startAt().limitToFirst()和orderByChild()来对项目进行分页，但是我需要从头开始加载列表...使用endAt()执行查询时，orderByChild（）似乎被忽略。

这是我的categories节点的JSON

{
  "-LY8EYaWHINB1khsIEvJ" : {
    "Recipies" : true,
    "username" : "User1"
  },
  "-LYDaIrnDKIWndMcLE-g" : {
    "Recipies" : true,
    "username" : "User4"
  },
  "-LY8Em4B6COFk3how5FC" : {
    "Buds" : true,
    "username" : "User2"
  },
  "-LY8Eq2E1muFcOBstODa" : {...},
  "-LY8Esi98QdhszIgvRRN" : {...},
  "-LY9OSc7u8wTNQaJ7BXL" : {...},
  "-LY8EymPGxK8Y_YnRfC0" : {...},
  "-LY8F0RrYbLNPpYwIuMX" : {...},
  "-LY8F3QfERAhOq3iW3jC" : {...},
}

这里是我的查询的样子（请注意，我需要从下往上获取）：

  const fetchCategoryImages = (category, currentImages, lastKey) => {
      if (!lastKey) {
        return () => {
          firebase.database().ref('/categories')
            .orderByChild('Recipies' //this is the category)
            .endAt(true)
            .limitToLast(4)
            .on('value', snapshot => {
              const arrayOfKeys = Object.keys(snapshot.val())
                 .sort()
                 .reverse();

              const results = arrayOfKeys
                .map((key) => snapshot.val()[key]);

              const createLastKey = arrayOfKeys[arrayOfKeys.length - 1];

              //just passing the initial data with redux here... (snapshot and lastKey...)
            });
        };
      } else {
         //subsequent fetch if there is a lastKey to reference start point
         return () => {
           firebase.database().ref('/categories')
             .orderByChild('Recipies' //this is the category)
             .endAt(true, '-LY9OSc7u8wTNQaJ7BXL' //this is the lastKey)
             .limitToLast(3)
             .on('value', snapshot => {
               const arrayOfKeys = Object.keys(snapshot.val())
                  .sort()
                  .reverse()
                  .slice(1);

               const results = arrayOfKeys
                  .map((key) => snapshot.val()[key]);

               const createLastKey = arrayOfKeys[arrayOfKeys.length - 1];
               const concatImages = _.concat(currentImages, results);

               //passing the new data with redux here... (snapshot and lasy ley...)

               }
           });
        };
     };

当我简单地将查询改为使用startAt().limitToFirst()和orderByChild()时，所有这些问题都消失了。

非常感谢我在此问题上能获得的所有帮助，干杯！

Answer 1

因此，在尝试了几乎所有内容之后，事实证明，我要做的就是也添加startAt(true)。我不确定为什么，但是可以。我还有其他查询，与没有查询的情况几乎相同，但我想知道为什么我需要这个才能使其正常工作。

这是我的工作代码：

 const fetchCategoryImages = (category, currentImages, lastKey) => {
      if (!lastKey) {
        return () => {
          firebase.database().ref('/categories')
            .orderByChild('Recipies' //this is the category)
            //THE SOLUTION
            .startAt(true)
            .endAt(true)
            .limitToLast(4)
            .on('value', snapshot => {
          const arrayOfKeys = Object.keys(snapshot.val())
             .sort()
             .reverse();

          const results = arrayOfKeys
            .map((key) => snapshot.val()[key]);

          const createLastKey = arrayOfKeys[arrayOfKeys.length - 1];

          //just passing the initial data with redux here... (snapshot and lastKey...)
        });
    };
  } else {
     //subsequent fetch if there is a lastKey to reference start point
     return () => {
       firebase.database().ref('/categories')
         .orderByChild('Recipies' //this is the category)
          //THE SOLUTION
          .startAt(true)
         .endAt(true, '-LY9OSc7u8wTNQaJ7BXL' //this is the lastKey)
         .limitToLast(3)
         .on('value', snapshot => {
           const arrayOfKeys = Object.keys(snapshot.val())
              .sort()
              .reverse()
              .slice(1);

           const results = arrayOfKeys
              .map((key) => snapshot.val()[key]);

           const createLastKey = arrayOfKeys[arrayOfKeys.length - 1];
           const concatImages = _.concat(currentImages, results);

           //passing the new data with redux here... (snapshot and lasy ley...)

           }
       });
    };
 };

Answer 2

我只是用此JSON进行测试：

{
  "-LY8EYaWHINB1khsIEvJ" : {
    "Recipies" : true,
    "username" : "User1"
  },
  "-LY8Em4B6COFk3how5FC" : {
    "Buds" : true,
    "username" : "User2"
  },
  "-LY8Eq2E1muFcOBstODa" : {
    "Buds" : true,
    "username" : "User3"
  },
  "-LY8Esi98QdhszIgvRRN" : {
    "Buds" : true,
    "username" : "User4"
  },
  "-LY8EymPGxK8Y_YnRfC0" : {
    "Buds" : true,
    "username" : "User5"
  },
  "-LY8F0RrYbLNPpYwIuMX" : {
    "Buds" : true,
    "username" : "User6"
  },
  "-LY8F3QfERAhOq3iW3jC" : {
    "Receipies" : true,
    "username" : "User7"
  },
  "-LY9OSc7u8wTNQaJ7BXL" : {
    "Buds" : true,
    "username" : "User8"
  },
  "-LYDaIrnDKIWndMcLE-g" : {
    "Recipies" : true,
    "username" : "User8"
  }
}

这段代码：

ref.orderByChild('Buds')
  .endAt(true, '-LY9OSc7u8wTNQaJ7BXL')
  .limitToLast(3)
  .once('value', snapshot => {
    snapshot.forEach(child => {
      console.log(child.key+": "+JSON.stringify(child.val()));
    })
  })

实时示例：https://jsbin.com/zigofok/edit?js,console

运行时打印：

“-LY8EymPGxK8Y_YnRfC0：{\” Buds \“：true，\” username \“：\” User5 \“}”

“-LY8F0RrYbLNPpYwIuMX：{\” Buds \“：true，\” username \“：\” User6 \“}”

“-LY9OSc7u8wTNQaJ7BXL：{\” Buds \“：true，\” username \“：\” User8 \“}”

请注意，由于该用户没有User7属性，因此如何跳过Buds。

分页/无限滚动时，firebase endAt（）与orderByChild（）的时髦行为

2 个答案: