10小时后删除Firebase Child

时间:2017-09-20 10:49:01

标签: javascript firebase firebase-realtime-database google-cloud-functions firebase-admin

我想要一个删除10个小时孩子的功能。到目前为止我有这个代码但是,如果我将其部署到Firebase云功能,它会立即从数据库中删除所有数据。我不确定我做错了什么。请帮忙!

exports.deleteOldItems = functions.database.ref('Rollerbanken/{pushId}')
.onWrite(event => {
  var ref = event.data.ref.parent; // reference to the items
  var now = Date.now();
  var cutoff = now - 10 * 60 * 60 * 1000;
  var oldItemsQuery = ref.orderByChild('timestamp').endAt(cutoff);
  return oldItemsQuery.once('value', function(snapshot) {
    // create a map with all children that need to be removed
    var updates = {};
    snapshot.forEach(function(child) {
      updates[child.key] = null
    });
    // execute all updates in one go and return the result to end the function
    return ref.update(updates);
  });
});

我的Firebase数据库结构:

{
  "Rollerbanken" : {
    "-Ku_Ywh8wElDdwNqa0KW" : {
      "Extrainformatie" : "",
      "Latitude" : "51.8306880305073",
      "Longitude" : "5.90483402833892",
      "Staater" : "Staat er nog steeds",
      "Staaternietmeer" : "",
      "Stad" : "Nijmegen",
      "Tijd" : "19:50",
      "TijdControle" : "19:50",
      "TijdControleniet" : "",
      "TypeControle" : "Rollerbank",
      "TypeControle2" : "Rollerbank",
      "postKey" : "-Ku_Ywh8wElDdwNqa0KW",
      "timestamp" : 1506016223000
        }
      }
    }

2 个答案:

答案 0 :(得分:0)

您的代码中有ref.orderByChild('timestamp'),但您的数据中没有timestamp字段。这被解释为具有null时间戳,其实际上具有Firebase强加的ordering中的最高优先级。因此,所有数据节点都在指定的cutoff之前排序,并被删除。

要解决此问题,您需要将timestamp字段添加到架构中。

答案 1 :(得分:0)

这是我24小时编码的方式。

 exports.deleteOldItems = functions.database.ref('/notification/{user_id}/{notification_id}').onWrite((change) => {

    const ref = change.after.ref.parent; // reference to the parent

    // For easier cross browser solution, get old time like this
    const yesterday = Date.now() - 86400000; // that is: 24 * 60 * 60 * 1000
    const oldItemsQuery = ref.orderByChild('time').endAt(yesterday);

    return oldItemsQuery.once('value').then((snapshot) => {
      // create a map with all children that need to be removed
      const updates = {};
          snapshot.forEach(child => {
                updates[child.key] = null;
          });
      return ref.update(updates);
      // execute all updates in one go and return the result to end the function
    });
  });

编码愉快!