firebase网站-更新找到特定值的数据库

问题描述

我的数据库如下所示:

我需要更新index_num为3的division_name.我尝试了以下代码，但没有用-

I need to update the division_name where the index_num is 3.I tried the following code but it did not work -

    var division_index_found="3";
    var division_name_given="new div";

    var query_update=firebase.database().ref("/divisions")
    .orderByChild('index_num').equalTo(division_index_found);

    query_update.once("child_added", function(snapshot) {
    snapshot.ref.update({ division_name: division_name_given });
    });

在这里采用什么方法?

我在Chrome控制台中收到警告:

I get warning in chrome console :

FIREBASE警告:使用未指定的索引.考虑在/divisions的安全规则中添加".indexOn":"index_num"以获得更好的性能

FIREBASE WARNING: Using an unspecified index. Consider adding ".indexOn": "index_num" at /divisions to your security rules for better performance

从Firebase数据库中，districts节点(?)看起来像:

From the firebase database the districts node (?) looks like :

 "districts" : {
    "-KbVYCSO8wrMoXD3vL81" : {
      "district_name" : "Rangpur",
      "division_index" : "3",
      "index_num" : 2
    },
    "-KbVYHgbWMDMtGsnmvei" : {
      "district_name" : "jessore",
      "division_index" : "3",
      "index_num" : 3
    },
    "-KbVYKtSnPMFDkx9z0cU" : {
      "district_name" : "district 1",
      "division_index" : "3",
      "index_num" : 4
    }
  }

现在我要为特定的index_num和division_index更新district_name.我使用以下代码:

Now I want to update district_name for a certain index_num and division_index. I use the following code :

var district_index="3";
var division_index="3"
var district_index_parsed = parseInt(district_index);

var query_update=firebase.database().ref("/districts")
.orderByChild('index_num').equalTo(district_index_parsed);


 query_update.once("child_added", function(snapshot) {

 snapshot.forEach(function(snapshot_indiv){

        if(parseInt(snapshot_indiv.division_index)==parseInt(division_index)){


    var district_name_again="new district name";

snapshot_indiv.ref.update({ district_name: district_name_again },function(error){

});

}// end of if division_index compare
    });// end of forEach



});// end of query_update once

但是控制台显示:

Uncaught Error: No index defined for index_num

at je.get (firebase-database.js:94)
...
...
 at edit_districts.php:359
...

最终提示我edit_districts.php文件中的以下代码行:

which ultimately hints at the following line of code in my edit_districts.php file:

snapshot.forEach(function(snapshot_indiv){

在query_update.once部分的内部

.forEach方法似乎引起了问题.

inside the query_update.once part.The forEach method seems to make the problem.

安全规则定义为

{
  "rules": {
    ".read": "auth != null",
    ".write": "auth != null",

    "divisions": {
      ".indexOn": "index_num"
    },

      "districts": {
      ".indexOn": "index_num"
    },

  }
}

如何摆脱更新数据库的错误?

推荐答案

您需要将索引添加到Firebase数据库安全规则中.我建议阅读有关安全规则的Firebase文档，尤其是.

You'll need to add an index to your Firebase Database security rules. I recommend reading the Firebase documentation on security rules, specifically the section on indexes.

文档中的第一个示例:

{
  "rules": {
    "dinosaurs": {
      ".indexOn": ["height", "length"]
    }
  }
}

修改为您的数据结构，看起来像这样:

Modified to your data structure, it'll look something like this:

{
  "rules": {
    "divisions": {
      ".indexOn": "index_num"
    }
  }
}

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