用户在Firebase上搜索信息 [英] Search information by user on firebase

问题描述

这可能是非常简单的，所以对不起...
我试图创建一个应用程序，我可以创建用户和存储Firebase的信息。这里有一个例子：

http://image.noelshack.com/fichiers/2015/11/1426182307-log.png

但现在我想要如果有人想创建一个新用户，请检查名称是否已经存在。

如何在不知道id的情况下抓取个人用户信息，如simplelogin：54？ / p>

我找到了这个主题通过名称属性使用Firebase获取用户但它不是一回事，因为在我的情况下，我不知道用户后的孩子

干杯， 像弗兰克说过的，你必须知道一些关于用户能够看到她/ p>

然而，这里是一个普遍的答案。我假设名称是指您创建的已识别属性。

建议

• 首先查看 Firebase查询文档

---

短回答

1. 要通过 identifiant 属性检查用户是否存在，您需要 orderByChild（identifiant）并使用 .equalTo（< identifient_here>）查询特定的用户。

例如，要检查使用identifyient =aaa，


var identifient =aaa;等等（识别）.once（价值，功能（快照）{
console.log（加载的用户由识别：，识别，snapshot.val）。
usersRef.orderByChild （））;
}）;

2. 如果您想通过键进行查询（例如 simplelogin：53 ），您可以使用 orderByKey（）而不是 orderByChild（） ...或者只是简单地将ref设置为用户的键：
   
   
   

     var userKey ='simplelogin：53'; 
    var userRef = new Firebase（https：//< YOUR-FIREBASE-APP> .firebaseio.com / Users+ userKey）; 
    userRef.once（value，function（snapshot）{
    console.log（Loaded user，snapshot.val（））; 
   }）; 
    
    
    
    
    
    $ b $ h $ Long（er）Answer  
    
    
    
    
   您可以使用用户工厂来处理这个问题（请参阅 Angular Providers文档）。
    
   使用 $ q 服务返回工厂中的承诺。
    li> 
    
   以下是<$ c $的 Angular API文档c> $ q  。
    
    
    
    UserFactory示例
    
    
    
    
    
   查看此工作 PLNKR示例 
    
   它与我的一个公共Firebase实例绑定。
    
    
   我创建了相同的 simplelogin：53  c $ c $>用户 / Users 就像你有。
    
   如果您搜索 identifient  code> =  aaa ，您将得到正确的用户。
    
    
    控制器i这里的补充仅仅是为了举例的目的，并没有真正做任何有价值的事情。  
           
    
    
    
    $ {
   Users：{
   simplelogin：53：{
   identifiant：aaa
   } 
   } 
    
    $ / code $ / pre 
    $ b 
    UserFactory 
    
    
    
     .factory（'UserFactory'，function（$ q，$ firebaseObject，fbUrl）{
    return function（userId）{
    var deferred = $ q.defer（）; 
    if（userId.isNotEmpty（））{
    var userRef = new Firebase（fbUrl +'/Users/').orderByChild(\"identifiant\" ).equalTo(userId ）; 
    userRef.once（value，
    function（dataSnapshot）{
    if（dataSnapshot.val（））{
    console.log（Loaded user，dataSnapshot .val（））; 
    deferred.resolve（dataSnapshot.val（））; 
   } else {
    console.info（通过id找不到用户，userId）; 
    deferred.reject（找不到用户识别符='+ userId + ）; 
   } 
   }，
   功能（误差）{
    console.error（ 错误加载用户，错误）; 
    deferred.reject（error）; 
   } 
   ）; 
   } else {
    deferred.reject（No id entered！）; 
   } 
    return deferred.promise; 
    
    
    
    
    $ h $ $ 
    
    
     .controller（'HomeController'，function（$ scope，UserFactory）{
    $ scope.identifient =''; $ 
    showUser（false）; 
    $ scope.error = errorMessage; 
   } else {
   删除$ scope.error; 
    
    
    var showUser = function（userObject）{
    if（userObject）{
    showError（false）; 
    $ scope.user = userObject; 
   } else {
    delete $ scope.user; 
   } 
   } 
    $ scope.loadUser = function（）{
    userPromise = new UserFactory（$ scope.identifient）; 
    userPromise.then（function（data）{
    showUser（data）; 
   }）。catch（function（error）{ 
    showError（error）; 
   }）; 
   } 
   }）
     
    
    
    
   模板
    
    
    
    < div ng-controller =HomeController> 
   < h2>主页模板< / h2> 
   < input ng-model =identifientplaceholder =identifient/> 
   < button ng-click =loadUser（）>查找用户< / button> 
   < hr /> 
   < div ng-if =user>用户：{{user}}< / div> 
   < div ng-if =error>错误：{{error}}< / div> 
   < / div> 
     
    
    
    
    
    
    
   希望有帮助。
    
   This is probably very simple ,so sorry... 
   I'm trying to create an application where i can create users and store informations by Firebase. There is an example here :
   
   http://image.noelshack.com/fichiers/2015/11/1426182307-log.png
   
   But now I want to check if the name already exists if someone wants to create a new user.
   
   How would I go about grabbing individual user information without knowing their id , like simplelogin:54?
   
   I've found this topic Get users by name property using Firebase but it is not the same thing because in my case I don't know the children after "Users"
   
   Cheers,
   解决方案
   Like Frank said, you must know something about the user to be able to look her/him up.
   
   However, here is a general answer. I am assuming that by "name" you mean the property "identifiant" that you've created.
   
   Suggestions
   
   
   Start by looking over the Firebase Query documentation.
   
   
   
   
   Short Answer
   
   
   To check if a user exists by the identifiant property, you'd orderByChild("identifiant") and query for a specific user with the .equalTo("<identifient_here>"). 
   For example, to check if a user with identifient="aaa",
   var usersRef = new Firebase("https://<YOUR-FIREBASE-APP>.firebaseio.com/Users");
   var identifient = "aaa";
   usersRef.orderByChild("identifiant").equalTo(identifient).once("value", function(snapshot) {
     console.log("Loaded user by identifient:",identifient,snapshot.val());
   });
   
   
   
   If instead you want to query by the key (such as simplelogin:53), you could query by using orderByKey() instead of orderByChild()... or just simply setting the ref to the user's key like so:
   
   
   var userKey = 'simplelogin:53';
   var userRef = new Firebase("https://<YOUR-FIREBASE-APP>.firebaseio.com/Users" + userKey);
   userRef.once("value", function(snapshot) {
     console.log("Loaded user",snapshot.val());
   });
   
   
   
   
   Long(er) Answer
   
   
   You can handle this with a user factory (see Angular Providers documentation).
   You return a promise in the factory using the $q service.
   Here is the  Angular API documentation for $q.
   
   
   Example with UserFactory
   
   
   Check out this working PLNKR example.
   It's tied to one of my public Firebase instances.
   I created the same simplelogin:53 user in /Users like you have.
   If you search for the identifient = aaa, you'll get the right user.
   The controller implementation here is for example purposes, and doesn't really do anything worth while. It's just for reference.
   
   
   The Data
   
   {
     "Users" : {
       "simplelogin:53" : {
         "identifiant" : "aaa"
       }
     }
   }
   
   
   UserFactory
   
   .factory('UserFactory', function($q, $firebaseObject, fbUrl){
     return function(userId){
       var deferred = $q.defer();
       if (userId.isNotEmpty()) {
         var userRef = new Firebase(fbUrl + '/Users/').orderByChild("identifiant").equalTo(userId);
         userRef.once("value", 
           function(dataSnapshot){
             if (dataSnapshot.val()) {
               console.log("Loaded user",dataSnapshot.val());
               deferred.resolve(dataSnapshot.val());
             } else {
               console.info("Couldn't find user by id",userId);
               deferred.reject("No user found by identifient = '"+userId+"'");
             } 
           },
           function(error){
             console.error("Error loading user",error);
             deferred.reject(error);
           }
         );
       } else {
         deferred.reject("No id entered!");
       }
       return deferred.promise;
     }
   })
   
   
   Controller
   
   .controller('HomeController',function($scope, UserFactory) {
     $scope.identifient = '';
     var showError = function(errorMessage) {
       if (errorMessage) {
         showUser(false);
         $scope.error = errorMessage;
       } else {
         delete $scope.error;
       }
     }
     var showUser = function (userObject) {
       if (userObject) {
         showError(false);
         $scope.user = userObject; 
       } else {
         delete $scope.user;
       }
     }
     $scope.loadUser = function() {
       var userPromise = new UserFactory($scope.identifient);
       userPromise.then(function(data){
         showUser(data);
       }).catch(function(error){
         showError(error);
       });
     }
    })
   
   
   Template
   
   <div ng-controller="HomeController">
     <h2>Home Template</h2>
     <input ng-model="identifient" placeholder="identifient"/>
     <button ng-click="loadUser()">Find User</button>
     <hr/>
     <div ng-if="user">User: {{user}}</div>
     <div ng-if="error">Error: {{error}}</div>
   </div>
   
   
   
   
   Hope that helps.
   
                           
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