Question

我尝试创建一个查询不同用户的问题和答案。我的目标是将不同的表连接在一起以创建我想要的结果。当前，SQL查询总是在注释列中，即从提出问题的用户处生成相同的内容。因此，根本不会输出UserA的答案。我将 INNER JOIN 更改为 LEFT / RIGHT ，然后可以使用，但是问题（评论）消失了。

.controller('pASSWORDGMCtrl', ['$scope', '$stateParams', 'aaaService',
 function ($scope, $stateParams, aaaService) {
   $scope.user = {
        email:"",
        pass:"",
    }
   $scope.signIn = function(){
    aaaService.login($scope.user.email, $scope.user.pass);
   }

aaaService.status();


$scope.signOut = function(){
    aaaService.signout();
}
}])

我还尝试了使用SQL语句“ UNION” 的不同方法，但是UserQ和UserA之间存在明显的分隔，这意味着当填充另一列时，其中一列始终为空

angular.module('firebaseAaa', ['firebase'])
.factory('aaaService', ['$firebaseArray','$firebaseAuth',
function($firebaseArray, $firebaseAuth){ 
var config = {
apiKey: "",
authDomain: "",
databaseURL: "",
projectId: "",
storageBucket: "",
messagingSenderId: ""
};
if(!firebase.apps.length){
firebase.initializeApp(config);
}

var ref;
var adminArray; 

return{     

 status: function(){
 firebase.auth().onAuthStateChanged(function(user) {

if (user) {

  // User is signed in.

  // if(loggingIn){
  alert("Login success.");

    ref = firebase.database().ref().child("Admin/" + user.uid);

    adminArray = $firebaseArray(ref); 


   }

       // loggingIn=false;
   else {

  // No user is signed in.

}

});
},

login: function(email, pass) {

  firebase.auth().signInWithEmailAndPassword('abc@gmail.com', 'aaa').catch(function(error) {

    // Handle Errors here.

    var errorCode = error.code;

    var errorMessage = error.message;

    alert("Error signing in user: ", errorCode + " " + errorMessage);

  });

},


  signout: function() {

  firebase.auth().signOut().then(function() {

    // Sign-out successful.

    alert("You are now signed out.");
     // loggingIn= false;

  }).catch(function(error) {

    // An error happened.

    // Handle Errors here.

    var errorCode = error.code;

    var errorMessage = error.message;

    alert("Error signout: ", errorCode + " " + errorMessage);

  });

  },
  }

  }])

Answer 1

您尝试过吗？

SELECT C.username UserQ,
       R.username UserA,
       C.commentID,
       D.Title,
       C.text Comment
  FROM Comments C 
  JOIN Detail D 
    ON C.vID = D.vId 
  JOIN Search S 
    ON C.vID = S.vID 
LEFT OUTER JOIN Replies R 
    ON C.commentID = R.CommentID_ParentID 
 WHERE D.categoryId = 25 
   AND S.searched_keyword = 'stackoverflow' 
 ORDER
    BY commentID,
       replyID

实际上，为什么不使用一些示例数据创建SQLfiddle？

Answer 2

INER JOIN，OUTER JOIN，FULL JOIN，LEFT JOIN

http://www.phpknowhow.com/mysql/joins/

联接多个表-评论和回复

2 个答案: