javascript - 使用 OnClick 函数 JavaScript 删除 Firebase 节点

Question

我有两个 JavaScript 函数，一个用于生成表，另一个用于通过从 Firebase 数据库中删除节点来删除行。我不断收到一条错误消息，指出 deleteRecord() 函数未定义。请帮忙...

            function generate_table(){

                        $('#emp_body').html('');    
                        console.log(dArr);

                        for (var i = 0; i < dArr.length; i++) {

                        var tr;
                        tr = $('<tr/>');

                        var strSleeve = "View Sleeve";
                        var sleeveLink = strSleeve.link(dArr[i][1].downloadURLSleeve);
                        var strAud = "View Audio";
                        var audioLink = strAud.link(dArr[i][1].downloadURLFile);
                                    tr.append("<td>" +  (i+1) + "</td>");
                                    tr.append("<td>" +  childKeys[i] + "</td>");
                                    tr.append("<td>" +  dArr[i][1].stageName + "</td>");
                                    tr.append("<td>" +  dArr[i][1].fullName + "</td>");
                                    tr.append("<td>" +  dArr[i][1].email + "</td>");
                                    tr.append("<td>" +  dArr[i][1].city + "</td>");
                                    tr.append("<td>" +  dArr[i][1].cell + "</td>");
                                    tr.append("<td>" +  sleeveLink + "</td>");
                                    tr.append("<td>" +  audioLink + "</td>");


                                    tr.append('<td>' + '<button id="deleteBtn" onclick="deleteRecord(\''+childKeys[i]+'\');" class="btn btn-danger">Delete Record</button>' + '</td>');
                                    $('#emp_body').append(tr);

                }

            }


                                        // DELETE FUNCTION
            function deleteRecord(key){     
                                var refDB = firebase.database().ref().child('submissions/'+key);
                                    refDB.once("value")
                                    .then(function(snapshot) {
                                        snapshot.ref.remove();
                                        alert("Record deleted..!");
                                }).catch(function(error) {alert("Data could not be deleted." + error);});
                        } 

Answer 1

您应该执行以下操作:

        function deleteRecord(key){     
                            var refDB = firebase.database().ref('submissions/' + key);
                            refDB.remove()
                            .then(function() {
                              console.log("Remove succeeded.")
                            })
                            .catch(function(error) {
                              console.log("Remove failed: " + error.message)
                            });
                    } 

请参阅此处相应的文档:https://firebase.google.com/docs/reference/js/firebase.database.Reference#remove

您不应使用用于读取数据的 once() 方法，如下所述:https://firebase.google.com/docs/reference/js/firebase.database.Reference#once