php - 数据未插入数据库

Question

数据没有被插入到数据库中

                </div>
            <div id="osx-modal-content" ng-app="newtask" ng-controller="taskcontroller">
                <div id="osx-modal-title">Adding a new task</div>
                <div class="close"><a href="#" class="simplemodal-close">x</a></div>
                <div id="osx-modal-data">
                    <form method="POST" action="">
                    Task Tittle: <input type="text" name="name" ng-model="taskname" >
                    <input class="add-input" name="description" placeholder="I need to..." type="text" ng-model="description"  ng-model-instant />
                    Due date : <input id="date" type="text" name="date" size="15" ng-model="date"/>
                    Priority : <select id="priority" name="priority" ng-model="priority">
                                <option name="high">High</option>
                                <option name="medium">Medium</option>
                                <option name="low">low</option>
                                 </select>
                    Date created: <input id="datec" type="text" name="datec" size="15"/>
                    <button type="submit" class="simplemodal-close" ng-click="insertdata()"><h5>Add</h5></button>
                    </form>
                </div>
                </div>

这是我用于 insertdata() 的 angularjs 函数:

var app=angular.module('newtask',[]);
    app.controller('taskcontroller',function($scope,$http){

        $scope.insertdata=function(){
            $http.post("insert.php",{'taskname':$scope.taskname,'description':$scope.description,'date':$scope.date,'priority':$scope.priority,'datec':$scope.datec})
                .success(function (data, status, headers, config) {
                    console.log("New Task Inserted")
                })
        }

});

这是用于解析 json 的 php 文件和用于将数据插入数据库的查询

$data = json_decode(file_get_contents("php://input"));
$taskname = mysql_real_escape_string($data->taskname);
$description = mysql_real_escape_string($data->description);
$date = mysql_real_escape_string($data->date);
$priority = mysql_real_escape_string($data->priority);
$datec = mysql_real_escape_string($data->datec);
mysql_connect("localhost","root","");
mysql_select_db("todo");
mysql_query("INSERT INTO tasks(`name` , `datecreated` , `duedate` , `priority` , `description`)
              VALUES('".$taskname."' , '".$datec."' , '".$date."' , '".$priority."' , '".$description."')");

Answer 1

在你的查询之后使用die(mysql_error());你会找到你的解决方案

 mysql_query("INSERT INTO tasks(`name` , `datecreated` , `duedate` , 
`priority` , `description`)
          VALUES('".$taskname."' , '".$datec."' , '".$date."' , 
 '".$priority."' , '".$description."')") or die(mysql_error());