如何一次性将所有选定记录从sqlite发布到mysql phonegap / cordova

Question

我有一些运行良好的代码，但仅从数据库发送了第一条记录。问题是如何按照select语句指示一次发送所有记录。 我已经尝试遵循Sending multiple objects with node.js, mysql and socket.io和send data from phone to server，但是它们的实现并不能真正解决我的问题，因为我希望发布使用该帖子定义的记录以进行序列化和跟踪。

Index.html

    <script type="text/javascript"> 
        function sendAll() {
         db.transaction(function(tx) {
            Squery = 'SELECT * FROM details';
                tx.executeSql(Squery,
                    null,
                    function(tx, results)
                    {
                            for(i=0; i<results.rows. length; i++){
                                row = results.rows.item(0);

                    var name = row['name'];
                    var age = row['age'];

                    var dataString = "name=" + name + "&age=" + age +"&sendAll=";
                    if ($.trim(name).length > 0 & $.trim(age).length) {

                      $.ajax({ type: "POST",
                            url: "http://my_location/send.php",
                            data: dataString,
                            crossDomain: true,
                            cache: false,
                            success: function(data) {
                                if (data == "success") {
                                    alert("Record sent");
                                    $("#sendAll").val('submit');
                                } else if (data == "error") {
                                    alert("error");
                                }
                            }
                        });
                     }
                    return false;   
                            }});
                    });
                    }
        </script>


        <form id="sendAll"  method="POST" data-ajax="false"  onsubmit="return false;" class="formular">
            <button class="button button-block" onClick="sendAll()" data-inline="true">Send</button>
        </form> 

send.php

     include "db.php";
     if(isset($_POST['sendAll']))
     {
     $name=$_POST['name'];
     $age=$_POST['age'];
     $price=$_POST['price'];
     $q=mysqli_query($con,"INSERT INTO `details` (`name`,`age`) VALUES ('$name','$age')");
     if($q)
      echo "success";
     else
      echo "error";
     }