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Php 要在数据库as3中插入计时器类时间吗_Php_Actionscript 3 - Fatal编程技术网

Php 要在数据库as3中插入计时器类时间吗

Php 要在数据库as3中插入计时器类时间吗,php,actionscript-3,Php,Actionscript 3,我想在我的数据库中插入计时器时间。我创建了一个计时器类,该类以分钟和秒为单位递增(在动态文本字段中)。我有一个主页按钮,我希望当我单击主页按钮时,计时器时间将插入到scoreu表的时间列中。所有代码工作正常。列正在更新,但时间未显示在数据库表中。请帮助我解决此问题。这是我的as3代码 var myCount:Number = 00; var seconds:Number = 00; var minutes:Number = 0; var myTime:Timer = new Timer(1000

我想在我的数据库中插入计时器时间。我创建了一个计时器类,该类以分钟和秒为单位递增(在动态文本字段中)。我有一个主页按钮,我希望当我单击主页按钮时,计时器时间将插入到scoreu表的时间列中。所有代码工作正常。列正在更新,但时间未显示在数据库表中。请帮助我解决此问题。这是我的as3代码

var myCount:Number = 00;
var seconds:Number = 00;
var minutes:Number = 0;
var myTime:Timer = new Timer(1000,myCount);
myTime.addEventListener(TimerEvent.TIMER, startCount);
my_time.text = "";
myTime.start();

function startCount(event:TimerEvent):void
{
     seconds +=1;
 if (seconds > 60) 
 {
  seconds =00;
  minutes +=1;
 } 
 my_time.text = minutes+":"+(seconds >= 10 ? seconds : "0"+seconds);
}
var variables:URLVariables = new URLVariables();
variables.time1 = my_time.text;
home_Btn.buttonMode = true;
home_Btn.addEventListener(MouseEvent.CLICK, indexBtn_click1);
function indexBtn_click1(event:MouseEvent):void
{
 myTime.stop();
var request:URLRequest = new URLRequest('http://localhost/timesend.php');
request.method = URLRequestMethod.POST;
request.data = variables;
var loader:URLLoader = new URLLoader(request);
loader.dataFormat = URLLoaderDataFormat.VARIABLES;
loader.load(request);
loader.addEventListener(Event.COMPLETE, dataOnLoad);

}

function dataOnLoad(event:Event) 
  {
    var variables:URLVariables = URLVariables(event.target.data);
    trace(variables.result);    // gives : System Updated
 }
这里是timesend.php

<?php

// data.php

include('connect1.php');

// if we are here, that's mean that we are already connected to mysql server
// and we don't need to do another connection

$etime = $_POST['time1'];


// $link is the same connection created in your connect.php script
if (mysqli_query($link, "INSERT INTO scoreu (user) VALUES('$etime')")) {

    echo 'result=System Updated';


} else {
    echo 'result=error';
}

mysqli_close($link);

表字段类型声明中出现问题..我将其声明为varchar..但现在我已将该类型修复为TEXT..现在它工作正常