Javascript打印JS值并插入数据库

时间:2014-02-15 12:43:51

标签: javascript php mysql database

我正在使用脚本在我的网站上拖动图像,图像的位置我希望用PHP存储在MySQL数据库中。我的问题是我不知道如何打印javascript位置值,以便我可以将其插入数据库。其他一切都很好。我正在遵循的指南是:http://draggabilly.desandro.com/您可以查看最后一个跟踪位置并存储在变量中的示例。提前谢谢。

 <div id="house_wall1">
            <img src="x" class="draggie" style="position:relative; left:200px;">
            <img src="x" class="draggie" style="position:relative;">
            <img src="x" class="draggie" style="position:relative;">


            <!--Cypyrights and many thanks to http://desandro.mit-license.org for element movements -->

        </div>
        <script type="text/javascript" src="js/draggabilly.pkgd.min.js"></script>


            <script>
    ( function() {
      var container = document.querySelector('#house_wall1');
      var elems = container.querySelectorAll('.draggie');
      for ( var i=0, len = elems.length; i < len; i++ ) {
        var elem = elems[i];
        new Draggabilly( elem, {
      containment: true,
      });
     }
   })();


function onDragMove( instance, event, pointer ) {
  console.log( 'dragMove on ' + event.type +
    pointer.pageX + ', ' + pointer.pageY +
    ' position at ' + instance.position.x + ', ' + instance.position.y );

}
// bind event listener
draggie.on( 'dragMove', onDragMove );
// un-bind event listener
draggie.off( 'dragMove', onDragMove );
// return true to trigger an event listener just once
draggie.once( 'dragMove', function() {
return true;

});



 </script>

    </div>

更新: 试图让ajax解析值并打印它以确保它有效:

function loadXMLDoc()
        {
        var xmlhttp;
        if (window.XMLHttpRequest)
          {// code for IE7+, Firefox, Chrome, Opera, Safari
          xmlhttp=new XMLHttpRequest();
          }
        else
          {// code for IE6, IE5
          xmlhttp=new ActiveXObject("Microsoft.XMLHTTP");
          }
        xmlhttp.onreadystatechange=function()
          {
          if (xmlhttp.readyState==4 && xmlhttp.status==200)
            {
            dragEnd: true
            }
          }
        xmlhttp.open("POST","post.php",true);
        xmlhttp.send();
        }

0 个答案:

没有答案
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