Question

我创建了一个列表视图，一旦从数据库中检索到数据，它就会显示在该列表视图中。我想刷一下listview项目以从数据库中删除数据，并希望更新的数据显示在listview中。我可以在这个..plz帮助中实现。

showRecords（），用于显示数据库中的数据

function showRecords() 
{
var selectAllStatement = "SELECT * FROM  list ORDER BY subject";

db.transaction(function (tx) {

    tx.executeSql(selectAllStatement, [], function (tx, result) {
    var dataset = result.rows.length;
    var combinelist='';

    for (var i = 0;  i < dataset; i++) 
    {

        res = result.rows.item(i);
        var Lid=res.id;
        var Lsubject=res.subject;
        var Ldesc=res.desc;
        var Ldate=res.date;
        combinelist+='<li id="lists" rel="'+Lid+'" style="border-bottom:solid; background:#CFF">' + Lsubject + '<br>' + Ldesc + '<br>' + Ldate + '<br>' + '<a href="#" data-ajax="false" data-role="button" onclick="loadRecord('+Lid+');" class="ui-btn ui-corner-all ui-btn-inline" id="editbtn"> Edit</a></li>';
            $("#showlist").html(combinelist).listview('refresh');
             }
          });
        });
      }

列表视图 ..

的代码

<center><h2> To-Do List</h2></center>
<ul id="showlist" data-role="listview">
</ul>

Pageinit

$(document).on("pageinit","#pageone",function(){
localStorage.Lid=Lid;
alert(Lid);
$( "ul_showlist" ).on( "swiperight", swipeHandler );
function swipeHandler(){
alert("hello");;

var sql='DELETE FROM list WHERE id='+Lid+'';
  db.transaction(function (tx) { tx.executeSql(sql,showRecords); alert("Delete Sucessfully"); });
}
});

Answer 1

感谢能够滑动并删除记录..

    $('#showlist #lists').on("swiperight",swiperight_list);

    function swiperight_list(event)
    {
    var Lid=$(this).attr('rel');
    var sql='DELETE FROM list WHERE id='+Lid+'';
    db.transaction(function (tx) { tx.executeSql(sql); 
    alert("Delete Sucessfully"); });
    }

从列表项滑动以从数据库中删除数据

1 个答案: