gzlx
gzlx
  • 发布:2016-02-23 13:16
  • 更新:2016-02-23 13:16
  • 阅读:1899

PopPicker 需要按两次以上才出来

分类:MUI

自定义的数据,从sqlite中获取数据,很不灵敏,需要按多次才弹出选择界面,选择界面出现过一次以后,就灵敏了,每次按都出来
function getcheckoutday(){

            var sql = 'select RecoId as value,Cname as text From basecheckoutday Where RecoType = "1" Order By BillNo; '  

            db.doQuery(sql, function(result) {  

                var userPicker = new mui.PopPicker();  
                userPicker.setData(result);  
                var showUserPickerButton = document.getElementById('checkoutday');  

                showUserPickerButton.addEventListener('tap', function(event) {  
                    userPicker.show(function(items) {  
                        showUserPickerButton.setAttribute("data-col", items[0].value);  
                        showUserPickerButton.value = items[0].text;  
                        userPicker.CLOSED;  

                        //返回 false 可以阻止选择框的关闭  
                        //return false;  
                    });  
                }, false);                    

            });                   

    }
2016-02-23 13:16 负责人:无 分享
已邀请:

该问题目前已经被锁定, 无法添加新回复