我正在使用jqGrid 4.0并遇到问题。
这是我的JavaScript代码:
$grid.jqGrid({
url:'parties-process.php?action=SELECT',
datatype: "json",
colNames:['Party ID', 'Party Name', 'Balance', 'Opening Balance', 'Opening Debit', 'Options'],
colModel:[
{name:'partyID',index:'partyID', width:55},
{name:'partyName',index:'partyName', width:150},
{name:'balance',index:'balance', width:50, align:'right'},
{name:'opening',index:'opening', hidden:true},
{name:'openingdr',index:'openingdr', hidden:true},
{name:'act',index:'act', width:150, sortable:false, align:'center', hidden:true}
],
scroll: true,
autowidth: true,
height: myLayout.panes.center.outerHeight()-183,
pager: jQuery('#pager'),
rowNum:999,
rowList:[10,20,30],
sortname: 'partyID',
viewrecords: true,
sortorder: "desc",
caption: "Parties",
gridComplete: function(){
var data = $grid.jqGrid('getDataIDs');
for(var i=0;i < data.length;i++){
var id = data[i];
editbtn = "<small><a class=\"editbtn\" onclick=\"edit('"+id+"');\">Edit</a></small> ";
deletebtn = "<small><a class=\"deletebtn\" onclick=\"del('"+id+"');\">Delete</a></small>";
$grid.jqGrid('setRowData',data[i],{act:editbtn+deletebtn});
}
$(".editbtn").button({
icons: { primary: "ui-icon-pencil" },
text: false
});
$(".deletebtn").button({
icons: { primary: "ui-icon-close" },
text: false
});
}
}).jqGrid('bindKeys');
问题是,当我按向上和向下箭头键时,选择了相应的行,bt它还会滚动整个网格。
答案 0 :(得分:2)
我这样做是为了解决问题:
$(grid).bind('keydown', function (e) {
if (e.keyCode == 38 || e.keyCode == 40) e.preventDefault();
});