在鼠标悬停时从jqgrid获取数据

Question

I'm trying to do a mouseover for the jquery and when the mouse is hovered over a certain row I can get the id from that row and populate information and display an image. However, I have been having the hardest time trying to do so.

Here is what I want to happen

Just like in the onSelectRow where I obtain the data using the following code

var ret = $('#list').jqGrid('getRowData', Id);

I want to use when I do a mouseover. However, I don't see a way of doing this. I tried the following under gridComplete

gridComplete: function() {'.jqgrow').mouseover(function(e) {
 var rowId = $('.jqgrow').parent(tr:first).attr('id');
 alert("You rolled over " + rowId.Id);
});
}

but it only gave me the ide number of the row of that table inside the jqgrid and I need the data from that row instead.

For instance, in my data I have Id, FirstName, LastName, FullName, Title, SortID

I would like to present a picture on the right side of my HTML page when hovered over certain rows by passing the Id to the HTML page, and querying through an array. If I can some how get the actual Id that's within my dataset I can do the rest.

Any help would be lovely.

I have given the entire code of my jqGrid at the bottom for reference.

jQuery("#list").jqGrid({
url: '/Providers/DynamicGridData/',
datatype: 'json',
mtype: 'GET',
colNames: ['Id', 'LastName', 'FirstName', 'FullName', 'Title', 'Url', 'SortId'],
colModel: [
{ name: 'Id', index: 'Id', width: 30, align: 'left', hidden: true },
{ name: 'LastName', index: 'LastName', width: 30, align: 'left', hidden: true },
{ name: 'FirstName', index: 'FirstName', width: 30, align: 'left', hidden: true },
{ name: 'FullName', index: 'FullName', width: 100, align: 'left' },
/*{ name: 'FirstName', index: 'FirstName', width: 100, align: 'left' },*/
{name: 'Title', index: 'Title', width: 200, align: 'left' },
{ name: 'Url', index: 'Url', width: 30, align: 'left', hidden: true },
{ name: 'SortId', index: 'SortId', width: 30, align: 'left', hidden: true}],
pager: jQuery('#pager'),
rowNum: 10,
rowList: [5, 10, 20, 50],
sortname: 'Id',
scrollOffset: 0,
width: '425',
altRows: 'true',
altClass: 'ui-priority-secondary',
autowidth: 'true',
height: '300',
altRows: 'true',
altClass: 'ui-priority-secondary',
viewrecords: true,
caption: 'Clinical Providers',
onSelectRow: function() {
var Id = $("#list").jqGrid('getGridParam', 'selrow');
if (Id) {
var ret = $('#list').jqGrid('getRowData', Id);
var url = ret.Url;
url.split(' ').join('');
//alert("id=" + ret.Id + "FullName=" + ret.FullName + "...");
window.location = "/" + url;
}
else { alert("Please select a row"); }
},
gridComplete: function() {
$('.jqgrow').mouseover(function(e) {
var rowId = $('.jqgrow').
alert("You rolled over " + rowId.Id);
});
}
});

Answer 1

I'm trying this code and it works fine:

gridComplete: function() {
    $('.jqgrow').mouseover(function(e) {
        var rowId = $(this).attr('id');
        alert('You rolled over ' + rowId);
    });
}

Answer 2

I am confused - when you say:

var rowId = $('.jqgrow').parent(tr:first).attr('id');

That should be returning the ID of the row. You can then pass rowID to the getRowData method to retrieve additional data for the row.

Answer 3

try this code..this code work... this code retrieve jqgrid row object ..

$('.jqgrow').mouseover(function(e) {
 var rowId = $(this).attr('id');
 var dataFromTheRow = jQuery("#list").jqGrid('getRowData',rowId);// this is your jqgrid row object;
 alert('your jqgrid row object id = ' + dataFromTheRow.id);
               });