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除了一個小問題，一切都正常工作，與我的jqgrid。我已經定義如下POSTDATA：基於用戶輸入過濾jqgrid
$(document).ready(function() {
$("#ctl00_ContentPlaceHolder2_drpUSite").change(function() {
site = ($("#ctl00_ContentPlaceHolder2_drpUSite").val());
loadusergrid();
});
var usrparams = new Object();
var site = ($("#ctl00_ContentPlaceHolder2_drpUSite").val());
//----grid code---------
$("#users").jqGrid({
prmNames: {
_search: "isSearch",
nd: null,
rows: "numRows",
page: "page",
sort: "sortField",
order: "sortOrder"
},
// add by default to avoid webmethod parameter conflicts
postData: { searchString: '', searchField: '', searchOper: '', sites: site },
datatype: function(postdata) {
mtype: "GET",
$.ajax({
url: 'Users.aspx/GetUsers',
type: "POST",
contentType: "application/json; charset=utf-8",
data: JSON.stringify(postdata),
dataType: "json",
success: function(data, st) {
if (st == "success") {
var grid = $("#users")[0];
var m = JSON.parse(data.d);
grid.addJSONData(m);
}
},
error: function() {
alert("Loading Failed!");
}
});
},
// this is what jqGrid is looking for in json callback
jsonReader: {
root: "rows",
page: "page",
total: "total",
records: "records",
cell: "cell",
id: "login",
repeatitems: true
},
colNames: ['Login', 'First Name', 'Last Name', 'Email', 'Site', 'Role', 'Room', 'UnitID', 'Supervisor', 'Super'],
colModel: [
{ name: 'login', index: 'login', width: 20 },
{ name: 'fname', index: 'fname', width: 20, hidden: true },
{ name: 'lname', index: 'lname', width: 60, align: "center", sortable: true, searchoptions: { sopt: ['eq', 'ne']} },
{ name: 'email', index: 'email', width: 20, align: "center", sortable: false },
{ name: 'site', index: 'site', width: 50, align: "center", sortable: true, searchoptions: { sopt: ['eq', 'ne']} },
{ name: 'role', index: 'role', width: 15, align: "center", sortable: true, searchoptions: { sopt: ['eq', 'ne']} },
{ name: 'room', index: 'room', width: 30, align: "center", sortable: true },
{ name: 'unitid', index: 'unitid', width: 10, align: "center", sortable: false },
{ name: 'super', index: 'super', width: 20 },
{ name: 'supername', index: 'supername', width: 10, align: "center", sortable: false },
],
pager: "#pageusers",
viewrecords: true,
caption: "Registered Users",
imgpath: 'themes/steel/images',
rowNum: 20,
rowList: [10, 20, 30, 40, 50],
sortname: "pname",
sortorder: "desc",
showpage: true,
gridModel: true, gridToolbar: true,
onSelectRow: function(id) {
var ret = jQuery("#users").getRowData(id);
accpara.id = ret.id;
accpara.pname = ret.pname;
accpara.pid = ret.pid;
accpara.bld = ret.bld;
accpara.cname = ret.cname;
accpara.amt = ret.amt;
accpara.status = ret.status;
accpara.notes = ret.notes;
accpara.lname = ret.lname;
}
});
jQuery("#users").navGrid('#pageusers', { view: false, del: false, add: false, edit: false },
{}, // default settings for edit
{}, // default settings for add
{}, // delete
{closeOnEscape: true, multipleSearch: true,
closeAfterSearch: true
}, // search options
{}
);
$("#users").setGridWidth(1300, true);
$("#users").setGridHeight(500, true);
jQuery("#users").jqGrid('filterToolbar');
//----grid code ends here
function loadusergrid() {
$("#users").setGridParam({ page: 1 }, { pgbuttons: true }, { pginput: true }, { postData: { "site": site} }).trigger("reloadGrid");
}
});
當頁面加載的第一時間，該工程..
現在我有4個下拉菜單，其過濾器的用戶。我已經寫了一個函數，當下拉更改時重新加載網格，但它沒有工作..我在這裏做錯了什麼？當我啓用回傳下拉菜單時，我會得到過濾結果。我想避免我的頁面回發:)。現在我已經添加了網站下拉菜單作爲過濾器。一旦開始工作生病添加剩餘的3.
螢火蟲顯示ajax調用成功觸發，但與一個空sitename。請注意，當頁面第一次加載時，網站下拉菜單會顯示一個空值。
在此先感謝