我正在尝试添加从表单中获取的数据并将其显示给JQGrid。
我的表格中有以下元素。
用户名
的文本框出生日期的Datepicker
Combobox用于选择国家。
两个按钮添加和清除按钮。
每当我单击Add按钮时,它都必须向JQGrid添加一行。然后单击它所具有的清除按钮来重置整个表。
目前,我正在尝试将表单中的数据显示到行中。
以下是我的努力。
<script>
$(function() {
$( "#pwd" ).datepicker();
var source = [{
text: "Australia",
value: 0
},
{
text: "India",
value: 1
},
{
text: "United States",
value: 2
},
{
text: "United Kingdom",
value: 3
}];
$("#jqxComboBox").jqxComboBox({
source: source,
theme: 'energyblue',
width: '240px',
height: '30px',
displayMember: 'text',
selectedIndex: 0,
valueMember: 'value'
});
$('#add').click(function(){
name=$('#name').val();
date=$('#pwd').val();
country=$('#jqxComboBox').val();
alert(name);
$('#jqGrid').jqGrid('addRowData',name,date,country);
}); });
</script>
<style type="text/css">
</head>
<body>
<div class="container">
<h2>Horizontal form</h2>
<form class="form-horizontal" role="form" id="add_form">
<input type="text" id="name"></input>
<input type="text" id="pwd"></input>
<div id="jqxComboBox"></div>
<input type="submit" value="add">
<input type="submit" value="reset">
</div>
</form><table id="jqGrid">
</table>
</body>
</html>
答案 0 :(得分:0)
可能你应该首先初始化jqGird并运行如下代码:
$(function() {
$("#jqGrid").jqGrid({
'datatype' : 'local',
// if there is no data at the beginning, just define an empty array [],
// or you can set init data with data option below
'data' : [ {
'name' : 'testUser',
'date' : '15/4/2016',
'country' : 'somewhere'
} ],
'colNames' : [ 'Name', 'Date', 'Country' ],
'colModel' : [ {'name' : 'name'}, {'name' : 'date'}, {'name' : 'country'} ],
'loadComplete' : function() { // You can add data manually using code below.
// You can define a datarow variable as single object and also an
// array of objects.
// array data example: var datarow = [{"name":"addedName",
// "date":"16/4/2016", "country":"any"},
// {"name":"addedName2", "date":"16/4/2016", "country":"any2"}];
var datarow = {
"name" : "addedName",
"date" : "16/4/2016",
"country" : "any"
};
// second parameter of method below is rowid just for generate id
// attribute of tr element.
// if keep rowid variable as undefined, jqGrid will generate a
// random id instead.
var rowid;
// last paremeter tell jqGrid add new data after last row.
$("#jqGrid").jqGrid("addRowData", rowid, datarow, "last");
}
});
});
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<script src="http://www.trirand.com/blog/jqgrid/js/jquery.js"></script>
<script src="http://www.trirand.com/blog/jqgrid/js/jquery.jqGrid.js"></script>
<table id="jqGrid"></table>
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