Javascript Ajax Jquery并行调用不提供数据
我正在尝试在我的aspx页面中实现一个google图表。 我有一个搜索按钮,它将显示两个饼图。我使用Ajax Jquery并行读取数据。这是我的密码Javascript Ajax Jquery并行调用不提供数据,javascript,jquery,ajax,google-visualization,Javascript,Jquery,Ajax,Google Visualization,我正在尝试在我的aspx页面中实现一个google图表。 我有一个搜索按钮,它将显示两个饼图。我使用Ajax Jquery并行读取数据。这是我的密码 <div> <div style="text-align: left; padding-left: 30px;"> <uc1:ucChartDateSelector ID="ucChartDateSelector1" runat="server"></uc1:ucCh
<div>
<div style="text-align: left; padding-left: 30px;">
<uc1:ucChartDateSelector ID="ucChartDateSelector1" runat="server"></uc1:ucChartDateSelector>
<asp:Button ID="btnSearch" runat="server" Text="Search" AutoPostBack="false"
OnClientClick="javascript:GenerateChartReport();" />
</div>
</div>
<!--Div that will hold the dashboard-->
<div id="dashboard_div">
<!--Divs that will hold each control and chart-->
<div id="chartLine_div" style="width: 600px; height: 400px;">
<img src="../images/ajax-loader.gif" alt="Loading Report" />
</div>
<div id="visualization" style="width: 600px; height: 400px;">
<img src="../images/ajax-loader.gif" alt="Loading Report" />
</div>
</div>
<script type="text/javascript">
var toDate = '<%=ucChartDateSelector1.ToDate%>';
var fromdate = '<%=ucChartDateSelector1.FromDate%>';
var reportFor = '<%=ucChartDateSelector1.ddlReportType.SelectedItem.Text%>';
var periodFor = '<%=ucChartDateSelector1.ddlDateSelection.SelectedItem.Text%>';
function GenerateChartReport() {
$.ajax({
type: 'POST',
url: 'Dashboard.aspx/GetData',
contentType: 'application/json',
dataType: 'JSON',
async: true,
success: function (response) {
TwoColumnReport(response.d, "visualization", "Google Charts Example");
},
error: function (error) {
console.log(error);
}
}),
$.ajax({
type: 'POST',
url: 'Dashboard.aspx/ReadBooking',
contentType: 'application/json',
dataType: 'json',
async: true,
data: JSON.stringify({ fromDate: fromdate, toDate: toDate, reportFor: reportFor, periodFor: periodFor }),
success: function (response) {
TwoColumnReport(response.d, "chartLine_div", "Google Dealy Example");
},
error: function (error) {
console.log(error);
}
});
}
function TwoColumnReport(dataValues, mainDivId, reportTitle) {
var data = new google.visualization.DataTable();
data.addColumn('string', 'Column Name');
data.addColumn('number', 'Column Value');
for (var i = 0; i < dataValues.length; i++) {
data.addRow([dataValues[i].ColumnName, dataValues[i].Value]);
}
new google.visualization.PieChart(document.getElementById(mainDivId)).
draw(data, { title: reportTitle });
}
</script>`
问题是回邮。
将JavaScript代码修改为返回false
function GenerateChartReport() {
return false;
}
并从asp:按钮中删除AutoPostBack=“false”
您可以更改WebMethod
以返回string
并将dataList
序列化为json:
[WebMethod]
public static string GetData()
{
int milliseconds = 2000;
Thread.Sleep(milliseconds);
List<Data> dataList = new List<Data>();
dataList.Add(new Data("Column 1", 100));
dataList.Add(new Data("Column 2", 200));
dataList.Add(new Data("Column 3", 300));
dataList.Add(new Data("Column 4", 400));
var jsonSerialiser = new JavaScriptSerializer();
var jsonString = jsonSerialiser.Serialize(dataList);
return jsonString;
}
[WebMethod]
公共静态字符串GetData()
{
整数毫秒=2000;
睡眠(毫秒);
List dataList=新列表();
添加(新数据(“第1栏”,100));
添加(新数据(“第2栏”,200));
添加(新数据(“第3栏”,300));
添加(新数据(“第4栏”,400));
var jsonSerialiser=新的JavaScriptSerializer();
var jsonString=jsonSerialiser.Serialize(dataList);
返回jsonString;
}
然后您的ajax帖子,在success:function(response){}
上,
响应
将在列表中列出
[WebMethod]
public static string GetData()
{
int milliseconds = 2000;
Thread.Sleep(milliseconds);
List<Data> dataList = new List<Data>();
dataList.Add(new Data("Column 1", 100));
dataList.Add(new Data("Column 2", 200));
dataList.Add(new Data("Column 3", 300));
dataList.Add(new Data("Column 4", 400));
var jsonSerialiser = new JavaScriptSerializer();
var jsonString = jsonSerialiser.Serialize(dataList);
return jsonString;
}