【发布时间】:2014-03-14 16:34:13
【问题描述】:
我正在通过 jQuery 读取 XML 文件,并希望将值 (var_address) 传输到 a 中的属性 - “数据地址”。但是,我无法做到这一点。我认为我的 jQuery 语句存在问题,因为“var_address”没有获取任何值。有人可以帮忙吗?这是我所拥有的:
非常感谢。
jQuery:
$(document).ready(function () {
$("#paneldetails").append("<ul></ul>");
$.ajax({
type: "GET",
url: "/datafile.xml",
dataType: "xml",
success: function (xml) {
$(xml).find('Table1').each(function () {
var BuildingNumber = $(this).find('BuildingNumber').text();
var StreetName = $(this).find('StreetName').text();
var StreetType = $(this).find('StreetType').text();
var City = $(this).find('City').text();
var State = $(this).find('State').text();
var ZipCode = $(this).find('ZipCode').text()
$('div#var_address').attr('data-address', BuildingNumber + '+' + StreetName + '+' + StreetType + '+' + ',' + '+' + City + '+' + ',' + State + '+' + ' ' + ZipCode); // transfer var0023 to HTML file
return false;
}
},
error: function () {
alert("Error found in the data associated with this record");
}
});
});
HTML:
<div class="client" id="var_address" data-address="theaddress" ></div>
XML
<?xml version="1.0" encoding="UTF-8"?>
<DocumentElement>
<Table1>
<Business_x0020_Category>BusinessEssentials</Business_x0020_Category>
<PanelID>0001</PanelID>
<Business_x0020_ID>1001</Business_x0020_ID>
<BusinessName>Acme Tech</BusinessName>
<BuildingNumber>2173</BuildingNumber>
<StreetName>Salk</StreetName>
<StreetType>Ave</StreetType>
<Suite-Unit_x0020_Number>Suite 600</Suite-Unit_x0020_Number>
<City>Carlsbad</City>
<State>CA</State>
<ZipCode>92008</ZipCode>
<Telephone>XXX-XXX-XXXX</Telephone>
<Email>XXXXX@XXXXXX.XXX</Email>
</Table>
【问题讨论】:
-
Table1是你的XML element吗? -
可以分享xml吗
-
请分享您通过xml的ajax响应
-
@RohanKumar 这是 xml
-
BusinessEssentials 0001 1001 Acme Tech 2173 Salk Ave Suite 600 Carlsbad CA 92008 XXX-XXX-XXXX XXXXX@XXXXXX.XXX