这是我的xml文件: -
<results value="1">
<result value="111">
<Country_Code value="IN"/>
<Country_Name value="India"/>
<Region_Name value="Gujarat"/>
<City value="Rajkot"/>
<loction lat="13.060422" lng="80.24958300000003"/>
</result>
<result value="222">
<Country_Code value="KE"/>
<Country_Name value="Kenya"/>
<Region_Name value="Vihiga"/>
<City value="Kakamega"/>
<loction lat="0.1182473" lng="34.7334515999997"/>
</result>
</results>
这是我的代码: -
<html>
<head>
<script src="http://ajax.googleapis.com/ajax/libs/jquery/1.8.3/jquery.min.js"></script>
<script>
var xml;
$.get(
"sea.xml",
null,
function (data) {
xml = data;
},
"xml"
);
function get_list(ls) {
var elName = $('#select').val();
var value = $('#value').val();
if(elName == "" || value == "")
{
}
else if (elName != "" || value != "")
{
var xPath = '//*[ @lat '+ ls +' "'+elName+'" and @lng '+ ls +' "'+value+'"]'+'/../City/@*';
var iterator = xml.evaluate(xPath, xml.documentElement, null,
XPathResult.UNORDERED_NODE_ITERATOR_TYPE, null);
var thisNode = iterator.iterateNext();
var str = '';
while (thisNode) {
if (str) {
str += ', ';
}
str += thisNode.textContent;
thisNode = iterator.iterateNext();
}
$("#result").text(str);
}
else
{}
}
</script>
</head>
<body>
<input type="text" id="select">
<input type="text" id="value">
<input type="button" name="button" value="Search" onclick="get_list('=')">
<div id="result">
</div>
</body>
</html>
使用此代码xpath返回属性值Rajkot
但我想在Xpath中更改以获取所有属性
在xPath的帮助下,如果xml文件上的值匹配,则仅返回city属性值
但我希望所有的属性值如: - IN,Inida,Gujarat,Rajkot,13.060422,80.24958300000003
答案 0 :(得分:1)
以下XPath:
//loction[@lat='13.060422' and @lng='80.24958300000003'][1]/preceding-sibling::City/@value
输出:
value="Rajkot"
以下XPath:
//loction[@lat='13.060422' and @lng='80.24958300000003'][1]/preceding-sibling::*/@*
输出:
value="IN"
-----------------------
value="India"
-----------------------
value="Gujarat"
-----------------------
value="Rajkot"