使用Linq-to-Xml和一些循环读取几个元素
本文关键字:几个 元素 读取 Linq-to-Xml 循环 使用 | 更新日期: 2023-09-27 18:15:13
使用以下代码,我目前正在读取每个<nv>
的<ns>CounterSales</ns>
(第三个元素),这很好。我将包括代码和XML。
这是代码,如果ElementAt
是硬编码的,效果很好:
var xDoc = XDocument.Parse(xmlStr);
var nvs = xDoc.Descendants("nv");
var nads = nvs.Select(nv => nv.Elements("nad").First().Value).ToList();
var thirdRs = nvs.Select(nv => nv.Elements("r").ElementAt(2).Value).ToList();
xmlStr
的值:
<ni>
<nss>20150927</nss>
<gp>Addon</gp>
<ns>CounterBlah1</ns>
<ns>CounterBlah2</ns>
<ns>CounterSales</ns>
<ns>CounterBlah4</ns>
<ns>CounterBlah5</ns>
<ns>CounterBlah6</ns>
<nv>
<nad>Style=1,Rfu=1,Id=132</nad>
<r>0</r>
<r>15</r>
<r>8</r>
<r>3</r>
<r>2</r>
<r>2</r>
</nv>
<nv>
<nad>Style=1,Rfu=1,Id=433</nad>
<r>0</r>
<r>15</r>
<r>30</r>
<r>3</r>
<r>2</r>
<r>2</r>
</nv>
<nv>
<nad>Style=1,Rfu=1,Id=665</nad>
<r>0</r>
<r>15</r>
<r>90</r>
<r>3</r>
<r>2</r>
<r>2</r>
</nv>
</ni>
但是假设我想读取多个计数器。我有一个数据表,目前有两行:[CounterSales]
&[CounterBlah5]
。我想以同样的方式处理另外两个计数器:
int index = 0;
DataTable results = GetCounters();
//results has two rows: [CounterSales] & [CounterBlah5]
foreach (DataRow row in results.Rows)
{
string counter = cellRow["counter"]; //counter equals "CounterSales"
/* Iterate through all <ns> to see where "CounterSales" is located.
In this case, index = 2. On next foreach, index equals 4 */
var xDoc = XDocument.Parse(xmlStr);
var nvs = xDoc.Descendants("nv");
var nads = nvs.Select(nv => nv.Elements("nad").First().Value).ToList();
var thirdRs = nvs.Select(nv => nv.Elements("r").ElementAt(index).Value).ToList();
}
任何帮助都是感激的。
谢谢。
如果我正确理解了你的问题,这就是你需要的
DataTable results = GetCounters();
//results has two rows: [CounterSales] & [CounterBlah5]
foreach (DataRow row in results.Rows)
{
string counter = cellRow["counter"]; //counter equals "CounterSales"
/* Iterate through all <ns> to see where "CounterSales" is located.
In this case, index = 2. On next foreach, index equals 4 */
var xDoc = XDocument.Parse(xmlStr);
//here we find the index
var requiredIndex = xDoc.Descendants("ns")
.Select((node, idx) => new { node, idx})
.First(x => x.node.Value.Equals(counter)).idx;
var nvs = xDoc.Descendants("nv");
var nads = nvs.Select(nv => nv.Elements("nad").First().Value).ToList();
var thirdRs = nvs.Select(nv => nv.Elements("r").ElementAt(requiredIndex).Value).ToList();
}
编辑:从这篇文章中找到List<T>
索引的更短语法。
var requiredIndex = xDoc.Descendants("ns").ToList().FindIndex(x => x.Value == counter);