Question

我想在Xml中读取xml数据，但问题是url有请求参数，这使我无法读取xml数据w。我所知道的是像xamarin这样的标准读取数据xml

var document = XDocument.Load("http://www.forex.se/ratesxml.asp? id=492");
var row = document.Descendants("row").FirstOrDefault();
if(row != null)
{
   var sellCash = row.Element("sell_cash")?.Value;
   USDSellLabel.Text = sellCash;
}

我的Request Xml是

<?xml version='1.0' encoding='UTF-8'?>
<Request param="name" param="name" param="name">
</Request>

所以我的问题是如何将该请求输入到我的代码中以便我可以获取/读取xml数据？

这是我在Postman中尝试从URL返回的xml

<?xml version="1.0" encoding="UTF-8"?>
<Response type="service-response">
    <ResponseCode>00</ResponseCode>
    <ResponseStatus>success</ResponseStatus>
    <ResponseMessage>
        <Accounts>
            <Account Name= 'Payments' AvBal ='0 USD' ResBal ='0.00 S$' Currency ='USD' LocalBalance ='0.00 MYR' defaultW ='false'  />
        </Accounts>
        <Accounts>
            <Account Name= 'Point' AvBal ='100 LP' ResBal ='0.00' Currency ='LP' LocalBalance ='100.00 MYR' defaultW ='false'  />
        </Accounts>

        <Accounts>
            <Account Name= 'AccountMoney' AvBal ='381000 USD' ResBal ='0.00 S$' Currency ='USD' LocalBalance ='116.08 MYR' defaultW ='true'  />
        </Accounts>


        <PdfUrl>http://123.123.123.123:7171/file?a123123.pdf</PdfUrl>
    </ResponseMessage>
</Response>

Answer 1

使用xml linq：

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Xml;
using System.Xml.Linq;
using System.IO;

namespace ConsoleApplication14
{
    class Program
    {
        const string FILENAME = @"c:\temp\test.xml";
        static void Main(string[] args)
        {
            string xml = File.ReadAllText(FILENAME);

            XDocument doc = XDocument.Parse(xml);

            var response = doc.Elements("Response").Select(x => new
            {
                type = (string)x.Attribute("type"),
                code = (int)x.Element("ResponseCode"),
                status = (string)x.Element("ResponseStatus"),
                message = x.Descendants("Account").Select(y => new
                {
                    name = (string)y.Attribute("Name"),
                    avBal = (string)y.Attribute("AvBal"),
                    resBal = (string)y.Attribute("ResBal"),
                    currency = (string)y.Attribute("Currency"),
                    localBalance = (string)y.Attribute("LocalBalance"),
                    defaultw = (Boolean)y.Attribute("defaultW")
                }).ToList()
            }).FirstOrDefault();
        }

    }
}

如何在Xamarin Forms中生成请求XML并使用它来读取数据Xml

1 个答案: