如何从 XML 中查询最小值和最大值？

发布于 2024-09-25 18:54:03 字数 1063 浏览 1 评论 0原文

<?xml version="1.0" encoding="utf-8"?>
<Customers>
   <Customer Id="1">
    <Name>rtertr</Name>
    <DOB>2010-12-12T00:00:00</DOB>
    <EMail>[email protected]</EMail>
  </Customer>
  <Customer Id="2">
    <Name>west</Name>
    <DOB>0001-01-01T00:00:00</DOB>
    <EMail>[email protected]</EMail>
   </Customer> 
  <Customer Id="3">
    <Name>west</Name>
    <DOB>0001-01-01T00:00:00</DOB>
    <EMail>[email protected]</EMail>
   </Customer> 
</Customers>

如何使用 LinQ to XML 从上述示例 XML 中查找最小和最大 CustomerId？例如，对于上面的示例，Min 应为 1，Max 应为 3

原文

<?xml version="1.0" encoding="utf-8"?>
<Customers>
   <Customer Id="1">
    <Name>rtertr</Name>
    <DOB>2010-12-12T00:00:00</DOB>
    <EMail>[email protected]</EMail>
  </Customer>
  <Customer Id="2">
    <Name>west</Name>
    <DOB>0001-01-01T00:00:00</DOB>
    <EMail>[email protected]</EMail>
   </Customer> 
  <Customer Id="3">
    <Name>west</Name>
    <DOB>0001-01-01T00:00:00</DOB>
    <EMail>[email protected]</EMail>
   </Customer> 
</Customers>

How to find Min and Max CustomerId from the above sample XML using LinQ to XML? For example for the above sample, Min should be 1 and Max should be 3

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巷子口的你 2024-10-02 18:54:03

var doc = XDocument.Parse(...);

//.ToArray() so we only iterate once
var ids = doc.Root.Elements("Customer").Select(c => (int)c.Attribute("Id")).ToArray();
var minId = ids.Min();
var maxId = ids.Max();

正如 dtb 的评论所说，这将迭代数组两次，但只迭代 xml 文档一次。我觉得这是可读性和性能之间的一个很好的折衷——至少当它没有在生产代码中使用时。迭代 int 数组将是 xml 文档迭代的一个小因素。最好的解决方案是仅迭代文档一次并跟踪 dtb 显示的值。

var doc = XDocument.Parse(...);

//.ToArray() so we only iterate once
var ids = doc.Root.Elements("Customer").Select(c => (int)c.Attribute("Id")).ToArray();
var minId = ids.Min();
var maxId = ids.Max();

As dtb´s comment says, this will iterate the array twice but only the xml document once. I feel this is a good compromise between readability and performance - at least when its not used in production code. Iterating the int array will be a small factor of the xml document iteration. The best solution is to only iterate the document once and keep track of the values along the way as dtb shows.

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强者自强 2024-10-02 18:54:03

您可以使用 Min 和 Max 扩展方法：

var doc = XDocument.Parse("<Customers>...</Customers>");

var minId = doc.Root.Elements().Min(e => (int)e.Attribute("Id"));
var maxId = doc.Root.Elements().Max(e => (int)e.Attribute("Id"));

或 Aggregate 扩展方法（仅迭代 XML 文档一次）：

var doc = XDocument.Parse("<Customers>...</Customers>");

var result = doc.Root
                .Elements()
                .Select(e => (int)e.Attribute("Id"))
                .Aggregate(
                    Tuple.Create(int.MinValue, int.MaxValue),
                    (t, x) => Tuple.Create(Math.Max(t.Item1, x),
                                           Math.Min(t.Item2, x)));

var maxId = result.Item1;
var minId = result.Item2;

You can use the Min and Max extension methods:

var doc = XDocument.Parse("<Customers>...</Customers>");

var minId = doc.Root.Elements().Min(e => (int)e.Attribute("Id"));
var maxId = doc.Root.Elements().Max(e => (int)e.Attribute("Id"));

or the Aggregate extension method (iterating the XML document only once):

var doc = XDocument.Parse("<Customers>...</Customers>");

var result = doc.Root
                .Elements()
                .Select(e => (int)e.Attribute("Id"))
                .Aggregate(
                    Tuple.Create(int.MinValue, int.MaxValue),
                    (t, x) => Tuple.Create(Math.Max(t.Item1, x),
                                           Math.Min(t.Item2, x)));

var maxId = result.Item1;
var minId = result.Item2;

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