使用LINQ到XML，如何基于序数位置连接两组数据?[英] Using LINQ to XML, how can I join two sets of data based on ordinal position?

问题描述

Using LINQ to XML, how can I join two sets of data based on ordinal position?

<document>
    <set1>
        <value>A</value>
        <value>B</value>
        <value>C</value>
    </set1>
    <set2>
        <value>1</value>
        <value>2</value>
        <value>3</value>
    </set2>
</document>

Based on the above fragment, I would like to join the two sets together such that "A" and "1" are in the same record, "B" and "2" are in the same record, and "C" and "3" are in the same record.

推荐答案

This is what the Enumerable.Zip extension does in .NET 4. You'd write it like so (assuming that this is the entire XDocument):

var set1Elements = document.Element("set1").Elements();
var set2Elements = document.Element("set2").Elements();
var results = set1Elements.Zip(set2Elements,
    (s1, s2) => new { Value1 = s1.Value, Value2 = s2.Value });

If you're using .NET 3.5 or earlier, it's not too difficult to write the Zip extension:

public static IEnumerable<TResult> Zip<TFirst, TSecond, TResult>(
    this IEnumerable<TFirst> first,
    IEnumerable<TSecond> second,
    Func<TFirst, TSecond, TResult> resultSelector)
{
    using (var firstEnumerator = first.GetEnumerator())
    using (var secondEnumerator = second.GetEnumerator())
    {
        while ((firstEnumerator.MoveNext() && secondEnumerator.MoveNext()))
        {
            yield return resultSelector(firstEnumerator.Current,
                secondEnumerator.Current);
        }
    }
}

其他推荐答案

Here's another method using the overload of Select that will include an element's index

XElement set1 = document.Root.Element("set1");
XElement set2 = document.Root.Element("set2");

var query = from value1 in set1.Descendants("value").Select((ele, idx) => new { Value = ele.Value, Index = idx })
            join value2 in set2.Descendants("value").Select((ele, idx) => new { Value = ele.Value, Index = idx })
            on value1.Index equals value2.Index
            select new { Value1 = value1.Value, Value2 = value2.Value };