blocks|key|525379|text|这个怎么样：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|525380|for(i+=+0;+i+<+_teams.Length;+i%2B%2B)
{
++++Console.Write(_teams[i]+%2B+"+");
++++Console.Write(_wins[i]);
++++if(i+<+_teams.Length+-+1)
++++++++Console.Write(",");
}|code-block|syntax|javascript|525381|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

How about this:

<pre><code>for(i = 0; i &lt; _teams.Length; i++)
{
 Console.Write(_teams[i] + " ");
 Console.Write(_wins[i]);
 if(i &lt; _teams.Length - 1)
 Console.Write(",");
}
</code></pre>

blocks|key|2134029|text|我建议在IEnumerable上查看zip函数。就像这样：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2134030|var+lst1+=+new+List<string>(){"One",+"Two","Three"};
var+lst2+=+new+List<string>(){"A",+"B","C"};
var+combined+=+lst1.Zip(lst2,+(fst,snd)+=>+{return+fst+%2B+":"+%2B+snd;});
foreach+(var+item+in+combined)
{
++++Console.WriteLine+(item);
}|code-block|syntax|javascript|2134031|Zip将采用两个独立的列表，并允许您在这两个视图中构建一个视图。|2134032|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

I would suggest looking into the zip function on IEnumerable. Something like this:

<pre><code>var lst1 = new List&lt;string&gt;(){"One", "Two","Three"};
var lst2 = new List&lt;string&gt;(){"A", "B","C"};
var combined = lst1.Zip(lst2, (fst,snd) =&gt; {return fst + ":" + snd;});
foreach (var item in combined)
{
 Console.WriteLine (item);
}
</code></pre>

Zip will take two seperate lists and allow you to build a single view into both.

blocks|key|98214|text|试试这个：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|98215|foreach+(var+a+in+_teams.Zip(_wins,+(t,+w)+=>+new+{+t,+w+}))
{
++++Console.WriteLine(a.t+%2B+"+"+%2B+a.w);
}|code-block|syntax|javascript|98216|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Try this:

<pre><code>foreach (var a in _teams.Zip(_wins, (t, w) =&gt; new { t, w }))
{
 Console.WriteLine(a.t + " " + a.w);
}
</code></pre>

blocks|key|525434|text|您可以使用下面的菜单化代码|type|unstyled|depth|inlineStyleRanges|entityRanges|data|525435|++++++++List<string>+_terms+=+new+List<string>();
++++++++List<string>+_wins+=+new+List<string>();
++++++++_terms.Add("Australia");
++++++++_wins.Add("5");
++++++++using+(var+e1+=+_terms.GetEnumerator())
++++++++using+(var+e2+=+_wins.GetEnumerator())
++++++++{
++++++++++++while+(e1.MoveNext()+&&+e2.MoveNext())
++++++++++++{
++++++++++++++++var+item1+=+e1.Current;
++++++++++++++++var+item2+=+e2.Current;

++++++++++++++++//+use+item1+and+item2
++++++++++++}
++++++++}|code-block|syntax|javascript|525436|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

you can use below menioned code 

<pre><code> List&lt;string&gt; _terms = new List&lt;string&gt;();
 List&lt;string&gt; _wins = new List&lt;string&gt;();
 _terms.Add("Australia");
 _wins.Add("5");
 using (var e1 = _terms.GetEnumerator())
 using (var e2 = _wins.GetEnumerator())
 {
 while (e1.MoveNext() &amp;&amp; e2.MoveNext())
 {
 var item1 = e1.Current;
 var item2 = e2.Current;

 // use item1 and item2
 }
 }
</code></pre>

blocks|key|2134097|text|你可以用一个for循环..。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2134098|for+(int+i+=+0;+i+<+teams.Count;+i%2B%2B)
{
++++Console.WriteLine(_teams[i]);
++++Console.WriteLine(_wins[i]);
}|code-block|syntax|javascript|2134099|..。但字典是一个更好的解决方案：|2134100|Dictionary<string,+int>+_teams+=+new+Dictionary<string,+int>();
_teams.Add("Australia",+5);
_teams.Add("England",+6);
...
foreach(+KeyValuePair<string,+int>+kvp+in+_teams+)
{
++++Console.WriteLine("{0}+{1}",+kvp.Key,+kvp.Value);
}|2134101|MSDN字典|offset|length|2134102|entityMap|0|LINK|mutability|MUTABLE|url|http://msdn.microsoft.com/it-it/library/xfhwa508%2528v=vs.110%2529.aspx^0|0|0|0|0|0|6|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|X|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Y|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|Z|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|10|8|@]|9|@$M|11|N|12|1|13]]|A|$]]|$1|O|3|-4|5|6|7|14|8|@]|9|@]|A|$]]]|P|$Q|$5|R|S|T|A|$U|V]]]]

You could use a for loop...

<pre><code>for (int i = 0; i &lt; teams.Count; i++)
{
 Console.WriteLine(_teams[i]);
 Console.WriteLine(_wins[i]);
}
</code></pre>

... but Dictionary is a better solution:

<pre><code>Dictionary&lt;string, int&gt; _teams = new Dictionary&lt;string, int&gt;();
_teams.Add("Australia", 5);
_teams.Add("England", 6);
...
foreach( KeyValuePair&lt;string, int&gt; kvp in _teams )
{
 Console.WriteLine("{0} {1}", kvp.Key, kvp.Value);
}
</code></pre>

<a href="http://msdn.microsoft.com/it-it/library/xfhwa508%28v=vs.110%29.aspx" rel="nofollow">MSDN Dictionary</a>

blocks|key|525476|text|试试这个：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|525477|_teams.ForEach(x+=>+Console.WriteLine(x+%2B"+"+%2B+_wins[_teams.IndexOf(x)]));|code-block|syntax|javascript|525478|我们使用IndexOf方法从_teams获取当前字符串的索引，并在_wins的索引处获取元素。|offset|length|style|CODE|525479|但是，如果这是团队和胜利之间的1比1的映射，那么Dictionary将是我首选的数据结构。|525480|entityMap^0|0|0|4|7|E|6|X|5|0|O|A|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|S|8|@$I|T|J|U|K|L]|$I|V|J|W|K|L]|$I|X|J|Y|K|L]]|9|@]|A|$]]|$1|M|3|N|5|6|7|Z|8|@$I|10|J|11|K|L]]|9|@]|A|$]]|$1|O|3|-4|5|6|7|12|8|@]|9|@]|A|$]]]|P|$]]

Try this:

<pre><code>_teams.ForEach(x =&gt; Console.WriteLine(x +" " + _wins[_teams.IndexOf(x)]));
</code></pre>

We use the <code>IndexOf</code> method to get index of the current string from <code>_teams</code>, and get the element at that index of <code>_wins</code>.

However, if this is a 1-to-1 map between team and it's wins, a <code>Dictionary</code> would be my preferred data structure.

I have two lists as

<pre><code>foreach (var a in _teams)
{
 Console.WriteLine(a);
}

foreach (var b in _wins)
{
 Console.WriteLine(b);
}
</code></pre>

Each list have same number of values. Right now i am printing each values separtely but how can i print each value of a or b together . 

_teams list return Australia, England , USA
_wins list return 5,6,7

I want to print like that Australia 5, England 6, USA 7

Updated *
i am creating xml nodes so basically i dont want to print it together. I want values like
Australia than 5 than England than 6 so on and i will create xml nodes like

<pre><code>&lt;Team&gt;Australia&lt;/Team&gt;
&lt;Win&gt;5&lt;/Win&gt;
&lt;Team&gt;England &lt;/Team&gt;
&lt;Win&gt;6&lt;/Win&gt;
</code></pre>

...so on

How to print two list results together

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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我有两份名单foreach (var a in _teams){    Console.WriteLine(a);}foreach (var b in _wins){    Console.WriteLine(b);}每个列表都有相同数量的值。现在，我正在分别打印每个值，但如何将a或b的每个值打印在一起。_teams列...

问如何将两个列表结果打印在一起
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问如何将两个列表结果打印在一起EN