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|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|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。

文章

问答

视频

学习中心

腾讯云实验室

直播

竞赛

腾讯云代码分析专区

腾讯iOA零信任安全管理系统专区

腾讯云架构师技术同盟交流圈

腾讯云数据库专区

腾讯云顾问专区

腾讯云原生专区

腾讯混元专区

腾讯云TCE专区

腾讯云Lighthouse专区

腾讯云HAI专区

腾讯云Edgeone专区

腾讯云存储专区

腾讯云智能专区

腾讯轻联专区 

腾讯云开发专区

TAPD专区

腾讯轻量云游戏服专区

腾讯云最具价值专家

腾讯云架构师技术同盟

腾讯云创作之星

腾讯云开发者先锋

腾讯云代码助手

云原生构建

TAPD 敏捷项目管理

Cloud Studio

SDK中心

API中心

命令行工具

涵盖代码开发、场景应用、自动测试全流程，助你从零构建专属AI助手

一站式MCP教程库，解锁AI应用新玩法

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列表返回澳大利亚，英国，美国_wins列表返回5,6,7我想印这样的澳大利亚5，英

问如何将两个列表结果打印在一起
EN

回答 6

Stack Overflow用户

Stack Overflow用户

Stack Overflow用户

社区

活动

圈层

关于

腾讯云开发者

热门产品

热门推荐

更多推荐

问如何将两个列表结果打印在一起EN

回答 6

Stack Overflow用户

Stack Overflow用户

Stack Overflow用户

社区

活动

圈层

关于

腾讯云开发者

热门产品

热门推荐

更多推荐

问如何将两个列表结果打印在一起
EN