与C#抗争,在我的生成器中读取XML

与C#抗争,在我的生成器中读取XML,c#,xml,arrays,random,C#,Xml,Arrays,Random,我放弃了在Excel中执行此操作,因此我创建了一个XML文件(名为“ideas.XML”),如下所示: <ideas> <verbs> <verb>Verb1</verb> <verb>Verb2</verb> <verb>Verb3</verb> <verb>Verb4</verb> <verb>Verb5</verb

我放弃了在Excel中执行此操作,因此我创建了一个XML文件(名为“ideas.XML”),如下所示:

<ideas>
<verbs>
    <verb>Verb1</verb>
    <verb>Verb2</verb>
    <verb>Verb3</verb>
    <verb>Verb4</verb>
    <verb>Verb5</verb>

</verbs>
<adjectives>
    <adjective>Adjective1</adjective>
    <adjective>Adjective2</adjective>
    <adjective>Adjective3</adjective>
    <adjective>Adjective4</adjective>
    <adjective>Adjective5</adjective>
</adjectives>
<nouns>
    <noun>Noun1</noun>
    <noun>Noun2</noun>
    <noun>Noun3</noun>
    <noun>Noun4</noun>
    <noun>Noun5</noun>
</nouns>
</ideas>
我正试图在输出中添加一个随机化器,但我似乎无法正确执行。另外,我打印出数组中的第一个值只是为了让它工作,任何关于如何在0和最后一个之间选择一个随机数的提示都非常有用


非常感谢您的任何帮助、提示或建议。

使用
r
变量(即
Random
类实例)生成索引,而不是像代码中那样硬编码的
0
s:

textBox1.Text = ((verbTag[r.Next(0, verbTag.Count - 1)].InnerText) + " " + (adjectiveTag[r.Next(0, adjectiveTag.Count - 1)].InnerText) + " " + (nounTag[r.Next(0, nounTag.Count - 1)].InnerText)) + ".";
textBox1.Text = ((verbTag[r.Next(0, verbTag.Count - 1)].InnerText) + " " + (adjectiveTag[r.Next(0, adjectiveTag.Count - 1)].InnerText) + " " + (nounTag[r.Next(0, nounTag.Count - 1)].InnerText)) + ".";