blocks|key|952952|text|为什么每个人都要创造6位数字？真正需要的是创建4和连接。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|952953|int+random+=+new+Random().Next(1000,+9999);
string+code+=+"00"+%2B+random.ToString();|code-block|syntax|javascript|952954|编辑：|952955|谢谢你标记为正确答案，但我的代码是错误的。如果不将最小值传递给下一个方法，则可以得到从1到999的数字，这是您不想要的。我编辑了我的答案。希望现在什么都不晚。|952956|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|P|8|@]|9|@]|A|$]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

why is everybody creating 6 digit ? what really is needed is create 4 and concat.

<pre><code>int random = new Random().Next(1000, 9999);
string code = "00" + random.ToString();
</code></pre>

EDIT:

Thanks for marking as correct answer but my code is wrong. If you don't pass minimum value to next method, you can get numbers from 1 to 999 which you don't want. I edited my answer. Hope it's not late for anything.

blocks|key|1378204|text|public+string+getRandomNumber()
{
++++Random+random+=+new+Random();
++++string+randomNumber+=+"00"%2Brandom.Next(10000).ToString("D4");
++++return+randomNumber;
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1378205|您不应该在随机中设置一个fix种子，因为它总是会生成相同的随机数序列。例如，如果您在循环中调用您的函数，它将始终返回相同的数字，因此不是随机的。|unstyled|1378206|另外，Next()函数中的参数是上界排他性的，所以使用10000代替9999，这样也可以得到9999作为随机数。|1378207|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|L|8|@]|9|@]|A|$]]|$1|G|3|H|5|F|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|F|7|N|8|@]|9|@]|A|$]]]|J|$]]

<pre><code>public string getRandomNumber()
{
 Random random = new Random();
 string randomNumber = "00"+random.Next(10000).ToString("D4");
 return randomNumber;
}
</code></pre>

You should not set a fix seed in the Random, because it will allways generate the same random number sequence. For example if you call your function in a loop it will allways return the same number, thus not being random. 

Also the parameter in the Next() function is the upper bound exclusive so use 10000 instead 9999 so you can get 9999 as a random number as well.

blocks|key|1378119|text|您编写了.Next(999999)，所以我假设您需要从0到999999的随机值。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1378120|首先，您应该编写1000000，因为.Next()确定一个从0到独占输入值的随机数。|1378121|那你应该|1378122|替换|1378123|string+randomNumber+=+random.Next(1000000).ToString("D6");|code-block|syntax|javascript|1378124|使用|1378125|string+randomNumber+=+random.Next(999999).ToString().PadLeft(6,+'0');|1378126|用0填充缺失的数字|1378127|entityMap^0|4|D|0|8|7|I|7|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Y|8|@$9|Z|A|10|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|11|8|@$9|12|A|13|B|C]|$9|14|A|15|B|C]]|D|@]|E|$]]|$1|H|3|I|5|6|7|16|8|@]|D|@]|E|$]]|$1|J|3|K|5|6|7|17|8|@]|D|@]|E|$]]|$1|L|3|M|5|N|7|18|8|@]|D|@]|E|$O|P]]|$1|Q|3|R|5|6|7|19|8|@]|D|@]|E|$]]|$1|S|3|T|5|N|7|1A|8|@]|D|@]|E|$O|P]]|$1|U|3|V|5|6|7|1B|8|@]|D|@]|E|$]]|$1|W|3|-4|5|6|7|1C|8|@]|D|@]|E|$]]]|X|$]]

You wrote <code>.Next(999999)</code> so i assume you want random values from 0 to 999999.

First you should write <code>1000000</code> because <code>.Next()</code> determines a random number from 0 to exclusive the input value.

Then you should

replace 

<pre><code>string randomNumber = random.Next(1000000).ToString("D6");
</code></pre>

with

<pre><code>string randomNumber = random.Next(999999).ToString().PadLeft(6, '0');
</code></pre>

to fill the missing digits with 0

blocks|key|1790409|text|另一种可能是"000000"格式字符串(六位数字，前导零是强制性的)：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1790410|random.Next(10000)+-最后4个数字是随机的|unordered-list-item|1790411|返回ToString("000000")+-6位数(至少2个前导零)|1790412|实现|1790413|+string+randomNumber+=+random.Next(10000).ToString("000000");|code-block|syntax|javascript|1790414|另一个问题是不应该在本地声明random：|1790415|+//+Simplest,+but+not+thread+safe
+private+static+random+=+new+Random();

+...

+public+static+string+getRandomNumber()
+{
+++++return+random.Next(10000).ToString("000000");
+}|1790416|entityMap^0|6|8|0|0|I|0|2|I|0|0|0|E|6|0|0^^$0|@$1|2|3|4|5|6|7|X|8|@$9|Y|A|Z|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|10|8|@$9|11|A|12|B|C]]|D|@]|E|$]]|$1|I|3|J|5|H|7|13|8|@$9|14|A|15|B|C]]|D|@]|E|$]]|$1|K|3|L|5|6|7|16|8|@]|D|@]|E|$]]|$1|M|3|N|5|O|7|17|8|@]|D|@]|E|$P|Q]]|$1|R|3|S|5|6|7|18|8|@$9|19|A|1A|B|C]]|D|@]|E|$]]|$1|T|3|U|5|O|7|1B|8|@]|D|@]|E|$P|Q]]|$1|V|3|-4|5|6|7|1C|8|@]|D|@]|E|$]]]|W|$]]

Another possibility is <code>"000000"</code> format string (six digits, leading zeros are mandatory):

<ul>
<li><code>random.Next(10000)</code> - last 4 digits are random</li>
<li><code>ToString("000000")</code> - 6 digits are returned (at least 2 leading zeros)</li>
</ul>

Implementation

<pre><code> string randomNumber = random.Next(10000).ToString("000000");
</code></pre>

Another issue is that you should not declare <code>random</code> locally:

<pre><code> // Simplest, but not thread safe
 private static random = new Random();

 ...

 public static string getRandomNumber()
 {
 return random.Next(10000).ToString("000000");
 }
</code></pre>

blocks|key|1790438|text|我会这样做：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1790439|var+rnd+=+new+Random();+//+the+internal+seed+is+good+enough
var+rndNums+=+string.Join("",+Enumerable.Range(0,+4).Select(x+=>+rnd.Next(10)));
return+"00"+%2B+rndNums;|code-block|syntax|javascript|1790440|然后，您可以轻松地更改您想要的金额，如下所示：|1790441|string+GetRandomBatch(int+numberOfRandomNumbers)
{
++++var+rnd+=+new+Random();+//+the+internal+seed+is+good+enough
++++var+rndNums+=+string.Join("",+Enumerable.Range(0,+numberOfRandomNumbers).Select(x+=>+rnd.Next(0,+9)));
++++return+"00"+%2B+rndNums;
}|1790442|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

I would do something like this:

<pre><code>var rnd = new Random(); // the internal seed is good enough
var rndNums = string.Join("", Enumerable.Range(0, 4).Select(x =&gt; rnd.Next(10)));
return "00" + rndNums;
</code></pre>

Then you can easily change the amount you want, like this:

<pre><code>string GetRandomBatch(int numberOfRandomNumbers)
{
 var rnd = new Random(); // the internal seed is good enough
 var rndNums = string.Join("", Enumerable.Range(0, numberOfRandomNumbers).Select(x =&gt; rnd.Next(0, 9)));
 return "00" + rndNums;
}
</code></pre>

blocks|key|1790457|text|使用内插字符串(C#+6.0)，您可以这样做：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1790458|Random+random+=+new+Random();
string+randomNumber+=+$"00{random.Next(9999)}";|code-block|syntax|javascript|1790459|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|$]]

With interpolated string (C# 6.0) you can do it like this : 

<pre><code>Random random = new Random();
string randomNumber = $"00{random.Next(9999)}";
</code></pre>

blocks|key|1790490|text|您的ToString("D6")完全可以实现您想要的结果，但是由于您正在为您的random.Next传递一个较大的上限，因此在许多情况下不会有任何需要处理的内容。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1790491|正如其他人所指出的，您可以在"00"前加上前缀，或者只需设置一个下限，因为"D6"会处理您的填充：|1790492|string+randomNumber+=+random.Next(9999).ToString("D6");|code-block|syntax|javascript|1790493|如果希望数字始终为4位，而不是(例如，"000123"+)，则还需要指定一个下限：|1790494|string+randomNumber+=+random.Next(1000,+9999).ToString("D6");|1790495|还请注意，如果您指定的是硬编码的种子(如您的示例中所示)，则始终会得到相同的随机数。|1790496|entityMap^0|2|E|13|B|0|E|4|11|4|0|0|J|8|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@$9|V|A|W|B|C]|$9|X|A|Y|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|Z|8|@$9|10|A|11|B|C]|$9|12|A|13|B|C]]|D|@]|E|$]]|$1|H|3|I|5|J|7|14|8|@]|D|@]|E|$K|L]]|$1|M|3|N|5|6|7|15|8|@$9|16|A|17|B|C]]|D|@]|E|$]]|$1|O|3|P|5|J|7|18|8|@]|D|@]|E|$K|L]]|$1|Q|3|R|5|6|7|19|8|@]|D|@]|E|$]]|$1|S|3|-4|5|6|7|1A|8|@]|D|@]|E|$]]]|T|$]]

Your <code>ToString("D6")</code> does exactly what it appears that you want, but since you are passing a large upper bound for your <code>random.Next</code>, there will in many scenarios not be anything to pad.

As others have indicated, you may prefix <code>"00"</code>, or you could simply set a lower upper bound, since the <code>"D6"</code> will take care of your padding:

<pre><code>string randomNumber = random.Next(9999).ToString("D6");
</code></pre>

If you want the number to always be 4 digits and not, say, <code>"000123"</code>, you'll need to specify a lower bound as well:

<pre><code>string randomNumber = random.Next(1000, 9999).ToString("D6");
</code></pre>

Also note that if you are specifying a hardcoded seed as in your example, you will always get the same random number back.

blocks|key|1790523|text|像这样的东西就行了。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1790524|public+static+void+Main()
{
++++Random+rnd+=+new+Random();
++++string+Value+=+rnd.Next(1,+9999).ToString("D6");
++++Console.WriteLine(Value);
}|code-block|syntax|javascript|1790525|或者用更低的可能性得到相同的号码|1790526|public+static+void+Main()
{
++++Random+rnd+=+new+Random(DateTime.Now.Millisecond);
++++string+Value+=+rnd.Next(1,+9999).ToString("D6");
++++Console.WriteLine(Value);
}|1790527|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

Something like this will do.

<pre><code>public static void Main()
{
 Random rnd = new Random();
 string Value = rnd.Next(1, 9999).ToString("D6");
 Console.WriteLine(Value);
}
</code></pre>

Or with lower possibility of getting the same number would be 

<pre><code>public static void Main()
{
 Random rnd = new Random(DateTime.Now.Millisecond);
 string Value = rnd.Next(1, 9999).ToString("D6");
 Console.WriteLine(Value);
}
</code></pre>

blocks|key|1378299|text|public+string+RandomNum()
{
++++return+"00"%2B+new+Random().Next(9999).ToString("D4");
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1378300|还可以为Next()函数提供最小和最大数目。比如：|unstyled|1378301|return+"00"%2B+new+Random().Next(1,9999).ToString("D4");|1378302|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|L|8|@]|9|@]|A|$]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$B|C]]|$1|I|3|-4|5|F|7|N|8|@]|9|@]|A|$]]]|J|$]]

<pre><code>public string RandomNum()
{
 return "00"+ new Random().Next(9999).ToString("D4");
}
</code></pre>

you can also give minimum and maximum number to Next() function. Like:

<pre><code>return "00"+ new Random().Next(1,9999).ToString("D4");
</code></pre>

I'm making a program where products are searched for by there batch number. I want to the batch number to be a random 6 digit code starting with 00, for example 002142; at the moment it is only making a random 6 digit code.
Here is the code I have tried:

<pre><code> public string getRandomNumber()
 {
 Random random = new Random(100000);
 string randomNumber = random.Next(999999).ToString("D6");
 return randomNumber;
 }
</code></pre>

C# needing a six digit number beginning with two 0's

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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我正在做一个程序，在那里搜索产品的批号。我希望批号是一个从00开始的随机6位代码，例如002142；目前它只生成一个随机的6位代码。下面是我尝试过的代码： public string getRandomNumber()    {        Random random = new Random(100000);        string randomNumber = random.Next(9

问C#需要一个以二0开头的六位数。
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回答 9

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问C#需要一个以二0开头的六位数。EN

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问C#需要一个以二0开头的六位数。
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