blocks|key|102604|text|DateTime结构只存储一个值，而不是值的范围。MinValue和MaxValue是静态字段，它们包含DateTime结构实例的可能值范围。这些字段是静态的，与DateTime的特定实例无关。它们与DateTime类型本身相关。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|102605|推荐阅读：static+(C#+Reference)|102606|更新:获取月份范围：|102607|DateTime+date+=+...
var+firstDayOfMonth+=+new+DateTime(date.Year,+date.Month,+1);
var+lastDayOfMonth+=+firstDayOfMonth.AddMonths(1).AddDays(-1);|code-block|syntax|javascript|102608|entityMap|0|LINK|mutability|MUTABLE|url|http://msdn.microsoft.com/en-us/library/System.DateTime%2528v=vs.110%2529.aspx|1|http://msdn.microsoft.com/en-us/library/98f28cdx.aspx^0|0|8|P|8|Y|8|1G|8|29|8|2S|8|0|8|0|0|5|L|1|0|0|0^^$0|@$1|2|3|4|5|6|7|Y|8|@$9|Z|A|10|B|C]|$9|11|A|12|B|C]|$9|13|A|14|B|C]|$9|15|A|16|B|C]|$9|17|A|18|B|C]|$9|19|A|1A|B|C]]|D|@$9|1B|A|1C|1|1D]]|E|$]]|$1|F|3|G|5|6|7|1E|8|@]|D|@$9|1F|A|1G|1|1H]]|E|$]]|$1|H|3|I|5|6|7|1I|8|@]|D|@]|E|$]]|$1|J|3|K|5|L|7|1J|8|@]|D|@]|E|$M|N]]|$1|O|3|-4|5|6|7|1K|8|@]|D|@]|E|$]]]|P|$Q|$5|R|S|T|E|$U|V]]|W|$5|R|S|T|E|$U|X]]]]

<a href="http://msdn.microsoft.com/en-us/library/System.DateTime%28v=vs.110%29.aspx" rel="noreferrer"><code>DateTime</code></a> structure stores only one value, not range of values. <code>MinValue</code> and <code>MaxValue</code> are static fields, which hold range of possible values for instances of <code>DateTime</code> structure. These fields are static and do not relate to particular instance of <code>DateTime</code>. They relate to <code>DateTime</code> type itself.

Suggested reading: <a href="http://msdn.microsoft.com/en-us/library/98f28cdx.aspx" rel="noreferrer">static (C# Reference)</a>

UPDATE: Getting month range:

<pre><code>DateTime date = ...
var firstDayOfMonth = new DateTime(date.Year, date.Month, 1);
var lastDayOfMonth = firstDayOfMonth.AddMonths(1).AddDays(-1);
</code></pre>

blocks|key|102622|text|使用.Net接口获取月份范围(只是另一种方式)：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|102623|DateTime+date+=+...
var+firstDayOfMonth+=+new+DateTime(date.Year,+date.Month,+1);
var+lastDayOfMonth+=+new+DateTime(date.Year,+date.Month,+DateTime.DaysInMonth(date.Year,+date.Month));|code-block|syntax|javascript|102624|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|$]]

Getting month range with .Net API (just another way):

<pre><code>DateTime date = ...
var firstDayOfMonth = new DateTime(date.Year, date.Month, 1);
var lastDayOfMonth = new DateTime(date.Year, date.Month, DateTime.DaysInMonth(date.Year, date.Month));
</code></pre>

blocks|key|102636|text|DateTime+dCalcDate+=+DateTime.Now;
dtpFromEffDate.Value+=+new+DateTime(dCalcDate.Year,+dCalcDate.Month,+1);
dptToEffDate.Value+=+new+DateTime(dCalcDate.Year,+dCalcDate.Month,+DateTime.DaysInMonth(dCalcDate.Year,+dCalcDate.Month));|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|102637|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>DateTime dCalcDate = DateTime.Now;
dtpFromEffDate.Value = new DateTime(dCalcDate.Year, dCalcDate.Month, 1);
dptToEffDate.Value = new DateTime(dCalcDate.Year, dCalcDate.Month, DateTime.DaysInMonth(dCalcDate.Year, dCalcDate.Month));
</code></pre>

blocks|key|4609302|text|这更多的是对@Sergey和@Steffen的答案的长篇评论。我自己以前也写过类似的代码，所以我决定检查一下什么是最performant的，同时也要记住清晰也很重要。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|4609303|结果|4609304|下面是一个1000万次迭代的示例测试运行结果：|4609305|2257+ms+for+FirstDayOfMonth_AddMethod()
2406+ms+for+FirstDayOfMonth_NewMethod()
6342+ms+for+LastDayOfMonth_AddMethod()
4037+ms+for+LastDayOfMonth_AddMethodWithDaysInMonth()
4160+ms+for+LastDayOfMonth_NewMethod()
4212+ms+for+LastDayOfMonth_NewMethodWithReuseOfExtMethod()
2491+ms+for+LastDayOfMonth_SpecialCase()|code-block|syntax|javascript|4609306|代码|4609307|我使用LINQPad+4+(在C#程序模式下)在打开编译器优化的情况下运行测试。以下是为清晰和方便起见而作为扩展方法分解的测试代码：|4609308|public+static+class+DateTimeDayOfMonthExtensions
{
++++public+static+DateTime+FirstDayOfMonth_AddMethod(this+DateTime+value)
++++{
++++++++return+value.Date.AddDays(1+-+value.Day);
++++}
++++
++++public+static+DateTime+FirstDayOfMonth_NewMethod(this+DateTime+value)
++++{
++++++++return+new+DateTime(value.Year,+value.Month,+1);
++++}
++++
++++public+static+DateTime+LastDayOfMonth_AddMethod(this+DateTime+value)
++++{
++++++++return+value.FirstDayOfMonth_AddMethod().AddMonths(1).AddDays(-1);
++++}
++++
++++public+static+DateTime+LastDayOfMonth_AddMethodWithDaysInMonth(this+DateTime+value)
++++{
++++++++return+value.Date.AddDays(DateTime.DaysInMonth(value.Year,+value.Month)+-+value.Day);
++++}
++++
++++public+static+DateTime+LastDayOfMonth_SpecialCase(this+DateTime+value)
++++{
++++++++return+value.AddDays(DateTime.DaysInMonth(value.Year,+value.Month)+-+1);
++++}
++++
++++public+static+int+DaysInMonth(this+DateTime+value)
++++{
++++++++return+DateTime.DaysInMonth(value.Year,+value.Month);
++++}
++++
++++public+static+DateTime+LastDayOfMonth_NewMethod(this+DateTime+value)
++++{
++++++++return+new+DateTime(value.Year,+value.Month,+DateTime.DaysInMonth(value.Year,+value.Month));
++++}

++++public+static+DateTime+LastDayOfMonth_NewMethodWithReuseOfExtMethod(this+DateTime+value)
++++{
++++++++return+new+DateTime(value.Year,+value.Month,+value.DaysInMonth());
++++}
}

void+Main()
{
++++Random+rnd+=+new+Random();
++++DateTime[]+sampleData+=+new+DateTime[10000000];
++++
++++for(int+i+=+0;+i+<+sampleData.Length;+i%2B%2B)+{
++++++++sampleData[i]+=+new+DateTime(1970,+1,+1).AddDays(rnd.Next(0,+365+*+50));
++++}
++++
++++GC.Collect();
++++System.Diagnostics.Stopwatch+sw+=+System.Diagnostics.Stopwatch.StartNew();
++++for(int+i+=+0;+i+<+sampleData.Length;+i%2B%2B)+{
++++++++DateTime+test+=+sampleData[i].FirstDayOfMonth_AddMethod();
++++}
++++string.Format("{0}+ms+for+FirstDayOfMonth_AddMethod()",+sw.ElapsedMilliseconds).Dump();
++++
++++GC.Collect();
++++sw.Restart();
++++for(int+i+=+0;+i+<+sampleData.Length;+i%2B%2B)+{
++++++++DateTime+test+=+sampleData[i].FirstDayOfMonth_NewMethod();
++++}
++++string.Format("{0}+ms+for+FirstDayOfMonth_NewMethod()",+sw.ElapsedMilliseconds).Dump();
++++
++++GC.Collect();
++++sw.Restart();
++++for(int+i+=+0;+i+<+sampleData.Length;+i%2B%2B)+{
++++++++DateTime+test+=+sampleData[i].LastDayOfMonth_AddMethod();
++++}
++++string.Format("{0}+ms+for+LastDayOfMonth_AddMethod()",+sw.ElapsedMilliseconds).Dump();

++++GC.Collect();
++++sw.Restart();
++++for(int+i+=+0;+i+<+sampleData.Length;+i%2B%2B)+{
++++++++DateTime+test+=+sampleData[i].LastDayOfMonth_AddMethodWithDaysInMonth();
++++}
++++string.Format("{0}+ms+for+LastDayOfMonth_AddMethodWithDaysInMonth()",+sw.ElapsedMilliseconds).Dump();

++++GC.Collect();
++++sw.Restart();
++++for(int+i+=+0;+i+<+sampleData.Length;+i%2B%2B)+{
++++++++DateTime+test+=+sampleData[i].LastDayOfMonth_NewMethod();
++++}
++++string.Format("{0}+ms+for+LastDayOfMonth_NewMethod()",+sw.ElapsedMilliseconds).Dump();

++++GC.Collect();
++++sw.Restart();
++++for(int+i+=+0;+i+<+sampleData.Length;+i%2B%2B)+{
++++++++DateTime+test+=+sampleData[i].LastDayOfMonth_NewMethodWithReuseOfExtMethod();
++++}
++++string.Format("{0}+ms+for+LastDayOfMonth_NewMethodWithReuseOfExtMethod()",+sw.ElapsedMilliseconds).Dump();

++++for(int+i+=+0;+i+<+sampleData.Length;+i%2B%2B)+{
++++++++sampleData[i]+=+sampleData[i].FirstDayOfMonth_AddMethod();
++++}
++++
++++GC.Collect();
++++sw.Restart();
++++for(int+i+=+0;+i+<+sampleData.Length;+i%2B%2B)+{
++++++++DateTime+test+=+sampleData[i].LastDayOfMonth_SpecialCase();
++++}
++++string.Format("{0}+ms+for+LastDayOfMonth_SpecialCase()",+sw.ElapsedMilliseconds).Dump();
++++
}|4609309|分析|4609310|我对其中一些结果感到惊讶。|4609311|尽管FirstDayOfMonth_AddMethod的功能不多，但在测试的大多数运行过程中，它都比FirstDayOfMonth_NewMethod稍微快一些。然而，我认为后者的意图稍微明确一些，所以我更倾向于这样做。|style|CODE|4609312|与LastDayOfMonth_AddMethodWithDaysInMonth、LastDayOfMonth_NewMethod和LastDayOfMonth_NewMethodWithReuseOfExtMethod相比，LastDayOfMonth_AddMethod是一个明显的输家。在最快的三个之间，没有太多的东西，所以归根结底是你的个人偏好。我选择了LastDayOfMonth_NewMethodWithReuseOfExtMethod的简洁性，因为它重用了另一个有用的扩展方法。我的意图更明确，我愿意接受较小的性能成本。|4609313|LastDayOfMonth_SpecialCase假设您在特殊情况下提供的是月份的第一天，在这种情况下您可能已经计算了该日期，并且它使用带有DateTime.DaysInMonth的add方法来获得结果。正如您所期望的那样，这比其他版本更快，但除非您迫切需要速度，否则我看不出在您的武器库中有这个特殊情况的意义。|4609314|结论|4609315|下面是一个带有我的选择的扩展方法类，我相信它与@Steffen大体一致：|4609316|public+static+class+DateTimeDayOfMonthExtensions
{
++++public+static+DateTime+FirstDayOfMonth(this+DateTime+value)
++++{
++++++++return+new+DateTime(value.Year,+value.Month,+1);
++++}
++++
++++public+static+int+DaysInMonth(this+DateTime+value)
++++{
++++++++return+DateTime.DaysInMonth(value.Year,+value.Month);
++++}
++++
++++public+static+DateTime+LastDayOfMonth(this+DateTime+value)
++++{
++++++++return+new+DateTime(value.Year,+value.Month,+value.DaysInMonth());
++++}
}|4609317|如果你已经走到这一步，谢谢你抽出时间来！这很有趣：()。如果您对这些算法有任何其他建议，请发表评论。|4609318|entityMap|0|LINK|mutability|MUTABLE|url|https://english.stackexchange.com/questions/38945/what-is-wrong-with-the-word-performant|1|https://www.linqpad.net/^0|1M|A|0|0|0|0|0|0|3|9|1|0|0|0|0|2|P|1E|P|0|1|13|15|O|1U|18|35|O|51|18|0|0|Q|1Z|K|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|1M|8|@]|9|@$A|1N|B|1O|1|1P]]|C|$]]|$1|D|3|E|5|6|7|1Q|8|@]|9|@]|C|$]]|$1|F|3|G|5|6|7|1R|8|@]|9|@]|C|$]]|$1|H|3|I|5|J|7|1S|8|@]|9|@]|C|$K|L]]|$1|M|3|N|5|6|7|1T|8|@]|9|@]|C|$]]|$1|O|3|P|5|6|7|1U|8|@]|9|@$A|1V|B|1W|1|1X]]|C|$]]|$1|Q|3|R|5|J|7|1Y|8|@]|9|@]|C|$K|L]]|$1|S|3|T|5|6|7|1Z|8|@]|9|@]|C|$]]|$1|U|3|V|5|6|7|20|8|@]|9|@]|C|$]]|$1|W|3|X|5|6|7|21|8|@$A|22|B|23|Y|Z]|$A|24|B|25|Y|Z]]|9|@]|C|$]]|$1|10|3|11|5|6|7|26|8|@$A|27|B|28|Y|Z]|$A|29|B|2A|Y|Z]|$A|2B|B|2C|Y|Z]|$A|2D|B|2E|Y|Z]|$A|2F|B|2G|Y|Z]]|9|@]|C|$]]|$1|12|3|13|5|6|7|2H|8|@$A|2I|B|2J|Y|Z]|$A|2K|B|2L|Y|Z]]|9|@]|C|$]]|$1|14|3|15|5|6|7|2M|8|@]|9|@]|C|$]]|$1|16|3|17|5|6|7|2N|8|@]|9|@]|C|$]]|$1|18|3|19|5|J|7|2O|8|@]|9|@]|C|$K|L]]|$1|1A|3|1B|5|6|7|2P|8|@]|9|@]|C|$]]|$1|1C|3|-4|5|6|7|2Q|8|@]|9|@]|C|$]]]|1D|$1E|$5|1F|1G|1H|C|$1I|1J]]|1K|$5|1F|1G|1H|C|$1I|1L]]]]

This is more a long comment on @Sergey and @Steffen's answers. Having written similar code myself in the past I decided to check what was most <a href="https://english.stackexchange.com/questions/38945/what-is-wrong-with-the-word-performant">performant</a> while remembering that clarity is important too.
<h2>Result</h2>
Here is an example test run result for 10 million iterations:
<pre class="lang-none prettyprint-override"><code>2257 ms for FirstDayOfMonth_AddMethod()
2406 ms for FirstDayOfMonth_NewMethod()
6342 ms for LastDayOfMonth_AddMethod()
4037 ms for LastDayOfMonth_AddMethodWithDaysInMonth()
4160 ms for LastDayOfMonth_NewMethod()
4212 ms for LastDayOfMonth_NewMethodWithReuseOfExtMethod()
2491 ms for LastDayOfMonth_SpecialCase()
</code></pre>
<h2>Code</h2>
I used <a href="https://www.linqpad.net/" rel="noreferrer">LINQPad 4</a> (in C# Program mode) to run the tests with compiler optimization turned on. Here is the tested code factored as Extension methods for clarity and convenience:
<pre><code>public static class DateTimeDayOfMonthExtensions
{
 public static DateTime FirstDayOfMonth_AddMethod(this DateTime value)
 {
 return value.Date.AddDays(1 - value.Day);
 }
 
 public static DateTime FirstDayOfMonth_NewMethod(this DateTime value)
 {
 return new DateTime(value.Year, value.Month, 1);
 }
 
 public static DateTime LastDayOfMonth_AddMethod(this DateTime value)
 {
 return value.FirstDayOfMonth_AddMethod().AddMonths(1).AddDays(-1);
 }
 
 public static DateTime LastDayOfMonth_AddMethodWithDaysInMonth(this DateTime value)
 {
 return value.Date.AddDays(DateTime.DaysInMonth(value.Year, value.Month) - value.Day);
 }
 
 public static DateTime LastDayOfMonth_SpecialCase(this DateTime value)
 {
 return value.AddDays(DateTime.DaysInMonth(value.Year, value.Month) - 1);
 }
 
 public static int DaysInMonth(this DateTime value)
 {
 return DateTime.DaysInMonth(value.Year, value.Month);
 }
 
 public static DateTime LastDayOfMonth_NewMethod(this DateTime value)
 {
 return new DateTime(value.Year, value.Month, DateTime.DaysInMonth(value.Year, value.Month));
 }

 public static DateTime LastDayOfMonth_NewMethodWithReuseOfExtMethod(this DateTime value)
 {
 return new DateTime(value.Year, value.Month, value.DaysInMonth());
 }
}

void Main()
{
 Random rnd = new Random();
 DateTime[] sampleData = new DateTime[10000000];
 
 for(int i = 0; i &lt; sampleData.Length; i++) {
 sampleData[i] = new DateTime(1970, 1, 1).AddDays(rnd.Next(0, 365 * 50));
 }
 
 GC.Collect();
 System.Diagnostics.Stopwatch sw = System.Diagnostics.Stopwatch.StartNew();
 for(int i = 0; i &lt; sampleData.Length; i++) {
 DateTime test = sampleData[i].FirstDayOfMonth_AddMethod();
 }
 string.Format(&quot;{0} ms for FirstDayOfMonth_AddMethod()&quot;, sw.ElapsedMilliseconds).Dump();
 
 GC.Collect();
 sw.Restart();
 for(int i = 0; i &lt; sampleData.Length; i++) {
 DateTime test = sampleData[i].FirstDayOfMonth_NewMethod();
 }
 string.Format(&quot;{0} ms for FirstDayOfMonth_NewMethod()&quot;, sw.ElapsedMilliseconds).Dump();
 
 GC.Collect();
 sw.Restart();
 for(int i = 0; i &lt; sampleData.Length; i++) {
 DateTime test = sampleData[i].LastDayOfMonth_AddMethod();
 }
 string.Format(&quot;{0} ms for LastDayOfMonth_AddMethod()&quot;, sw.ElapsedMilliseconds).Dump();

 GC.Collect();
 sw.Restart();
 for(int i = 0; i &lt; sampleData.Length; i++) {
 DateTime test = sampleData[i].LastDayOfMonth_AddMethodWithDaysInMonth();
 }
 string.Format(&quot;{0} ms for LastDayOfMonth_AddMethodWithDaysInMonth()&quot;, sw.ElapsedMilliseconds).Dump();

 GC.Collect();
 sw.Restart();
 for(int i = 0; i &lt; sampleData.Length; i++) {
 DateTime test = sampleData[i].LastDayOfMonth_NewMethod();
 }
 string.Format(&quot;{0} ms for LastDayOfMonth_NewMethod()&quot;, sw.ElapsedMilliseconds).Dump();

 GC.Collect();
 sw.Restart();
 for(int i = 0; i &lt; sampleData.Length; i++) {
 DateTime test = sampleData[i].LastDayOfMonth_NewMethodWithReuseOfExtMethod();
 }
 string.Format(&quot;{0} ms for LastDayOfMonth_NewMethodWithReuseOfExtMethod()&quot;, sw.ElapsedMilliseconds).Dump();

 for(int i = 0; i &lt; sampleData.Length; i++) {
 sampleData[i] = sampleData[i].FirstDayOfMonth_AddMethod();
 }
 
 GC.Collect();
 sw.Restart();
 for(int i = 0; i &lt; sampleData.Length; i++) {
 DateTime test = sampleData[i].LastDayOfMonth_SpecialCase();
 }
 string.Format(&quot;{0} ms for LastDayOfMonth_SpecialCase()&quot;, sw.ElapsedMilliseconds).Dump();
 
}
</code></pre>
<h2>Analysis</h2>
I was surprised by some of these results.
Although there is not much in it the <code>FirstDayOfMonth_AddMethod</code> was slightly faster than <code>FirstDayOfMonth_NewMethod</code> in most runs of the test. However, I think the latter has a slightly clearer intent and so I have a preference for that.
<code>LastDayOfMonth_AddMethod</code> was a clear loser against <code>LastDayOfMonth_AddMethodWithDaysInMonth</code>, <code>LastDayOfMonth_NewMethod</code> and <code>LastDayOfMonth_NewMethodWithReuseOfExtMethod</code>. Between the fastest three there is nothing much in it and so it comes down to your personal preference. I choose the clarity of <code>LastDayOfMonth_NewMethodWithReuseOfExtMethod</code> with its reuse of another useful extension method. IMHO its intent is clearer and I am willing to accept the small performance cost.
<code>LastDayOfMonth_SpecialCase</code> assumes you are providing the first of the month in the special case where you may have already calculated that date and it uses the add method with <code>DateTime.DaysInMonth</code> to get the result. This is faster than the other versions, as you would expect, but unless you are in a desperate need for speed I don't see the point of having this special case in your arsenal.
<h2>Conclusion</h2>
Here is an extension method class with my choices and in general agreement with @Steffen I believe:
<pre><code>public static class DateTimeDayOfMonthExtensions
{
 public static DateTime FirstDayOfMonth(this DateTime value)
 {
 return new DateTime(value.Year, value.Month, 1);
 }
 
 public static int DaysInMonth(this DateTime value)
 {
 return DateTime.DaysInMonth(value.Year, value.Month);
 }
 
 public static DateTime LastDayOfMonth(this DateTime value)
 {
 return new DateTime(value.Year, value.Month, value.DaysInMonth());
 }
}
</code></pre>
If you have got this far, thank you for time! Its been fun :¬). Please comment if you have any other suggestions for these algorithms.

blocks|key|102703|text|一种简单的方法|type|unstyled|depth|inlineStyleRanges|entityRanges|data|102704|Begin+=+new+DateTime(DateTime.Now.Year,+DateTime.Now.Month,1).ToShortDateString();
End+=+new+DataFim.Text+=+new+DateTime(DateTime.Now.Year,+DateTime.Now.Month,+DateTime.DaysInMonth(DateTime.Now.Year,+DateTime.Now.Month)).ToShortDateString();|code-block|syntax|javascript|102705|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|$]]

easy way to do it

<pre><code>Begin = new DateTime(DateTime.Now.Year, DateTime.Now.Month,1).ToShortDateString();
End = new DataFim.Text = new DateTime(DateTime.Now.Year, DateTime.Now.Month, DateTime.DaysInMonth(DateTime.Now.Year, DateTime.Now.Month)).ToShortDateString();
</code></pre>

blocks|key|102726|text|在这里，您可以为当月的第一天添加一个月，而不是从该天删除1天。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|102727|DateTime+now+=+DateTime.Now;
var+startDate+=+new+DateTime(now.Year,+now.Month,+1);
var+endDate+=+startDate.AddMonths(1).AddDays(-1);|code-block|syntax|javascript|102728|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|$]]

Here you can add one month for the first day of current month than delete 1 day from that day.

<pre><code>DateTime now = DateTime.Now;
var startDate = new DateTime(now.Year, now.Month, 1);
var endDate = startDate.AddMonths(1).AddDays(-1);
</code></pre>

blocks|key|102766|text|这里接受的答案没有考虑DateTime实例的类型。例如，如果您的原始DateTime实例是UTC类型的，那么通过创建一个新的DateTime实例，您将创建一个未知的类型实例，然后根据服务器设置将其视为本地时间。因此，获取该月的第一个和最后一个日期的更合适的方法是：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|102767|var+now+=+DateTime.UtcNow;
var+first+=+now.Date.AddDays(-(now.Date.Day+-+1));
var+last+=+first.AddMonths(1).AddTicks(-1);|code-block|syntax|javascript|102768|这样就保留了DateTime实例的原始类型。|102769|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|$]]

The accepted answer here does not take into account the Kind of the DateTime instance. For example if your original DateTime instance was a UTC Kind then by making a new DateTime instance you will be making an Unknown Kind instance which will then be treated as local time based on server settings. Therefore the more proper way to get the first and last date of the month would be this:

<pre><code>var now = DateTime.UtcNow;
var first = now.Date.AddDays(-(now.Date.Day - 1));
var last = first.AddMonths(1).AddTicks(-1);
</code></pre>

This way the original Kind of the DateTime instance is preserved.

blocks|key|102801|text|如果你只关心日期|type|unstyled|depth|inlineStyleRanges|entityRanges|data|102802|var+firstDay+=+new+DateTime(date.Year,+date.Month,+1,+0,+0,+0,+date.Kind);
var+lastDay+=+new+DateTime(date.Year,+date.Month,+1,+0,+0,+0,+date.Kind).AddMonths(1).AddDays(-1);|code-block|syntax|javascript|102803|如果你想节省时间|102804|var+firstDay+=+new+DateTime(date.Year,+date.Month,+1,+date.Hour,+date.Minute,+date.Second,+date.Kind);
var+lastDay+=+new+DateTime(date.Year,+date.Month,+1,+date.Hour,+date.Minute,+date.Second,+date.Kind).AddMonths(1).AddDays(-1);|102805|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|$]]

If you only care about the date 

<pre><code>var firstDay = new DateTime(date.Year, date.Month, 1, 0, 0, 0, date.Kind);
var lastDay = new DateTime(date.Year, date.Month, 1, 0, 0, 0, date.Kind).AddMonths(1).AddDays(-1);
</code></pre>

If you want to preserve time 

<pre><code>var firstDay = new DateTime(date.Year, date.Month, 1, date.Hour, date.Minute, date.Second, date.Kind);
var lastDay = new DateTime(date.Year, date.Month, 1, date.Hour, date.Minute, date.Second, date.Kind).AddMonths(1).AddDays(-1);
</code></pre>

blocks|key|4609405|text|试试这个：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4609406|string+strDate+=+DateTime.Now.ToString("MM/01/yyyy");|code-block|syntax|javascript|4609407|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 one:

<pre><code>string strDate = DateTime.Now.ToString("MM/01/yyyy");
</code></pre>

blocks|key|102829|text|波斯文化|type|unstyled|depth|inlineStyleRanges|entityRanges|data|102830|PersianCalendar+pc+=+new+PersianCalendar();++++++++++++

var+today+=+pc.GetDayOfMonth(DateTime.Now);
var+firstDayOfMonth+=+pc.GetDayOfMonth(DateTime.Now.AddDays(-(today-1)));
var+lastDayOfMonth+=+pc.GetDayOfMonth(DateTime.Now.AddMonths(1).AddDays(-today));++++++++++++
Console.WriteLine("First+day+"%2B+firstDayOfMonth);
Console.WriteLine("Last+day+"+%2B+lastDayOfMonth);|code-block|syntax|javascript|102831|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|$]]

For Persian culture 

<pre><code>PersianCalendar pc = new PersianCalendar(); 

var today = pc.GetDayOfMonth(DateTime.Now);
var firstDayOfMonth = pc.GetDayOfMonth(DateTime.Now.AddDays(-(today-1)));
var lastDayOfMonth = pc.GetDayOfMonth(DateTime.Now.AddMonths(1).AddDays(-today)); 
Console.WriteLine("First day "+ firstDayOfMonth);
Console.WriteLine("Last day " + lastDayOfMonth);
</code></pre>

blocks|key|102862|text|我在我的脚本中使用了这一点(对我有效)，但我需要一个完整的日期，而不需要将其修剪为仅日期而不是时间。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|102863|public+DateTime+GetLastDayOfTheMonth()
{
++++int+daysFromNow+=+DateTime.DaysInMonth(DateTime.Now.Year,+DateTime.Now.Month)+-+(int)DateTime.Now.Day;
++++return+DateTime.Now.AddDays(daysFromNow);
}|code-block|syntax|javascript|102864|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|$]]

I used this in my script(works for me) but I needed a full date without the need of trimming it to only the date and no time.

<pre><code>public DateTime GetLastDayOfTheMonth()
{
 int daysFromNow = DateTime.DaysInMonth(DateTime.Now.Year, DateTime.Now.Month) - (int)DateTime.Now.Day;
 return DateTime.Now.AddDays(daysFromNow);
}
</code></pre>

blocks|key|4609465|text|DateTime+dCalcDate+=+DateTime.Now;
var+startDate+=+new+DateTime(Convert.ToInt32(Year),+Convert.ToInt32(Month),+1);
var+endDate+=+new+DateTime(Convert.ToInt32(Year),+Convert.ToInt32(Month),+DateTime.DaysInMonth((Convert.ToInt32(Year)),+Convert.ToInt32(Month)));|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|4609466|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>DateTime dCalcDate = DateTime.Now;
var startDate = new DateTime(Convert.ToInt32(Year), Convert.ToInt32(Month), 1);
var endDate = new DateTime(Convert.ToInt32(Year), Convert.ToInt32(Month), DateTime.DaysInMonth((Convert.ToInt32(Year)), Convert.ToInt32(Month)));
</code></pre>

blocks|key|4609474|text|你可以的|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4609475|DateTime+dt+=+DateTime.Now;+
DateTime+firstDayOfMonth+=+new+DateTime(dt.Year,+date.Month,+1);
DateTime+lastDayOfMonth+=+firstDayOfMonth.AddMonths(1).AddDays(-1);|code-block|syntax|javascript|4609476|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 do it

<pre><code>DateTime dt = DateTime.Now; 
DateTime firstDayOfMonth = new DateTime(dt.Year, date.Month, 1);
DateTime lastDayOfMonth = firstDayOfMonth.AddMonths(1).AddDays(-1);
</code></pre>

blocks|key|4609498|text|你可以尝试获取当月的第一天；|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4609499|DateTime.Now.AddDays(-(DateTime.Now.Day-1))|code-block|syntax|javascript|4609500|并为其赋值。|4609501|如下所示：|4609502|dateEndEdit.EditValue+=+DateTime.Now;
dateStartEdit.EditValue+=+DateTime.Now.AddDays(-(DateTime.Now.Day-1));|4609503|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|R|8|@]|9|@]|A|$]]|$1|K|3|L|5|D|7|S|8|@]|9|@]|A|$E|F]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

You can try this for get current month first day;
<pre><code>DateTime.Now.AddDays(-(DateTime.Now.Day-1))
</code></pre>
and assign it a value.
Like this:
<pre><code>dateEndEdit.EditValue = DateTime.Now;
dateStartEdit.EditValue = DateTime.Now.AddDays(-(DateTime.Now.Day-1));
</code></pre>

blocks|key|102926|text|创建DateTime类的实例|type|unstyled|depth|inlineStyleRanges|entityRanges|data|102927|DateTime+dateTime+=+DateTime.Now;|code-block|syntax|javascript|102928|如果你想得到这个月的最后一天，你可以这样做|102929|int+lastDayOfMonth+=+DateTime.DaysInMonth(caducidadPuntos.Year,+caducidadPuntos.Month);|102930|如果你想得到一个月的第一天，你可以这样做|102931|DateTime+firstDayMonth+=+new+DateTime(dateTime.Year,+dateTime.Month,+1);|102932|entityMap^0|0|0|0|0|0|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]]|$1|G|3|H|5|6|7|S|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|U|8|@]|9|@]|A|$]]|$1|M|3|N|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|O|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|P|$]]

Create an instance of DateTime class
<pre><code>DateTime dateTime = DateTime.Now;
</code></pre>
If you want to get the last day of the month you can do this
<pre><code>int lastDayOfMonth = DateTime.DaysInMonth(caducidadPuntos.Year, caducidadPuntos.Month);
</code></pre>
If you want to get the first day of the month, you can do this
<pre><code>DateTime firstDayMonth = new DateTime(dateTime.Year, dateTime.Month, 1);
</code></pre>

blocks|key|4609579|text|我们的要求是能够获得给定日期和月份的开始和结束，包括时间。我们最终利用了前面提到的解决方案，非常感谢这里的每个人，并将其合并到一个util类中，以便能够获得给定月份和年份组合的开始和结束，直到最后一毫秒。包括我们在帮助他人的情况下所取得的进展。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4609580|util：|4609581|++++public+class+DateUtil
++++{
++++++++public+static+(DateTime+startOfMonth,+DateTime+endOfMonth)+GetStartAndEndOfMonth(int+month,+int+year)
++++++++{
++++++++++++DateTime+startOfMonth+=+GetStartOfMonth(month,+year);
++++++++++++DateTime+endOfMonth+=+GetEndOfMonth(month,+year);
++++++++++++return+(startOfMonth,+endOfMonth);
++++++++}

++++++++public+static+DateTime+GetStartOfMonth(int+month,+int+year)
++++++++{
++++++++++++return+new+DateTime(year,+month,+1).Date;
++++++++}

++++++++public+static+DateTime+GetEndOfMonth(int+month,+int+year)
++++++++{
++++++++++++return+new+DateTime(year,+month,+1).Date.AddMonths(1).AddMilliseconds(-1);
++++++++}
++++}|code-block|syntax|javascript|4609582|用法：|4609583|(DateTime+startOfMonth,+DateTime+endOfMonth)+=+DateUtil.GetStartAndEndOfMonth(2,+2021);+//+February,+2021|4609584|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|P|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|Q|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|R|8|@]|9|@]|A|$]]|$1|K|3|L|5|F|7|S|8|@]|9|@]|A|$G|H]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

We had the requirement of being able to get the start and end of a given dates month, including times, inclusively. We ended up utilizing the aforementioned solutions, huge thanks to everyone here, and combined it into a util class to be able to get the start and end for a given month and year number combination up to the last millisecond. Including what we moved forward with in the event it helps someone else.
The util:
<pre><code> public class DateUtil
 {
 public static (DateTime startOfMonth, DateTime endOfMonth) GetStartAndEndOfMonth(int month, int year)
 {
 DateTime startOfMonth = GetStartOfMonth(month, year);
 DateTime endOfMonth = GetEndOfMonth(month, year);
 return (startOfMonth, endOfMonth);
 }

 public static DateTime GetStartOfMonth(int month, int year)
 {
 return new DateTime(year, month, 1).Date;
 }

 public static DateTime GetEndOfMonth(int month, int year)
 {
 return new DateTime(year, month, 1).Date.AddMonths(1).AddMilliseconds(-1);
 }
 }
</code></pre>
Usage:
<pre><code>(DateTime startOfMonth, DateTime endOfMonth) = DateUtil.GetStartAndEndOfMonth(2, 2021); // February, 2021
</code></pre>

I want to get the first day and last day of the month where a given date lies in. The date comes from a value in a UI field.

If I'm using a time picker I could say

<pre><code>var maxDay = dtpAttendance.MaxDate.Day;
</code></pre>

But I'm trying to get it from a DateTime object. So if I have this...

<pre><code>DateTime dt = DateTime.today;
</code></pre>

How to get first day and last day of the month from <code>dt</code>?

Getting the first and last day of a month, using a given DateTime object

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我想要得到给定日期所在的月份的第一天和最后一天。日期来自UI字段中的值。如果我使用的是时间选择器，我可以说var maxDay = dtpAttendance.MaxDate.Day;但我正在尝试从DateTime对象中获取它。如果我有这个..。DateTime dt = DateTime.today;如何从dt获取每...

问使用给定的DateTime对象获取一个月的第一天和最后一天
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