blocks|key|1038980|text|如果您只想打印小数点后有两位数的double，请使用如下所示：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1038981|double+value+=+200.3456;
System.out.printf("Value:+%25.2f",+value);|code-block|syntax|javascript|1038982|如果您希望将结果显示在String中，而不是打印到控制台，请使用带有相同参数的String.format()：|1038983|String+result+=+String.format("%25.2f",+value);|1038984|或者使用类DecimalFormat|1038985|DecimalFormat+df+=+new+DecimalFormat("####0.00");
System.out.println("Value:+"+%2B+df.format(value));|1038986|entityMap^0|G|6|0|0|B|6|13|F|0|0|5|D|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@$9|V|A|W|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|X|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|Y|8|@$9|Z|A|10|B|C]|$9|11|A|12|B|C]]|D|@]|E|$]]|$1|M|3|N|5|H|7|13|8|@]|D|@]|E|$I|J]]|$1|O|3|P|5|6|7|14|8|@$9|15|A|16|B|C]]|D|@]|E|$]]|$1|Q|3|R|5|H|7|17|8|@]|D|@]|E|$I|J]]|$1|S|3|-4|5|6|7|18|8|@]|D|@]|E|$]]]|T|$]]

If you just want to print a <code>double</code> with two digits after the decimal point, use something like this:

<pre><code>double value = 200.3456;
System.out.printf("Value: %.2f", value);
</code></pre>

If you want to have the result in a <code>String</code> instead of being printed to the console, use <code>String.format()</code> with the same arguments:

<pre><code>String result = String.format("%.2f", value);
</code></pre>

Or use class <code>DecimalFormat</code>:

<pre><code>DecimalFormat df = new DecimalFormat("####0.00");
System.out.println("Value: " + df.format(value));
</code></pre>

blocks|key|1254033|text|我认为这更简单：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1254034|double+time+=+200.3456;
DecimalFormat+df+=+new+DecimalFormat("#.##");++++++
time+=+Double.valueOf(df.format(time));

System.out.println(time);+//+200.35|code-block|syntax|javascript|1254035|请注意，这实际上将为您进行舍入，而不仅仅是格式化。|1254036|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 think this is easier:

<pre><code>double time = 200.3456;
DecimalFormat df = new DecimalFormat("#.##"); 
time = Double.valueOf(df.format(time));

System.out.println(time); // 200.35
</code></pre>

Note that this will actually do the rounding for you, not just formatting.

blocks|key|1253887|text|最简单的方法就是像这样做一个小把戏；|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1253888|double+val+=+....;
val+=+val*100;
val+=+Math.round(val);
val+=+val+/100;|code-block|syntax|javascript|1253889|如果val从200.3456开始，则为20034.56，然后四舍五入为20035，然后除以得到200.34。|1253890|如果你想总是四舍五入，我们总是可以通过转换为int来截断：|1253891|double+val+=+....;
val+=+val*100;
val+=+(double)((int)+val);
val+=+val+/100;|1253892|此技术适用于大多数情况，因为对于非常大的双精度(正或负)，它可能会溢出。但是，如果您知道您的值将在适当的范围内，那么这对您来说应该是有效的。|1253893|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]]|$1|G|3|H|5|6|7|S|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|T|8|@]|9|@]|A|$]]|$1|K|3|L|5|D|7|U|8|@]|9|@]|A|$E|F]]|$1|M|3|N|5|6|7|V|8|@]|9|@]|A|$]]|$1|O|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|P|$]]

The easiest way, would be to do a trick like this;

<pre><code>double val = ....;
val = val*100;
val = Math.round(val);
val = val /100;
</code></pre>

if val starts at 200.3456 then it goes to 20034.56 then it gets rounded to 20035 then we divide it to get 200.34.

if you wanted to always round down we could always truncate by casting to an int:

<pre><code>double val = ....;
val = val*100;
val = (double)((int) val);
val = val /100;
</code></pre>

This technique will work for most cases because for very large doubles (positive or negative) it may overflow. but if you know that your values will be in an appropriate range then this should work for you.

blocks|key|1039352|text|请使用Apache+commons+math|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1039353|Precision.round(10.4567,+2)|code-block|syntax|javascript|1039354|entityMap|0|LINK|mutability|MUTABLE|url|https://commons.apache.org/proper/commons-math/javadocs/api-3.3/org/apache/commons/math3/util/Precision.html#round(double,%2520int))^0|3|J|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@$A|R|B|S|1|T]]|C|$]]|$1|D|3|E|5|F|7|U|8|@]|9|@]|C|$G|H]]|$1|I|3|-4|5|6|7|V|8|@]|9|@]|C|$]]]|J|$K|$5|L|M|N|C|$O|P]]]]

Please use <a href="https://commons.apache.org/proper/commons-math/javadocs/api-3.3/org/apache/commons/math3/util/Precision.html#round(double,%20int)" rel="noreferrer">Apache commons math</a>:

<pre><code>Precision.round(10.4567, 2)
</code></pre>

blocks|key|1039322|text|function+Double+round2(Double+val)+{
++++return+new+BigDecimal(val.toString()).setScale(2,RoundingMode.HALF_UP).doubleValue();
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1039323|注意toString()！|unstyled|1039324|这是因为BigDecimal转换了double的二进制形式！|1039325|这些是各种建议的方法及其失败的案例。|1039326|//+Always+Good!
new+BigDecimal(val.toString()).setScale(2,RoundingMode.HALF_UP).doubleValue()+

Double+val+=+260.775d;+//EXPECTED+260.78
260.77+-+WRONG+-+new+BigDecimal(val).setScale(2,RoundingMode.HALF_UP).doubleValue()

Double+val+=+260.775d;+//EXPECTED+260.78
260.77+-+TRY+AGAIN+-+Math.round(val+*+100.d)+/+100.0d++

Double+val+=+256.025d;+//EXPECTED+256.03d
256.02+-+OOPS+-+new+DecimalFormat("0.00").format(val)+
//+By+default+use+half+even,+works+if+you+change+mode+to+half_up+

Double+val+=+256.025d;+//EXPECTED+256.03d
256.02+-+FAIL+-+(int)(val+*+100+%2B+0.5)+/+100.0;|1039327|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|P|8|@]|9|@]|A|$]]|$1|G|3|H|5|F|7|Q|8|@]|9|@]|A|$]]|$1|I|3|J|5|F|7|R|8|@]|9|@]|A|$]]|$1|K|3|L|5|6|7|S|8|@]|9|@]|A|$B|C]]|$1|M|3|-4|5|F|7|T|8|@]|9|@]|A|$]]]|N|$]]

<pre><code>function Double round2(Double val) {
 return new BigDecimal(val.toString()).setScale(2,RoundingMode.HALF_UP).doubleValue();
}
</code></pre>

Note the toString()!!!!

This is because BigDecimal converts the exact binary form of the double!!!

These are the various suggested methods and their fail cases.

<pre><code>// Always Good!
new BigDecimal(val.toString()).setScale(2,RoundingMode.HALF_UP).doubleValue() 

Double val = 260.775d; //EXPECTED 260.78
260.77 - WRONG - new BigDecimal(val).setScale(2,RoundingMode.HALF_UP).doubleValue()

Double val = 260.775d; //EXPECTED 260.78
260.77 - TRY AGAIN - Math.round(val * 100.d) / 100.0d 

Double val = 256.025d; //EXPECTED 256.03d
256.02 - OOPS - new DecimalFormat("0.00").format(val) 
// By default use half even, works if you change mode to half_up 

Double val = 256.025d; //EXPECTED 256.03d
256.02 - FAIL - (int)(val * 100 + 0.5) / 100.0;
</code></pre>

blocks|key|1254203|text|double+value=+200.3456;
DecimalFormat+df+=+new+DecimalFormat("0.00");++++++
System.out.println(df.format(value));|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1254204|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>double value= 200.3456;
DecimalFormat df = new DecimalFormat("0.00"); 
System.out.println(df.format(value));
</code></pre>

blocks|key|1039223|text|如果你真的想要同样的双精度，但是按照你想要的方式四舍五入，你可以使用BigDecimal，例如|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1039224|new+BigDecimal(myValue).setScale(2,+RoundingMode.HALF_UP).doubleValue();|code-block|syntax|javascript|1039225|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|$]]

If you really want the same double, but rounded in the way you want you can use BigDecimal, for example

<pre><code>new BigDecimal(myValue).setScale(2, RoundingMode.HALF_UP).doubleValue();
</code></pre>

blocks|key|1039242|text|double+d+=+28786.079999999998;
String+str+=+String.format("%251.2f",+d);
d+=+Double.valueOf(str);|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1039243|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>double d = 28786.079999999998;
String str = String.format("%1.2f", d);
d = Double.valueOf(str);
</code></pre>

blocks|key|1039178|text|两个四舍五入的数字。非常简单，您基本上是在更新变量，而不是像DecimalFormat那样只是出于显示目的。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1039179|x+=+Math.floor(x+*+100)+/+100;|offset|length|style|CODE|1039180|entityMap^0|0|0|U|0^^$0|@$1|2|3|4|5|6|7|J|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|K|8|@$D|L|E|M|F|G]]|9|@]|A|$]]|$1|H|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|I|$]]

For two rounding digits. Very simple and you are basically updating the variable instead of just display purposes which DecimalFormat does.

<code>x = Math.floor(x * 100) / 100;</code>

blocks|key|1038857|text|四舍五入通常不是人们想要的。相反，请使用String.format()以所需的格式表示它。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1038858|entityMap|0|LINK|mutability|MUTABLE|url|http://java.sun.com/j2se/1.5.0/docs/api/java/lang/String.html#format%2528java.lang.String,%2520java.lang.Object...%2529^0|K|F|K|F|0|0^^$0|@$1|2|3|4|5|6|7|N|8|@$9|O|A|P|B|C]]|D|@$9|Q|A|R|1|S]]|E|$]]|$1|F|3|-4|5|6|7|T|8|@]|D|@]|E|$]]]|G|$H|$5|I|J|K|E|$L|M]]]]

Rounding a double is usually not what one wants. Instead, use <a href="http://java.sun.com/j2se/1.5.0/docs/api/java/lang/String.html#format%28java.lang.String,%20java.lang.Object...%29" rel="noreferrer"><code>String.format()</code></a> to represent it in the desired format.

blocks|key|1254001|text|在你的问题中，你似乎也想避免对数字进行舍入？我想.format()会用半上半舍五入，afaik？|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1254002|所以如果你想舍入，200.3456应该是200.35，精度是2。但是在你的例子中，如果你只想要前2，然后丢弃剩下的？|1254003|您可以将其乘以100，然后转换为一个整数(或取该数字的底数)，然后再除以100。|1254004|200.3456+*+100+=+20034.56;++
(int)+20034.56+=+20034;++
20034/100.0+=+200.34;|code-block|syntax|javascript|1254005|你可能会在接近边界的时候遇到非常非常大的数字的问题。在这种情况下，转换为字符串并子字符串将同样容易地工作。|1254006|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|P|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|Q|8|@]|9|@]|A|$]]|$1|F|3|G|5|H|7|R|8|@]|9|@]|A|$I|J]]|$1|K|3|L|5|6|7|S|8|@]|9|@]|A|$]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

In your question, it seems that you want to avoid rounding the numbers as well? I think .format() will round the numbers using half-up, afaik? 
so if you want to round, 200.3456 should be 200.35 for a precision of 2. but in your case, if you just want the first 2 and then discard the rest?

You could multiply it by 100 and then cast to an int (or taking the floor of the number), before dividing by 100 again. 

<pre><code>200.3456 * 100 = 20034.56; 
(int) 20034.56 = 20034; 
20034/100.0 = 200.34;
</code></pre>

You might have issues with really really big numbers close to the boundary though. In which case converting to a string and substring'ing it would work just as easily.

blocks|key|1254171|text|value+=+(int)(value+*+100+%2B+0.5)+/+100.0;|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1254172|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>value = (int)(value * 100 + 0.5) / 100.0;
</code></pre>

If the value is <code>200.3456</code>, it should be formatted to <code>200.34</code>.
If it is <code>200</code>, then it should be <code>200.00</code>.

Round a double to 2 decimal places

如果值为200.3456，则应将其格式化为200.34。如果它是200，那么它应该是200.00。

问将双精度数舍入到小数点后两位
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回答 12

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