blocks|key|1428821|text|使用substring方法，如下所示：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1428822|int+n+=+8;
String+s+=+"Hello,+World!";
System.out.println(s.substring(0,n);|code-block|syntax|javascript|1428823|如果n大于字符串的长度，这将抛出异常，正如一位评论者所指出的那样。一个简单的解决方案是在else子句的条件if(s.length()<n)中包装所有这些内容，您可以选择是只打印/返回整个字符串，还是以其他方式处理它。|offset|length|style|CODE|1428824|entityMap^0|0|0|18|4|1H|G|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]]|$1|G|3|H|5|6|7|Q|8|@$I|R|J|S|K|L]|$I|T|J|U|K|L]]|9|@]|A|$]]|$1|M|3|-4|5|6|7|V|8|@]|9|@]|A|$]]]|N|$]]

Use the substring method, as follows:

<pre><code>int n = 8;
String s = "Hello, World!";
System.out.println(s.substring(0,n);
</code></pre>

If n is greater than the length of the string, this will throw an exception, as one commenter has pointed out. one simple solution is to wrap all this in the condition <code>if(s.length()&lt;n)</code> in your <code>else</code> clause, you can choose whether you just want to print/return the whole String or handle it another way.

blocks|key|1214076|text|有一类问题，所以有时不太合理，这个问题非常接近:-)|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1214077|也许你可以解释一下你不喜欢使用你排除的两种方法中的一种。|1214078|如果只是因为您不想在代码中填充if语句或异常捕获代码，一种解决方案是使用一个帮助函数来为您处理它，如下所示：|offset|length|style|CODE|1214079|static+String+substring_safe+(String+s,+int+start,+int+len)+{+...+}|code-block|syntax|javascript|1214080|它将预先检查长度并采取相应行动(返回较小的字符串或填充空格)。|1214081|然后你就完全不用在你的代码中担心它了，只要调用：|1214082|String+s2+=+substring_safe+(s,+10,+7);|1214083|而不是：|1214084|String+s2+=+s.substring+(10,7);|1214085|这将在你似乎担心的情况下工作(基于你对其他答案的评论)，在做大量字符串构建的事情时不会破坏代码流。|1214086|entityMap^0|0|0|F|2|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|12|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|13|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|14|8|@$F|15|G|16|H|I]]|9|@]|A|$]]|$1|J|3|K|5|L|7|17|8|@]|9|@]|A|$M|N]]|$1|O|3|P|5|6|7|18|8|@]|9|@]|A|$]]|$1|Q|3|R|5|6|7|19|8|@]|9|@]|A|$]]|$1|S|3|T|5|L|7|1A|8|@]|9|@]|A|$M|N]]|$1|U|3|V|5|6|7|1B|8|@]|9|@]|A|$]]|$1|W|3|X|5|L|7|1C|8|@]|9|@]|A|$M|N]]|$1|Y|3|Z|5|6|7|1D|8|@]|9|@]|A|$]]|$1|10|3|-4|5|6|7|1E|8|@]|9|@]|A|$]]]|11|$]]

There's a class of question on SO that sometimes make less than perfect sense, this one is perilously close :-)

Perhaps you could explain your aversion to using one of the two methods you ruled out.

If it's just because you don't want to pepper your code with <code>if</code> statements or exception catching code, one solution is to use a helper function that will take care of it for you, something like:

<pre><code>static String substring_safe (String s, int start, int len) { ... }
</code></pre>

which will check lengths beforehand and act accordingly (either return smaller string or pad with spaces).

Then you don't have to worry about it in your code at all, just call:

<pre><code>String s2 = substring_safe (s, 10, 7);
</code></pre>

instead of:

<pre><code>String s2 = s.substring (10,7);
</code></pre>

This would work in the case that you seem to be worried about (based on your comments to other answers), not breaking the flow of the code when doing lots of string building stuff.

blocks|key|1214115|text|String+upToNCharacters+=+String.format("%25."%2B+n+%2B"s",+str);|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1214116|如果n是一个变量，那就很糟糕了(所以你必须构造格式字符串)，但是如果是一个常量，那就很清楚了：|unstyled|offset|length|style|CODE|1214117|String+upToNCharacters+=+String.format("%25.10s",+str);|1214118|docs|1214119|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.oracle.com/javase/6/docs/api/java/lang/String.html#format%2528java.lang.String,%2520java.lang.Object...%2529^0|0|2|1|0|0|0|4|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|X|8|@$G|Y|H|Z|I|J]]|9|@]|A|$]]|$1|K|3|L|5|6|7|10|8|@]|9|@]|A|$B|C]]|$1|M|3|N|5|F|7|11|8|@]|9|@$G|12|H|13|1|14]]|A|$]]|$1|O|3|-4|5|F|7|15|8|@]|9|@]|A|$]]]|P|$Q|$5|R|S|T|A|$U|V]]]]

<pre><code>String upToNCharacters = String.format("%."+ n +"s", str);
</code></pre>

Awful if <code>n</code> is a variable (so you must construct the format string), but pretty clear if a constant:

<pre><code>String upToNCharacters = String.format("%.10s", str);
</code></pre>

<a href="http://docs.oracle.com/javase/6/docs/api/java/lang/String.html#format%28java.lang.String,%20java.lang.Object...%29" rel="noreferrer">docs</a>

blocks|key|1428978|text|为此，Apache+Commons+Lang有一个StringUtils.left方法。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1428979|String+upToNCharacters+=+StringUtils.left(s,+n);|code-block|syntax|javascript|1428980|entityMap|0|LINK|mutability|MUTABLE|url|http://commons.apache.org/proper/commons-lang/|1|http://commons.apache.org/proper/commons-lang/javadocs/api-3.1/org/apache/commons/lang3/StringUtils.html#left(java.lang.String,%2520int)^0|P|G|3|J|0|P|G|1|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@$9|V|A|W|B|C]]|D|@$9|X|A|Y|1|Z]|$9|10|A|11|1|12]]|E|$]]|$1|F|3|G|5|H|7|13|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|14|8|@]|D|@]|E|$]]]|L|$M|$5|N|O|P|E|$Q|R]]|S|$5|N|O|P|E|$Q|T]]]]

<a href="http://commons.apache.org/proper/commons-lang/">Apache Commons Lang</a> has a <a href="http://commons.apache.org/proper/commons-lang/javadocs/api-3.1/org/apache/commons/lang3/StringUtils.html#left(java.lang.String,%20int)"><code>StringUtils.left</code></a> method for this.

<pre><code>String upToNCharacters = StringUtils.left(s, n);
</code></pre>

blocks|key|1214235|text|ApacheCommons让我大吃一惊，StringUtils.abbreviate(String+str,+int+maxWidth)加上了"...“没有更改后缀的选项。WordUtils.abbreviate(String+str,+int+lower,+int+upper,+String+appendToEnd)查找下一个空格。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1214236|我就把这个留在这里：|1214237|public+static+String+abbreviate(String+s,+int+maxLength,+String+appendToEnd)+{
++++String+result+=+s;
++++appendToEnd+=+appendToEnd+==+null+?+""+:+appendToEnd;
++++if+(maxLength+>=+appendToEnd.length())+{
++++++++if+(s.length()>maxLength)+{
++++++++++++result+=+s.substring(0,+Math.min(s.length(),+maxLength+-+appendToEnd.length()))+%2B+appendToEnd;
++++++++}
++++}+else+{
++++++++throw+new+StringIndexOutOfBoundsException("maxLength+can+not+be+smaller+than+appendToEnd+parameter+length.");
++++}
++++return+result;
}|code-block|syntax|javascript|1214238|entityMap^0|K|1C|2E|22|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@$9|P|A|Q|B|C]|$9|R|A|S|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|T|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|U|8|@]|D|@]|E|$K|L]]|$1|M|3|-4|5|6|7|V|8|@]|D|@]|E|$]]]|N|$]]

ApacheCommons surprised me, 
<code>StringUtils.abbreviate(String str, int maxWidth)</code> appends "..." there is no option to change postfix.
<code>WordUtils.abbreviate(String str, int lower, int upper, String appendToEnd)</code> looks up to next empty space.

I’m just going to leave this here:

<pre><code>public static String abbreviate(String s, int maxLength, String appendToEnd) {
 String result = s;
 appendToEnd = appendToEnd == null ? "" : appendToEnd;
 if (maxLength &gt;= appendToEnd.length()) {
 if (s.length()&gt;maxLength) {
 result = s.substring(0, Math.min(s.length(), maxLength - appendToEnd.length())) + appendToEnd;
 }
 } else {
 throw new StringIndexOutOfBoundsException("maxLength can not be smaller than appendToEnd parameter length.");
 }
 return result;
}
</code></pre>

How do I get up to the first <code>n</code> characters of a string in Java without doing a size check first (inline is acceptable) or risking an <code>IndexOutOfBoundsException</code>?

How do I get the first n characters of a string without checking the size or going out of bounds?

Java

在Java语言中，如何在不进行大小检查(内联是可接受的)或冒着IndexOutOfBoundsException风险的情况下获得字符串的前几个n字符

问如何在不检查字符串大小或超出范围的情况下获得字符串的前n个字符？
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