blocks|key|2011432|text|这样如何：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2011433|String+numbers+=+text.substring(text.length()+-+7);|code-block|syntax|javascript|2011434|当然，这是假设结尾有7个字符。如果你传给它"12345“，它会抛出异常。你可以这样解决这个问题：|2011435|String+numbers+=+text.substring(Math.max(0,+text.length()+-+7));|2011436|或|2011437|String+numbers+=+text.length()+<=+7+?+text+:+text.substring(text.length()+-+7);|2011438|请注意，这仍然不会验证结果字符串是否包含数字-如果text为null，它仍然会抛出异常。|offset|length|style|CODE|2011439|entityMap^0|0|0|0|0|0|0|P|4|0^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|X|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Y|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|Z|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|10|8|@]|9|@]|A|$]]|$1|M|3|N|5|D|7|11|8|@]|9|@]|A|$E|F]]|$1|O|3|P|5|6|7|12|8|@$Q|13|R|14|S|T]]|9|@]|A|$]]|$1|U|3|-4|5|6|7|15|8|@]|9|@]|A|$]]]|V|$]]

How about:

<pre><code>String numbers = text.substring(text.length() - 7);
</code></pre>

That assumes that there are 7 characters at the end, of course. It will throw an exception if you pass it "12345". You could address that this way:

<pre><code>String numbers = text.substring(Math.max(0, text.length() - 7));
</code></pre>

or

<pre><code>String numbers = text.length() &lt;= 7 ? text : text.substring(text.length() - 7);
</code></pre>

Note that this still isn't doing any validation that the resulting string contains numbers - and it will still throw an exception if <code>text</code> is null.

blocks|key|2011497|text|这应该是可行的|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2011498|+Integer+i=+Integer.parseInt(text.substring(text.length()+-+7));|code-block|syntax|javascript|2011499|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|$]]

This should work

<pre><code> Integer i= Integer.parseInt(text.substring(text.length() - 7));
</code></pre>

blocks|key|2011508|text|你可以做的事情有很多。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2011509|s.substring(s.lastIndexOf(':')+%2B+1);|code-block|syntax|javascript|2011510|将获取最后一个冒号之后的所有内容。|2011511|s.substring(s.lastIndexOf('+')+%2B+1);|2011512|最后一个空格之后的所有内容。|2011513|String+numbers[]+=+s.split("[%5E0-9]%2B");|2011514|分割所有的数字序列；numbers数组的最后一个元素可能就是您想要的。|2011515|entityMap^0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|U|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|W|8|@]|9|@]|A|$]]|$1|M|3|N|5|D|7|X|8|@]|9|@]|A|$E|F]]|$1|O|3|P|5|6|7|Y|8|@]|9|@]|A|$]]|$1|Q|3|-4|5|6|7|Z|8|@]|9|@]|A|$]]]|R|$]]

Lots of things you could do.

<pre><code>s.substring(s.lastIndexOf(':') + 1);
</code></pre>

will get everything after the last colon.

<pre><code>s.substring(s.lastIndexOf(' ') + 1);
</code></pre>

everything after the last space.

<pre><code>String numbers[] = s.split("[^0-9]+");
</code></pre>

splits off all sequences of digits; the last element of the numbers array is probably what you want.

blocks|key|1797874|text|这段代码非常适合我：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1797875|String+numbers+=+text.substring(Math.max(0,+text.length()+-+7));|code-block|syntax|javascript|1797876|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|$]]

This code works for me perfectly:

<pre><code>String numbers = text.substring(Math.max(0, text.length() - 7));
</code></pre>

blocks|key|2011544|text|String+inputstr+=+"abcd:+efg:+1006746"
int+startindex+=+inputstr.length()+-+10;
String+outputtendigitstr+=+inputstr.substring(startindex);|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|2011545|确保检查字符串长度大于10。|unstyled|2011546|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|J|8|@]|9|@]|A|$]]|$1|G|3|-4|5|F|7|K|8|@]|9|@]|A|$]]]|H|$]]

<pre><code>String inputstr = "abcd: efg: 1006746"
int startindex = inputstr.length() - 10;
String outputtendigitstr = inputstr.substring(startindex);
</code></pre>

Make sure you check string length is more than 10.

blocks|key|2011551|text|org.apache.commons.lang3.StringUtils.substring(s,+-7)+|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|2011552|给你答案。如果小于7，则返回输入；如果s+==为null，则返回null。它从不抛出异常。|unstyled|2011553|请参阅https://commons.apache.org/proper/commons-lang/apidocs/org/apache/commons/lang3/StringUtils.html#substring-java.lang.String-int-int-|offset|length|2011554|entityMap|0|LINK|mutability|MUTABLE|url|https://commons.apache.org/proper/commons-lang/apidocs/org/apache/commons/lang3/StringUtils.html#substring-java.lang.String-int-int-^0|0|0|3|3O|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|T|8|@]|9|@]|A|$]]|$1|G|3|H|5|F|7|U|8|@]|9|@$I|V|J|W|1|X]]|A|$]]|$1|K|3|-4|5|F|7|Y|8|@]|9|@]|A|$]]]|L|$M|$5|N|O|P|A|$Q|R]]]]

<pre><code>org.apache.commons.lang3.StringUtils.substring(s, -7) 
</code></pre>

gives you the answer. It returns the input if it is shorter than 7, and null if s == null. It never throws an exception.

See <a href="https://commons.apache.org/proper/commons-lang/apidocs/org/apache/commons/lang3/StringUtils.html#substring-java.lang.String-int-int-" rel="nofollow noreferrer">https://commons.apache.org/proper/commons-lang/apidocs/org/apache/commons/lang3/StringUtils.html#substring-java.lang.String-int-int-</a>

blocks|key|1797915|text|例如："abcd:+efg:+1006746“"bhddy:+nshhf36:+1006754”"hfquv:+nd:+5894254“|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1797916|-Step+1：String+firstString+=++"abcd:+efg:+1006746";|offset|length|style|CODE|1797917|-Step+2：String+lastSevenNumber+=+firstString.substring(firstString.lastIndexOf(':'),+firstString.length()).trim();|1797918|entityMap^0|0|8|17|0|8|2Y|0^^$0|@$1|2|3|4|5|6|7|L|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|M|8|@$D|N|E|O|F|G]]|9|@]|A|$]]|$1|H|3|I|5|6|7|P|8|@$D|Q|E|R|F|G]]|9|@]|A|$]]|$1|J|3|-4|5|6|7|S|8|@]|9|@]|A|$]]]|K|$]]

E.g : "abcd: efg: 1006746"
"bhddy: nshhf36: 1006754"
"hfquv: nd: 5894254"

-Step 1: <code>String firstString = "abcd: efg: 1006746";</code>

-Step 2: <code>String lastSevenNumber = firstString.substring(firstString.lastIndexOf(':'), firstString.length()).trim();</code>

I'm looking for a way to pull the last characters from a String, regardless of size. Lets take these strings into example:

<pre><code>"abcd: efg: 1006746"
"bhddy: nshhf36: 1006754"
"hfquv: nd: 5894254"
</code></pre>

As you can see, completely random strings, but they have 7 numbers at the end. How would I be able to take those 7 numbers?

Edit:

I just realized that <code>String[] string = s.split(": ");</code> would work great here, as long as I call string[2] for the numbers and string[1] for anything in the middle.

How to get the last characters in a String in Java, regardless of String size

我正在寻找一种从字符串中提取最后一个字符的方法，而不管字符串的大小。让我们以这些字符串为例："abcd: efg: 1006746""bhddy: nshhf36: 1006754""hfquv: nd: 5894254"如你所见，完全随机的字符串，但它们的末尾有7个数字。我怎么才能取这7个数字呢？编辑：我刚刚意识到，...

问如何在Java中获取字符串的最后一个字符，而不考虑字符串的大小
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