blocks|key|563794|text|replace将替换letter的所有实例。您所需要做的就是使用substring()|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|563795|public+String+method(String+str)+{
++++if+(str+!=+null+&&+str.length()+>+0+&&+str.charAt(str.length()+-+1)+==+'x')+{
++++++++str+=+str.substring(0,+str.length()+-+1);
++++}
++++return+str;
}|code-block|syntax|javascript|563796|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.oracle.com/javase/7/docs/api/java/lang/String.html#substring(int,%2520int))^0|0|7|W|B|W|B|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@$9|T|A|U|B|C]|$9|V|A|W|B|C]]|D|@$9|X|A|Y|1|Z]]|E|$]]|$1|F|3|G|5|H|7|10|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|11|8|@]|D|@]|E|$]]]|L|$M|$5|N|O|P|E|$Q|R]]]]

<code>replace</code> will replace all instances of a letter. All you need to do is use <a href="http://docs.oracle.com/javase/7/docs/api/java/lang/String.html#substring(int,%20int)" rel="noreferrer"><code>substring()</code></a>:

<pre><code>public String method(String str) {
 if (str != null &amp;&amp; str.length() &gt; 0 &amp;&amp; str.charAt(str.length() - 1) == 'x') {
 str = str.substring(0, str.length() - 1);
 }
 return str;
}
</code></pre>

blocks|key|780228|text|public+String+removeLastChar(String+s)+{
++++if+(s+==+null+%7C%7C+s.length()+==+0)+{
++++++++return+s;
++++}
++++return+s.substring(0,+s.length()-1);
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|780229|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>public String removeLastChar(String s) {
 if (s == null || s.length() == 0) {
 return s;
 }
 return s.substring(0, s.length()-1);
}
</code></pre>

blocks|key|563849|text|为什么不只有一个班轮呢？|type|unstyled|depth|inlineStyleRanges|entityRanges|data|563850|public+static+String+removeLastChar(String+str)+{
++++return+removeLastChars(str,+1);
}

public+static+String+removeLastChars(String+str,+int+chars)+{
++++return+str.substring(0,+str.length()+-+chars);
}|code-block|syntax|javascript|563851|完整代码|563852|public+class+Main+{
++++public+static+void+main+(String[]+args)+throws+java.lang.Exception+{
++++++++String+s1+=+"Remove+Last+CharacterY";
++++++++String+s2+=+"Remove+Last+Character2";
++++++++System.out.println("After+removing+s1=="+%2B+removeLastChar(s1)+%2B+"==");
++++++++System.out.println("After+removing+s2=="+%2B+removeLastChar(s2)+%2B+"==");
++++}
++++
++++public+static+String+removeLastChar(String+str)+{
++++++++return+removeLastChars(str,+1);
++++}

++++public+static+String+removeLastChars(String+str,+int+chars)+{
++++++++return+str.substring(0,+str.length()+-+chars);
++++}
}|563853|Demo|offset|length|563854|entityMap|0|LINK|mutability|MUTABLE|url|http://ideone.com/pdoGoj^0|0|0|0|0|0|4|0|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]]|$1|K|3|L|5|6|7|10|8|@]|9|@$M|11|N|12|1|13]]|A|$]]|$1|O|3|-4|5|6|7|14|8|@]|9|@]|A|$]]]|P|$Q|$5|R|S|T|A|$U|V]]]]

Why not just one liner?
<pre><code>public static String removeLastChar(String str) {
 return removeLastChars(str, 1);
}

public static String removeLastChars(String str, int chars) {
 return str.substring(0, str.length() - chars);
}
</code></pre>
Full Code
<pre><code>public class Main {
 public static void main (String[] args) throws java.lang.Exception {
 String s1 = &quot;Remove Last CharacterY&quot;;
 String s2 = &quot;Remove Last Character2&quot;;
 System.out.println(&quot;After removing s1==&quot; + removeLastChar(s1) + &quot;==&quot;);
 System.out.println(&quot;After removing s2==&quot; + removeLastChar(s2) + &quot;==&quot;);
 }
 
 public static String removeLastChar(String str) {
 return removeLastChars(str, 1);
 }

 public static String removeLastChars(String str, int chars) {
 return str.substring(0, str.length() - chars);
 }
}
</code></pre>
<h2><a href="http://ideone.com/pdoGoj" rel="noreferrer">Demo</a></h2>

blocks|key|780292|text|所描述的问题和建议的解决方案有时与删除分隔符有关。如果这是您的情况，那么看看Apache+Commons+StringUtils，它有一个名为removeEnd的方法，它非常优雅。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|780293|示例：|780294|StringUtils.removeEnd("string+1%7Cstring+2%7Cstring+3%7C",+"%7C");|code-block|syntax|javascript|780295|将导致：“字符串1%7C字符串2%7C字符串3”|780296|entityMap|0|LINK|mutability|MUTABLE|url|https://commons.apache.org/proper/commons-lang/apidocs/org/apache/commons/lang3/StringUtils.html#removeEnd-java.lang.String-java.lang.String-^0|1Z|9|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@$A|V|B|W|1|X]]|C|$]]|$1|D|3|E|5|6|7|Y|8|@]|9|@]|C|$]]|$1|F|3|G|5|H|7|Z|8|@]|9|@]|C|$I|J]]|$1|K|3|L|5|6|7|10|8|@]|9|@]|C|$]]|$1|M|3|-4|5|6|7|11|8|@]|9|@]|C|$]]]|N|$O|$5|P|Q|R|C|$S|T]]]]

The described problem and proposed solutions sometimes relate to removing separators. If this is your case, then have a look at Apache Commons StringUtils, it has a method called <a href="https://commons.apache.org/proper/commons-lang/apidocs/org/apache/commons/lang3/StringUtils.html#removeEnd-java.lang.String-java.lang.String-" rel="noreferrer">removeEnd</a> which is very elegant.

Example:

<pre><code>StringUtils.removeEnd("string 1|string 2|string 3|", "|");
</code></pre>

Would result in:
"string 1|string 2|string 3"

blocks|key|780316|text|由于我们讨论的是一个主题，因此也可以使用正则表达式|type|unstyled|depth|inlineStyleRanges|entityRanges|data|780317|"aaabcd".replaceFirst(".$","");+//=>+aaabc++|code-block|syntax|javascript|780318|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|$]]

Since we're on a subject, one can use regular expressions too

<pre><code>"aaabcd".replaceFirst(".$",""); //=&gt; aaabc 
</code></pre>

blocks|key|563921|text|使用以下命令：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|563922|+if(string.endsWith("x"))+{

++++string=+string.substring(0,+string.length()+-+1);
+}|code-block|syntax|javascript|563923|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|$]]

Use this:

<pre><code> if(string.endsWith("x")) {

 string= string.substring(0, string.length() - 1);
 }
</code></pre>

blocks|key|780381|text|不要试图重复发明轮子，而其他人已经编写了库来执行字符串操作：org.apache.commons.lang3.StringUtils.chop()|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|780382|entityMap|0|LINK|mutability|MUTABLE|url|https://commons.apache.org/proper/commons-lang/javadocs/api-2.6/org/apache/commons/lang/StringUtils.html#chop%2528java.lang.String%2529^0|U|17|U|17|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]]]]

Don't try to reinvent the wheel, while others have already written libraries to perform string manipulation: 
<a href="https://commons.apache.org/proper/commons-lang/javadocs/api-2.6/org/apache/commons/lang/StringUtils.html#chop%28java.lang.String%29" rel="noreferrer"><code>org.apache.commons.lang3.StringUtils.chop()</code></a>

blocks|key|564006|text|public+String+removeLastChar(String+s)+{
++++if+(!Util.isEmpty(s))+{
++++++++s+=+s.substring(0,+s.length()-1);
++++}
++++return+s;
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|564007|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>public String removeLastChar(String s) {
 if (!Util.isEmpty(s)) {
 s = s.substring(0, s.length()-1);
 }
 return s;
}
</code></pre>

blocks|key|780501|text|删除上次出现的“xxx”：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|780502|++++System.out.println("aaa+xxx+aaa+xxx+".replaceAll("xxx([%5Exxx]*)$",+"$1"));|code-block|syntax|javascript|780503|删除'xxx‘的最后一个匹配项，如果它是最后一个：|780504|++++System.out.println("aaa+xxx+aaa++".replaceAll("xxx\\s*$",+""));|780505|你可以在你想要的字符上替换'xxx‘，但要注意特殊字符|780506|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]]|$1|G|3|H|5|6|7|Q|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|S|8|@]|9|@]|A|$]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

removes last occurence of the 'xxx':

<pre><code> System.out.println("aaa xxx aaa xxx ".replaceAll("xxx([^xxx]*)$", "$1"));
</code></pre>

removes last occurrence of the 'xxx' if it is last:

<pre><code> System.out.println("aaa xxx aaa ".replaceAll("xxx\\s*$", ""));
</code></pre>

you can replace the 'xxx' on what you want but watch out on special chars

blocks|key|564067|text|//+Remove+n+last+characters++
//+System.out.println(removeLast("Hello!!!333",3));

public+String+removeLast(String+mes,+int+n)+{
++++return+mes+!=+null+&&+!mes.isEmpty()+&&+mes.length()>n
+++++++++?+mes.substring(0,+mes.length()-n):+mes;
}

//+Leave+substring+before+character/string++
//+System.out.println(leaveBeforeChar("Hello!!!123",+"1"));

public+String+leaveBeforeChar(String+mes,+String+last)+{
++++return+mes+!=+null+&&+!mes.isEmpty()+&&+mes.lastIndexOf(last)!=-1
+++++++++?+mes.substring(0,+mes.lastIndexOf(last)):+mes;
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|564068|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>// Remove n last characters 
// System.out.println(removeLast("Hello!!!333",3));

public String removeLast(String mes, int n) {
 return mes != null &amp;&amp; !mes.isEmpty() &amp;&amp; mes.length()&gt;n
 ? mes.substring(0, mes.length()-n): mes;
}

// Leave substring before character/string 
// System.out.println(leaveBeforeChar("Hello!!!123", "1"));

public String leaveBeforeChar(String mes, String last) {
 return mes != null &amp;&amp; !mes.isEmpty() &amp;&amp; mes.lastIndexOf(last)!=-1
 ? mes.substring(0, mes.lastIndexOf(last)): mes;
}
</code></pre>

blocks|key|780540|text|看看StringBuilder类：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|780541|++++StringBuilder+sb=new+StringBuilder("toto,");
++++System.out.println(sb.deleteCharAt(sb.length()-1));//display+"toto"|code-block|syntax|javascript|780542|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|$]]

Look to StringBuilder Class :

<pre><code> StringBuilder sb=new StringBuilder("toto,");
 System.out.println(sb.deleteCharAt(sb.length()-1));//display "toto"
</code></pre>

blocks|key|564134|text|如果在json中有像；这样的特殊字符，只需使用String.replace(";"，"")，否则必须重写字符串中减去最后一个字符的所有字符。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|564135|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

if you have special character like ; in json just use String.replace(";", "") otherwise you must rewrite all character in string minus the last.

blocks|key|564186|text|这是删除字符串中最后一个字符的一种方法：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|564187|Scanner+in+=+new+Scanner(System.in);
String+s+=+in.nextLine();
char+array[]+=+s.toCharArray();
int+l+=+array.length;
for+(int+i+=+0;+i+<+l-1;+i%2B%2B)+{
++++System.out.print(array[i]);
}|code-block|syntax|javascript|564188|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 is the one way to remove the last character in the string:

<pre><code>Scanner in = new Scanner(System.in);
String s = in.nextLine();
char array[] = s.toCharArray();
int l = array.length;
for (int i = 0; i &lt; l-1; i++) {
 System.out.print(array[i]);
}
</code></pre>

blocks|key|564233|text|为什么不使用escape+sequence+...！|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|564234|System.out.println(str+%2B+'\b');|code-block|syntax|javascript|564235|现在生活轻松多了。XD！~一个可读的一行程序|564236|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.oracle.com/javase/tutorial/java/data/characters.html^0|6|F|0|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@$A|T|B|U|1|V]]|C|$]]|$1|D|3|E|5|F|7|W|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|6|7|X|8|@]|9|@]|C|$]]|$1|K|3|-4|5|6|7|Y|8|@]|9|@]|C|$]]]|L|$M|$5|N|O|P|C|$Q|R]]]]

Why not use the <a href="https://docs.oracle.com/javase/tutorial/java/data/characters.html" rel="nofollow noreferrer">escape sequence</a> ... !

<pre><code>System.out.println(str + '\b');
</code></pre>

Life is much easier now . XD ! ~ A readable one-liner

blocks|key|564280|text|就可读性而言，我发现这是最简洁的|type|unstyled|depth|inlineStyleRanges|entityRanges|data|564281|StringUtils.substring("string",+0,+-1);|code-block|syntax|javascript|564282|负索引可以在Apache的StringUtils实用程序中使用。所有负数都从字符串末尾的偏移量开始处理。|564283|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|$]]

As far as the readability is concerned, I find this to be the most concise

<pre><code>StringUtils.substring("string", 0, -1);
</code></pre>

The negative indexes can be used in Apache's StringUtils utility.
All negative numbers are treated from offset from the end of the string.

blocks|key|564348|text|我不得不为类似的问题编写代码。我能够解决这个问题的一种方法是使用递归编码方法。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|564349|static+String+removeChar(String+word,+char+charToRemove)
{
++++for(int+i+=+0;+i+<+word.lenght();+i%2B%2B)
++++{
++++++++if(word.charAt(i)+==+charToRemove)
++++++++{
++++++++++++String+newWord+=+word.substring(0,+i)+%2B+word.substring(i+%2B+1);
++++++++++++return+removeChar(newWord,+charToRemove);
++++++++}
++++}

++++return+word;
}|code-block|syntax|javascript|564350|我在这个主题上看到的大多数代码都没有使用递归，所以希望我能帮助你或有同样问题的人。|564351|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 had to write code for a similar problem. One way that I was able to solve it used a recursive method of coding.

<pre><code>static String removeChar(String word, char charToRemove)
{
 for(int i = 0; i &lt; word.lenght(); i++)
 {
 if(word.charAt(i) == charToRemove)
 {
 String newWord = word.substring(0, i) + word.substring(i + 1);
 return removeChar(newWord, charToRemove);
 }
 }

 return word;
}
</code></pre>

Most of the code I've seen on this topic doesn't use recursion so hopefully I can help you or someone who has the same problem.

blocks|key|780821|text|Java+8|type|unstyled|depth|inlineStyleRanges|entityRanges|data|780822|import+java.util.Optional;

public+class+Test
{
++public+static+void+main(String[]+args)+throws+InterruptedException
++{
++++System.out.println(removeLastChar("test-abc"));
++}

++public+static+String+removeLastChar(String+s)+{
++++return+Optional.ofNullable(s)
++++++.filter(str+->+str.length()+!=+0)
++++++.map(str+->+str.substring(0,+str.length()+-+1))
++++++.orElse(s);
++++}
}|code-block|syntax|javascript|780823|输出:+test-ab|780824|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|$]]

Java 8 

<pre><code>import java.util.Optional;

public class Test
{
 public static void main(String[] args) throws InterruptedException
 {
 System.out.println(removeLastChar("test-abc"));
 }

 public static String removeLastChar(String s) {
 return Optional.ofNullable(s)
 .filter(str -&gt; str.length() != 0)
 .map(str -&gt; str.substring(0, str.length() - 1))
 .orElse(s);
 }
}
</code></pre>

Output : test-ab

blocks|key|564425|text|如何在递归的末尾生成char：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|564426|public+static+String++removeChar(String+word,+char+charToRemove)
++++{
++++++++String+char_toremove=Character.toString(charToRemove);
++++++++for(int+i+=+0;+i+<+word.length();+i%2B%2B)
++++++++{
++++++++++++if(word.charAt(i)+==+charToRemove)
++++++++++++{
++++++++++++++++String+newWord+=+word.substring(0,+i)+%2B+word.substring(i+%2B+1);
++++++++++++++++return+removeChar(newWord,charToRemove);
++++++++++++}
++++++++}
++++++++System.out.println(word);
++++++++return+word;
++++}|code-block|syntax|javascript|564427|例如：|564428|removeChar+("hello+world,+let's+go!",'l')+→+"heo+word,+et's+go!llll"
removeChar("you+should+not+go",'o')+→+"yu+shuld+nt+goooo"|564429|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|$]]

How to make the char in the recursion at the end:

<pre><code>public static String removeChar(String word, char charToRemove)
 {
 String char_toremove=Character.toString(charToRemove);
 for(int i = 0; i &lt; word.length(); i++)
 {
 if(word.charAt(i) == charToRemove)
 {
 String newWord = word.substring(0, i) + word.substring(i + 1);
 return removeChar(newWord,charToRemove);
 }
 }
 System.out.println(word);
 return word;
 }
</code></pre>

for exemple:

<pre><code>removeChar ("hello world, let's go!",'l') → "heo word, et's go!llll"
removeChar("you should not go",'o') → "yu shuld nt goooo"
</code></pre>

blocks|key|564457|text|这是一个适用于基本多语言平面(Java+8%2B)之外的代码点的答案。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|564458|使用流：|564459|public+String+method(String+str)+{
++++return+str.codePoints()
++++++++++++.limit(str.codePoints().count()+-+1)
++++++++++++.mapToObj(i->new+String(Character.toChars(i)))
++++++++++++.collect(Collectors.joining());
}|code-block|syntax|javascript|564460|也许更高效：|564461|public+String+method(String+str)+{
++++return+str.isEmpty()?+"":+str.substring(0,+str.length()+-+Character.charCount(str.codePointBefore(str.length())));
}|564462|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|$]]

Here's an answer that works with codepoints outside of the Basic Multilingual Plane (Java 8+).

Using streams:

<pre><code>public String method(String str) {
 return str.codePoints()
 .limit(str.codePoints().count() - 1)
 .mapToObj(i-&gt;new String(Character.toChars(i)))
 .collect(Collectors.joining());
}
</code></pre>

More efficient maybe:

<pre><code>public String method(String str) {
 return str.isEmpty()? "": str.substring(0, str.length() - Character.charCount(str.codePointBefore(str.length())));
}
</code></pre>

blocks|key|564491|text|我们可以使用子字符串。这是一个例子，|type|unstyled|depth|inlineStyleRanges|entityRanges|data|564492|public+class+RemoveStringChar+
{
++++public+static+void+main(String[]+args)+
++++{+++
++++++++String+strGiven+=+"Java";
++++++++System.out.println("Before+removing+string+character+-+"+%2B+strGiven);
++++++++System.out.println("After+removing+string+character+-+"+%2B+removeCharacter(strGiven,+3));
++++}

++++public+static+String+removeCharacter(String+strRemove,+int+position)
++++{+++
++++++++return+strRemove.substring(0,+position)+%2B+strRemove.substring(position+%2B+1);++++
++++}
}|code-block|syntax|javascript|564493|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|$]]

We can use substring. Here's the example,

<pre><code>public class RemoveStringChar 
{
 public static void main(String[] args) 
 { 
 String strGiven = "Java";
 System.out.println("Before removing string character - " + strGiven);
 System.out.println("After removing string character - " + removeCharacter(strGiven, 3));
 }

 public static String removeCharacter(String strRemove, int position)
 { 
 return strRemove.substring(0, position) + strRemove.substring(position + 1); 
 }
}
</code></pre>

blocks|key|780945|text|+//+creating+StringBuilder
+StringBuilder+builder+=+new+StringBuilder(requestString);
+//+removing+last+character+from+String
+builder.deleteCharAt(requestString.length()+-+1);|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|780946|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> // creating StringBuilder
 StringBuilder builder = new StringBuilder(requestString);
 // removing last character from String
 builder.deleteCharAt(requestString.length() - 1);
</code></pre>

blocks|key|564555|text|在Kotlin中，您可以使用string类的dropLast()方法。它将从字符串中删除给定的数字，并返回一个新的字符串|type|unstyled|depth|inlineStyleRanges|entityRanges|data|564556|var+string1+=+"Some+Text"
string1+=+string1.dropLast(1)|code-block|syntax|javascript|564557|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|$]]

In Kotlin you can used dropLast() method of the string class.
It will drop the given number from string, return a new string

<pre><code>var string1 = "Some Text"
string1 = string1.dropLast(1)
</code></pre>

blocks|key|781042|text|U可以做i+hereString+=+hereString.replace(hereString.chatAt(hereString.length()+-+1)，‘白色和平’)；|type|unstyled|depth|inlineStyleRanges|entityRanges|data|781043|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

u can do i
hereString = hereString.replace(hereString.chatAt(hereString.length() - 1) ,' whitespeace');

blocks|key|781097|text|只需像这样替换"if“的条件：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|781098|if(a.substring(a.length()-1).equals("x"))'|code-block|syntax|javascript|781099|这将为您做这个把戏。|781100|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|$]]

just replace the condition of "if" like this: 

<pre><code>if(a.substring(a.length()-1).equals("x"))'
</code></pre>

this will do the trick for you.

blocks|key|564692|text|假设我的string=24的总长度我想把位置14之后的最后一个字符切到末尾，这意味着我想要开始的14在那里。因此，我采用了以下解决方案。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|564693|String+date+=+"2019-07-31T22:00:00.000Z";
String+result+=+date.substring(0,+date.length()+-+14);|code-block|syntax|javascript|564694|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|$]]

Suppose total length of my string=24
I want to cut last character after position 14 to end, mean I want starting 14 to be there.
So I apply following solution.

<pre><code>String date = "2019-07-31T22:00:00.000Z";
String result = date.substring(0, date.length() - 14);
</code></pre>

blocks|key|781149|text|public+String+removeLastCharacter(String+str){
+++++++String+result+=+null;
++++++++if+((str+!=+null)+&&+(str.length()+>+0))+{
++++++++++return+str.substring(0,+str.length()+-+1);
++++++++}
++++++++else{
++++++++++++return+"";
++++++++}

}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|781150|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>public String removeLastCharacter(String str){
 String result = null;
 if ((str != null) &amp;&amp; (str.length() &gt; 0)) {
 return str.substring(0, str.length() - 1);
 }
 else{
 return "";
 }

}
</code></pre>

blocks|key|564726|text|简单易懂：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|564727|StringBuilder+sb=+new+StringBuilder();
for(Entry<String,String>+entry+:+map.entrySet())+{
++++++++sb.append(entry.getKey()+%2B+"_"+%2B+entry.getValue()+%2B+"%7C");
}
String+requiredString+=+sb.substring(0,+sb.length()+-+1);|code-block|syntax|javascript|564728|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 Peasy:

<pre><code>StringBuilder sb= new StringBuilder();
for(Entry&lt;String,String&gt; entry : map.entrySet()) {
 sb.append(entry.getKey() + "_" + entry.getValue() + "|");
}
String requiredString = sb.substring(0, sb.length() - 1);
</code></pre>

blocks|key|781225|text|对于kotlin，请查看|type|unstyled|depth|inlineStyleRanges|entityRanges|data|781226|++++val+string+=+"<<<First+Grade>>>"
++++println(string.drop(6))+//+st+Grade>>>
++++println(string.dropLast(6))+//+<<<First+Gr
++++println(string.dropWhile+{+!it.isLetter()+})+//+First+Grade>>>
++++println(string.dropLastWhile+{+!it.isLetter()+})+//+<<<First+Grade|code-block|syntax|javascript|781227|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 kotlin check out

<pre><code> val string = "&lt;&lt;&lt;First Grade&gt;&gt;&gt;"
 println(string.drop(6)) // st Grade&gt;&gt;&gt;
 println(string.dropLast(6)) // &lt;&lt;&lt;First Gr
 println(string.dropWhile { !it.isLetter() }) // First Grade&gt;&gt;&gt;
 println(string.dropLastWhile { !it.isLetter() }) // &lt;&lt;&lt;First Grade
</code></pre>

blocks|key|564923|text|如果你想删除结尾的特定字符，你可以使用：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|564924|myString.removeSuffix("x")|code-block|syntax|javascript|564925|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 want to remove specific character at the end you could use:
<pre><code>myString.removeSuffix(&quot;x&quot;)
</code></pre>

blocks|key|781331|text|一个简单的任务怎么会变得复杂呢？我的解决方案是：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|781332|public+String+removeLastChar(String+s)+{
++++return+s[0..-1]
}|code-block|syntax|javascript|781333|或|781334|public+String+removeLastChar(String+s)+{
++++if+(s.length()+>+0)+{
++++++++return+s[0..-1]
++++}
++++return+s
}|781335|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|$]]

How can a simple task be made complicated. My solution is:
<pre><code>public String removeLastChar(String s) {
 return s[0..-1]
}
</code></pre>
or
<pre><code>public String removeLastChar(String s) {
 if (s.length() &gt; 0) {
 return s[0..-1]
 }
 return s
}
</code></pre>

blocks|key|565020|text|使用StringUtils.Chop(Str)它还会处理null和空字符串，您需要导入common-io：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|565021|++++<dependency>
++++++++<groupId>commons-io</groupId>
++++++++<artifactId>commons-io</artifactId>
++++++++<version>2.8.0</version>
++++</dependency>|code-block|syntax|javascript|565022|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|$]]

Use StringUtils.Chop(Str) it also takes care of null and empty strings, u need to import common-io:
<pre><code> &lt;dependency&gt;
 &lt;groupId&gt;commons-io&lt;/groupId&gt;
 &lt;artifactId&gt;commons-io&lt;/artifactId&gt;
 &lt;version&gt;2.8.0&lt;/version&gt;
 &lt;/dependency&gt;
</code></pre>

blocks|key|565049|text|如果要删除给定文件的文件扩展名，|type|unstyled|depth|inlineStyleRanges|entityRanges|data|565050|**示例代码|565051|+public+static+String+removeNCharactersFromLast(String+str,int+n){
++++if+(str+!=+null+&&+(str.length()+>+0))+{
++++++++return+str.substring(0,+str.length()+-+n);
++++}

++++return+"";

}|code-block|syntax|javascript|565052|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|L|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|M|8|@]|9|@]|A|$G|H]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

if we want to remove file extension of the given file,
** Sample code
<pre><code> public static String removeNCharactersFromLast(String str,int n){
 if (str != null &amp;&amp; (str.length() &gt; 0)) {
 return str.substring(0, str.length() - n);
 }

 return &quot;&quot;;

}
</code></pre>

blocks|key|781424|text|简单的Kotlin方式：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|781425|if+(inputString.isNotEmpty())+{
++++inputString.dropLast(1)+//+You+can+define+how+many+chars+you+want+to+remove
}|code-block|syntax|javascript|781426|请参考官方文档here|781427|entityMap|0|LINK|mutability|MUTABLE|url|https://kotlinlang.org/api/latest/jvm/stdlib/kotlin.text/drop-last.html^0|3|6|0|0|7|4|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|@]|D|@$9|Z|A|10|1|11]]|E|$]]|$1|M|3|-4|5|6|7|12|8|@]|D|@]|E|$]]]|N|$O|$5|P|Q|R|E|$S|T]]]]

Simple <code>Kotlin</code> way:
<pre><code>if (inputString.isNotEmpty()) {
 inputString.dropLast(1) // You can define how many chars you want to remove
}
</code></pre>
Refer to the official doc <a href="https://kotlinlang.org/api/latest/jvm/stdlib/kotlin.text/drop-last.html" rel="nofollow noreferrer">here</a>

I want to remove the last character from a string. I've tried doing this:

<pre><code>public String method(String str) {
 if (str.charAt(str.length()-1)=='x'){
 str = str.replace(str.substring(str.length()-1), "");
 return str;
 } else{
 return str;
 }
}
</code></pre>

Getting the length of the string - 1 and replacing the last letter with nothing (deleting it), but every time I run the program, it deletes middle letters that are the same as the last letter.

For example, the word is "admirer"; after I run the method, I get "admie." I want it to return the word admire.

How to remove the last character from a string?

Kotlin

Java

我想删除字符串中的最后一个字符。我试过这样做：public String method(String str) {    if (str.charAt(str.length()-1)=='x'){        str = str.replace(str.substring(str.length()-1), ""); ...

问如何删除字符串中的最后一个字符？
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