blocks|key|755594|text|您可以很容易地使用String.replace()|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|755595|String+helloWorld+=+"Hello+World!";
String+hellWrld+=+helloWorld.replace("o","");|code-block|syntax|javascript|755596|entityMap|0|LINK|mutability|MUTABLE|url|http://download.oracle.com/javase/6/docs/api/java/lang/String.html#replace%2528java.lang.CharSequence,%2520java.lang.CharSequence%2529^0|9|G|9|G|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@$9|T|A|U|B|C]]|D|@$9|V|A|W|1|X]]|E|$]]|$1|F|3|G|5|H|7|Y|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Z|8|@]|D|@]|E|$]]]|L|$M|$5|N|O|P|E|$Q|R]]]]

You could easily use <a href="http://download.oracle.com/javase/6/docs/api/java/lang/String.html#replace%28java.lang.CharSequence,%20java.lang.CharSequence%29" rel="noreferrer"><code>String.replace()</code></a>:

<pre><code>String helloWorld = "Hello World!";
String hellWrld = helloWorld.replace("o","");
</code></pre>

blocks|key|538998|text|replace('regex',+'replacement');
replaceAll('regex',+'replacement');|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|538999|在您的示例中，|unstyled|539000|String+hi+=+"Hello+World!"
String+no_o+=+hi.replaceAll("o",+"");|539001|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|L|8|@]|9|@]|A|$]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$B|C]]|$1|I|3|-4|5|F|7|N|8|@]|9|@]|A|$]]]|J|$]]

<pre><code>replace('regex', 'replacement');
replaceAll('regex', 'replacement');
</code></pre>

In your example,

<pre><code>String hi = "Hello World!"
String no_o = hi.replaceAll("o", "");
</code></pre>

blocks|key|539047|text|您应该查看允许您在指定offset处删除、插入和替换字符的StringBuilder/StringBuffer。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|539048|entityMap|0|LINK|mutability|MUTABLE|url|http://download.oracle.com/javase/6/docs/api/java/lang/StringBuilder.html^0|T|Q|B|6|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]]]]

You should have to look at <code>StringBuilder/StringBuffer</code> which allow you to delete, insert, replace char(s) at specified <a href="http://download.oracle.com/javase/6/docs/api/java/lang/StringBuilder.html" rel="nofollow noreferrer">offset</a>.

blocks|key|539050|text|您可以使用StringBuffer|type|unstyled|depth|inlineStyleRanges|entityRanges|data|539051|StringBuffer+text+=+new+StringBuffer("Hello+World");
text.replace(+StartIndex+,EndIndex+,String);|code-block|syntax|javascript|539052|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 use StringBuffer 

<pre><code>StringBuffer text = new StringBuffer("Hello World");
text.replace( StartIndex ,EndIndex ,String);
</code></pre>

blocks|key|539102|text|这对我来说很好。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|539103|String+hi+=+"Hello+World!"
String+no_o+=+hi.replaceAll("o",+"");|code-block|syntax|javascript|539104|或者您可以使用|539105|String+no_o+=+hi.replace("o",+"");|539106|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|$]]

This works good for me.

<pre><code>String hi = "Hello World!"
String no_o = hi.replaceAll("o", "");
</code></pre>

or you can use 

<pre><code>String no_o = hi.replace("o", "");
</code></pre>

blocks|key|755734|text|private+static+void+replaceChar()+{
++++String+str+=+"hello+world";
++++final+String[]+res+=+Arrays.stream(str.split(""))
++++++++++++.filter(s+->+!s.equalsIgnoreCase("o"))
++++++++++++.toArray(String[]::new);
++++System.out.println(String.join("",+res));
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|755735|如果您有一些复杂的逻辑来过滤字符，那么只需使用另一种方法而不是replace()。|unstyled|offset|length|style|CODE|755736|entityMap^0|0|V|9|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|N|8|@$G|O|H|P|I|J]]|9|@]|A|$]]|$1|K|3|-4|5|F|7|Q|8|@]|9|@]|A|$]]]|L|$]]

<pre><code>private static void replaceChar() {
 String str = "hello world";
 final String[] res = Arrays.stream(str.split(""))
 .filter(s -&gt; !s.equalsIgnoreCase("o"))
 .toArray(String[]::new);
 System.out.println(String.join("", res));
}
</code></pre>

In case you have some complicated logic to filter the char, just another way instead of <code>replace()</code>.

blocks|key|755778|text|下面是从给定字符串中删除所有子字符串的实现|type|unstyled|depth|inlineStyleRanges|entityRanges|data|755779|public+static+String+deleteAll(String+str,+String+pattern)
{
++++for(int+index+=+isSubstring(str,+pattern);+index+!=+-1;+index+=+isSubstring(str,+pattern))
++++++++str+=+deleteSubstring(str,+pattern,+index);

++++return+str;
}

public+static+String+deleteSubstring(String+str,+String+pattern,+int+index)
{
++++int+start_index+=+index;
++++int+end_index+=+start_index+%2B+pattern.length()+-+1;
++++int+dest_index+=+0;
++++char[]+result+=+new+char[str.length()];


++++for(int+i+=+0;+i<+str.length()+-+1;+i%2B%2B)
++++++++if(i+<+start_index+%7C%7C+i+>+end_index)
++++++++++++result[dest_index%2B%2B]+=+str.charAt(i);

++++return+new+String(result,+0,+dest_index+%2B+1);
}|code-block|syntax|javascript|755780|isSubstring()方法的实现是here|offset|length|755781|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/a/53113110/4773290^0|0|0|J|4|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]]|$1|G|3|H|5|6|7|U|8|@]|9|@$I|V|J|W|1|X]]|A|$]]|$1|K|3|-4|5|6|7|Y|8|@]|9|@]|A|$]]]|L|$M|$5|N|O|P|A|$Q|R]]]]

Here is the implementation to delete all the substrings from the given string

<pre><code>public static String deleteAll(String str, String pattern)
{
 for(int index = isSubstring(str, pattern); index != -1; index = isSubstring(str, pattern))
 str = deleteSubstring(str, pattern, index);

 return str;
}

public static String deleteSubstring(String str, String pattern, int index)
{
 int start_index = index;
 int end_index = start_index + pattern.length() - 1;
 int dest_index = 0;
 char[] result = new char[str.length()];


 for(int i = 0; i&lt; str.length() - 1; i++)
 if(i &lt; start_index || i &gt; end_index)
 result[dest_index++] = str.charAt(i);

 return new String(result, 0, dest_index + 1);
}
</code></pre>

The implementation of isSubstring() method is <a href="https://stackoverflow.com/a/53113110/4773290">here</a>

blocks|key|539230|text|replaceAll(String+regex,+String+replacement)|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|539231|上面的方法将有助于得到答案。|unstyled|539232|String+check+=+"Hello+World";
check+=+check.replaceAll("o","");|539233|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|L|8|@]|9|@]|A|$]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$B|C]]|$1|I|3|-4|5|F|7|N|8|@]|9|@]|A|$]]]|J|$]]

<pre><code>replaceAll(String regex, String replacement)
</code></pre>

Above method will help to get the answer.

<pre><code>String check = "Hello World";
check = check.replaceAll("o","");
</code></pre>

blocks|key|539268|text|您还可以使用Substring替换为现有的字符串：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|539269|var+str+=+"abc+awwwa";
var+Index+=+str.indexOf('awwwa');
str+=+str.substring(0,+Index);|code-block|syntax|javascript|539270|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 use Substring also for replacing with existing string:

<pre><code>var str = "abc awwwa";
var Index = str.indexOf('awwwa');
str = str.substring(0, Index);
</code></pre>

blocks|key|755882|text|如果你知道开始和结束索引，你可以使用下面的代码|type|unstyled|depth|inlineStyleRanges|entityRanges|data|755883|string+=+string.substring(0,+start_index)+%2B+string.substring(end_index,+string.length());|code-block|syntax|javascript|755884|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 know the start and end index you may use this

<pre><code>string = string.substring(0, start_index) + string.substring(end_index, string.length());
</code></pre>

Is there an easy way to remove substring from a given <code>String</code> in Java? 

Example: <code>"Hello World!"</code>, removing <code>"o"</code> → <code>"Hell Wrld!"</code>

How can I remove a substring from a given String?

在java语言中，有没有一种简单的方法从给定的String中删除子字符串？示例："Hello World!"，removing "o" "Hell Wrld!"→

问如何从给定字符串中删除子字符串？
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