根据字符串长度修剪字符串

Question

I want to trim a string if the length exceeds 10 characters.

Suppose if the string length is 12 ( String s="abcdafghijkl" ), then the new trimmed string will contain "abcdefgh.." .

How can I achieve this?

Answer 1

s = s.substring(0, Math.min(s.length(), 10));

Using Math.min like this avoids an exception in the case where the string is already shorter than 10 .

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Notes:

1. The above does real trimming. If you actually want to replace the last characters with three dots if the string is too long, use Apache Commons StringUtils.abbreviate ; see @H6's solution . If you want to use the Unicode horizontal ellipsis character, see @Basil's solution .

2. For typical implementations of String , s.substring(0, s.length()) will return s rather than allocating a new String .

3. This may behave incorrectly ¹ if your String contains Unicode codepoints outside of the BMP; eg Emojis. For a (more complicated) solution that works correctly for all Unicode code-points, see @sibnick's solution .

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^{1 - A Unicode codepoint that is not on plane 0 (the BMP) is represented as a "surrogate pair" (ie two char values) in the String . By ignoring this, we might trim the string to fewer than 10 code points, or (worse) truncate it in the middle of a surrogate pair. On the other hand, String.length() is not a good measure of Unicode text length, so trimming based on that property may be the wrong thing to do.}

Answer 2

StringUtils.abbreviate from Apache Commons Lang library could be your friend:

StringUtils.abbreviate("abcdefg", 6) = "abc..."
StringUtils.abbreviate("abcdefg", 7) = "abcdefg"
StringUtils.abbreviate("abcdefg", 8) = "abcdefg"
StringUtils.abbreviate("abcdefg", 4) = "a..."

Commons Lang3 even allow to set a custom String as replacement marker. With this you can for example set a single character ellipsis.

StringUtils.abbreviate("abcdefg", "\u2026", 6) = "abcde…"

Answer 3

There is a Apache Commons StringUtils function which does this.

s = StringUtils.left(s, 10)

If len characters are not available, or the String is null, the String will be returned without an exception. An empty String is returned if len is negative.

StringUtils.left(null, ) = null
StringUtils.left( , -ve) = ""
StringUtils.left("", *) = ""
StringUtils.left("abc", 0) = ""
StringUtils.left("abc", 2) = "ab"
StringUtils.left("abc", 4) = "abc"

StringUtils.Left JavaDocs

Courtesy:Steeve McCauley

Answer 4

As usual nobody cares about UTF-16 surrogate pairs. See about them: What are the most common non-BMP Unicode characters in actual use? Even authors of org.apache.commons/commons-lang3

You can see difference between correct code and usual code in this sample:

public static void main(String[] args) {
    //string with FACE WITH TEARS OF JOY symbol
    String s = "abcdafghi\uD83D\uDE02cdefg";
    int maxWidth = 10;
    System.out.println(s);
    //do not care about UTF-16 surrogate pairs
    System.out.println(s.substring(0, Math.min(s.length(), maxWidth)));
    //correctly process UTF-16 surrogate pairs
    if(s.length()>maxWidth){
        int correctedMaxWidth = (Character.isLowSurrogate(s.charAt(maxWidth)))&&maxWidth>0 ? maxWidth-1 : maxWidth;
        System.out.println(s.substring(0, Math.min(s.length(), correctedMaxWidth)));
    }
}

Answer 5

s = s.length() > 10 ? s.substring(0, 9) : s;

Answer 6

Or you can just use this method in case you don't have StringUtils on hand:

public static String abbreviateString(String input, int maxLength) {
    if (input.length() <= maxLength) 
        return input;
    else 
        return input.substring(0, maxLength-2) + "..";
}

Answer 7

以防万一您正在寻找一种方法来修剪和保留字符串的最后 10 个字符。

s = s.substring(Math.max(s.length(),10) - 10);

Answer 8

The question is asked on Java, but it was back in 2014.
In case you use Kotlin now, it is as simple as:

yourString.take(10)

Returns a string containing the first n characters from this string, or the entire string if this string is shorter.

Documentation

Answer 9

tl;dr

You seem to be asking for an ellipsis ( … ) character in the last place, when truncating. Here is a one-liner to manipulate your input string.

String input = "abcdefghijkl";
String output = ( input.length () > 10 ) ? input.substring ( 0 , 10 - 1 ).concat ( "…" ) : input;

See this code run live at IdeOne.com.

abcdefghi…

Ternary operator

We can make a one-liner by using the ternary operator .

String input = "abcdefghijkl" ;

String output = 
    ( input.length() > 10 )          // If too long…
    ?                                
    input     
    .substring( 0 , 10 - 1 )         // Take just the first part, adjusting by 1 to replace that last character with an ellipsis.
    .concat( "…" )                   // Add the ellipsis character.
    :                                // Or, if not too long…
    input                            // Just return original string.
;

See this code run live at IdeOne.com.

abcdefghi…

Java streams

The Java Streams facility makes this interesting, as of Java 9 and later. Interesting, but maybe not the best approach.

We use code points rather than char values. The char type is legacy, and is limited to the a subset of all possible Unicode characters.

String input = "abcdefghijkl" ;
int limit = 10 ;
String output =
        input
                .codePoints()
                .limit( limit )
                .collect(                                    // Collect the results of processing each code point.
                        StringBuilder::new,                  // Supplier<R> supplier
                        StringBuilder::appendCodePoint,      // ObjIntConsumer<R> accumulator
                        StringBuilder::append                // BiConsumer<R,​R> combiner
                )
                .toString()
        ;

If we had excess characters truncated, replace the last character with an ellipsis .

if ( input.length () > limit )
{
    output = output.substring ( 0 , output.length () - 1 ) + "…";
}

If only I could think of a way to put together the stream line with the "if over limit, do ellipsis" part.

Answer 10

str==null ? str : str.substring(0, Math.min(str.length(), 10))

or,

str==null ? "" : str.substring(0, Math.min(str.length(), 10))

Works with null.

Answer 11

// this is how you shorten the length of the string with .. // add following method to your class

private String abbreviate(String s){
  if(s.length() <= 10) return s;
  return s.substring(0, 8) + ".." ;
}

Answer 12

Here is the Kotlin solution

One line,

if (yourString?.length!! >= 10) yourString?.take(90).plus("...") else yourString

Traditional,

if (yourString?.length!! >= 10) {
  yourString?.take(10).plus("...")
 } else {
  yourString
 }

Answer 13

I want to trim a string if the length exceeds 10 characters.

Suppose if the string length is 12 ( String s="abcdafghijkl" ), then the new trimmed string will contain "abcdefgh.." .

How can I achieve this?