Limiting the number of characters in a string, and chopping off the rest

Question

I need to create a summary table at the end of a log with some values that are obtained inside a class. The table needs to be printed in fixed-width format. I have the code to do this already, but I need to limit Strings, doubles and ints to a fixed-width size that is hard-coded in the code.

So, suppose I want to print a fixed-width table with

    int,string,double,string
    int,string,double,string
    int,string,double,string
    int,string,double,string

    and the fixed widths are: 4, 5, 6, 6.

If a value exceeds this width, the last characters need to be cut off. So for example:

    124891, difference, 22.348, montreal

the strings that need to be printed ought to be:

    1248 diffe 22.348 montre

I am thinking I need to do something in the constructor that forces a string not to exceed a certain number of characters. I will probably cast the doubles and ints to a string, so I can enforce the maximum width requirements.

I don't know which method does this or if a string can be instantiated to behave taht way. Using the formatter only helps with the fixed-with formatting for printing the string, but it does not actually chop characters that exceed the maximum length.

if ( ff.length()>7 ) ff = ff.substring(0,7); – Fattie May 24 '14 at 12:16 — Fattie, May 24 '14 at 12:16

score 192 · Answer 1 · edited Mar 08 '19 at 19:03

192

You can also use String.format("%3.3s", "abcdefgh"). The first digit is the minimum length (the string will be left padded if it's shorter), the second digit is the maxiumum length and the string will be truncated if it's longer. So

System.out.printf("'%3.3s' '%3.3s'", "abcdefgh", "a");

will produce

'abc' '  a'

(you can remove quotes, obviously).

edited Mar 08 '19 at 19:03

Pshemo

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answered Nov 21 '14 at 11:45

Jaksa

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use [`String.format(String format, Object... args)`](https://docs.oracle.com/javase/7/docs/api/java/lang/String.html#format(java.lang.String,%20java.lang.Object...)) as a more common approach – radistao Mar 27 '17 at 07:51
1

Whats String Syntax to get the last n character of a string? – Shijil Aug 08 '18 at 12:28
Probably slower than `String::substring(0, max)` however. – Laurent Grégoire Jul 11 '19 at 08:37

score 103 · Accepted Answer · edited Apr 29 '13 at 12:21

103

Use this to cut off the non needed characters:

String.substring(0, maxLength);

Example:

String aString ="123456789";
String cutString = aString.substring(0, 4);
// Output is: "1234"

To ensure you are not getting an IndexOutOfBoundsException when the input string is less than the expected length do the following instead:

int maxLength = (inputString.length() < MAX_CHAR)?inputString.length():MAX_CHAR;
inputString = inputString.substring(0, maxLength);

If you want your integers and doubles to have a certain length then I suggest you use NumberFormat to format your numbers instead of cutting off their string representation.

edited Apr 29 '13 at 12:21

Claes Mogren

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answered Dec 03 '11 at 18:05

GETah

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Hmm. It would be great if "Precision" works for String as well. – Sandeep Jindal Oct 30 '14 at 22:04
44

I think `inputString.substring(0, Math.min(MAX_CHAR, inputString.length()));` is much more readable. – pvorb Nov 29 '15 at 12:50
1

or `inputString.substring(0, (inputString.length() > MAX_CHAR) ? MAX_CHAR : inputString.length())` – CervEd Mar 06 '18 at 22:24
this coupled with the String.transform() function introduced in JDK 12 is the best :) – tazmaniax Nov 29 '20 at 13:23
This answer may produce invalid string if it contains surrogate pairs. – bumfo Jan 18 '22 at 09:17
Performance of @pvorb answer is good, because `String.substring()` implementation has a special check to return `this` if `beginIndex==0 && endIndex==value.length)` – Sam Barnum Jul 27 '22 at 17:51

score 36 · Answer 3 · answered Jun 04 '14 at 23:41

For readability, I prefer this:

if (inputString.length() > maxLength) {
    inputString = inputString.substring(0, maxLength);
}

over the accepted answer.

int maxLength = (inputString.length() < MAX_CHAR)?inputString.length():MAX_CHAR;
inputString = inputString.substring(0, maxLength);

score 28 · Answer 4 · edited Mar 27 '17 at 07:55

28

You can use the Apache Commons StringUtils.substring(String str, int start, int end) static method, which is also null safe.

See: http://commons.apache.org/proper/commons-lang/javadocs/api-2.6/org/apache/commons/lang/StringUtils.html#substring%28java.lang.String,%20int,%20int%29

and http://commons.apache.org/proper/commons-lang/javadocs/api-2.6/src-html/org/apache/commons/lang/StringUtils.html#line.1961

edited Mar 27 '17 at 07:55

radistao

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answered Dec 13 '13 at 15:40

khriskooper

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Even easier: use the [left method](https://commons.apache.org/proper/commons-lang/apidocs/org/apache/commons/lang3/StringUtils.html#left-java.lang.String-int-). – beat Feb 02 '17 at 10:26
4

Very excessive to add a library to truncate a string. Caution for this solution would be to only use this if you already are using commons – Patrick Michaelsen Apr 03 '18 at 17:28

score 25 · Answer 5 · answered Nov 16 '15 at 17:27

25

You can achieve this easily using

    shortString = longString.substring(0, Math.min(s.length(), MAX_LENGTH));

answered Nov 16 '15 at 17:27

Arade

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score 14 · Answer 6 · answered Jun 26 '19 at 19:33

14

If you just want a maximum length, use StringUtils.left! No if or ternary ?: needed.

int maxLength = 5;
StringUtils.left(string, maxLength);

Output:

      null -> null
        "" -> ""
       "a" -> "a"
"abcd1234" -> "abcd1"

Left Documentation

answered Jun 26 '19 at 19:33

Gibolt

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score 3 · Answer 7 · answered Sep 11 '15 at 18:18

The solution may be java.lang.String.format("%" + maxlength + "s", string).trim(), like this:

int maxlength = 20;
String longString = "Any string you want which length is greather than 'maxlength'";
String shortString = "Anything short";
String resultForLong = java.lang.String.format("%" + maxlength + "s", longString).trim();
String resultForShort = java.lang.String.format("%" + maxlength + "s", shortString).trim();
System.out.println(resultForLong);
System.out.println(resultForShort);

ouput:

Any string you want w

Anything short

Using "%.20s" instead of "%20s" prevents the need for the trim — Robbert, Mar 19 '20 at 09:07

score 2 · Answer 8 · answered Dec 03 '11 at 18:10

2

Ideally you should try not to modify the internal data representation for the purpose of creating the table. Whats the problem with String.format()? It will return you new string with required width.

answered Dec 03 '11 at 18:10

Ashwinee K Jha

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Well, maybe I am using it wrong. but it does not actually change the length of my string. I have used the formatter with no effect. It does print stuff nicely, but it doesn't change string's lengths: Formatter fmt = new Formatter(); // Corresponding formats for declared members String hardCodedFormat = "%5s%5s"; // %20s String m_Project; String name = "mynameis"; String name2 = "ted123"; fmt.format(hardCodedFormat, name, name2); System.out.println(fmt); – Flethuseo Dec 03 '11 at 18:30
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@Flethuseo You should be able to truncate the strings. Try this: System.err.format("%1$.5s", "A long string"); – Roger Lindsjö Dec 03 '11 at 22:45

Limiting the number of characters in a string, and chopping off the rest

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