Trim characters in Java

Question

How can I trim characters in Java?
e.g.

String j = “\joe\jill\”.Trim(new char[] {“\”});

j should be

"joe\jill"

String j = “jack\joe\jill\”.Trim("jack");

j should be

"\joe\jill\"

etc

Answer 1

I don't think there is any built in function to trim based on a passed in string. Here is a small example of how to do this. This is not likely the most efficient solution, but it is probably fast enough for most situations, evaluate and adapt to your needs. I recommend testing performance and optimizing as needed for any code snippet that will be used regularly. Below, I've included some timing information as an example.

public String trim( String stringToTrim, String stringToRemove )
{
    String answer = stringToTrim;

    while( answer.startsWith( stringToRemove ) )
    {
        answer = answer.substring( stringToRemove.length() );
    }

    while( answer.endsWith( stringToRemove ) )
    {
        answer = answer.substring( 0, answer.length() - stringToRemove.length() );
    }

    return answer;
}

This answer assumes that the characters to be trimmed are a string. For example, passing in "abc" will trim out "abc" but not "bbc" or "cba", etc.

Some performance times for running each of the following 10 million times.

" mile ".trim(); runs in 248 ms included as a reference implementation for performance comparisons.

trim( "smiles", "s" ); runs in 547 ms - approximately 2 times as long as java's String.trim() method.

"smiles".replaceAll("s$|^s",""); runs in 12,306 ms - approximately 48 times as long as java's String.trim() method.

And using a compiled regex pattern Pattern pattern = Pattern.compile("s$|^s"); pattern.matcher("smiles").replaceAll(""); runs in 7,804 ms - approximately 31 times as long as java's String.trim() method.

Answer 2

Here's how I would do it.

I think it's about as efficient as it reasonably can be. It optimizes the single character case and avoids creating multiple substrings for each subsequence removed.

Note that the corner case of passing an empty string to trim is handled (some of the other answers would go into an infinite loop).

/** Trim all occurrences of the string <code>rmvval</code> from the left and right of <code>src</code>.  Note that <code>rmvval</code> constitutes an entire string which must match using <code>String.startsWith</code> and <code>String.endsWith</code>. */
static public String trim(String src, String rmvval) {
    return trim(src,rmvval,rmvval,true);
    }

/** Trim all occurrences of the string <code>lftval</code> from the left and <code>rgtval</code> from the right of <code>src</code>.  Note that the values to remove constitute strings which must match using <code>String.startsWith</code> and <code>String.endsWith</code>. */
static public String trim(String src, String lftval, String rgtval, boolean igncas) {
    int                                 str=0,end=src.length();

    if(lftval.length()==1) {                                                    // optimize for common use - trimming a single character from left
        char chr=lftval.charAt(0);
        while(str<end && src.charAt(str)==chr) { str++; }
        }
    else if(lftval.length()>1) {                                                // handle repeated removal of a specific character sequence from left
        int vallen=lftval.length(),newstr;
        while((newstr=(str+vallen))<=end && src.regionMatches(igncas,str,lftval,0,vallen)) { str=newstr; }
        }

    if(rgtval.length()==1) {                                                    // optimize for common use - trimming a single character from right
        char chr=rgtval.charAt(0);
        while(str<end && src.charAt(end-1)==chr) { end--; }
        }
    else if(rgtval.length()>1) {                                                // handle repeated removal of a specific character sequence from right
        int vallen=rgtval.length(),newend;
        while(str<=(newend=(end-vallen)) && src.regionMatches(igncas,newend,rgtval,0,vallen)) { end=newend; }
        }

    if(str!=0 || end!=src.length()) {
        if(str<end) { src=src.substring(str,end); }                            // str is inclusive, end is exclusive
        else        { src="";                     }
        }

    return src;
    }

Answer 3

public static String trim(String value, char c) {

    if (c <= 32) return value.trim();

    int len = value.length();
    int st = 0;
    char[] val = value.toCharArray();    /* avoid getfield opcode */

    while ((st < len) && (val[st] == c)) {
        st++;
    }
    while ((st < len) && (val[len - 1] == c)) {
        len--;
    }
    return ((st > 0) || (len < value.length())) ? value.substring(st, len) : value;
}

Answer 4

EDIT: Amended by answer to replace just the first and last '\' character.

System.err.println("\\joe\\jill\\".replaceAll("^\\\\|\\\\$", ""));

Answer 5

You could use removeStart and removeEnd from Apache Commons Lang StringUtils

Answer 6

10 year old question but felt most of the answers were a bit convoluted or didn't quite work the way that was asked. Also the most upvoted answer here didn't provide any examples. Here's a simple class I made:

https://gist.github.com/Maxdw/d71afd11db2df4f1297ad3722d6392ec

Usage:

Trim.left("\joe\jill\", "\") == "joe\jill\"

Trim.left("jack\joe\jill\", "jack") == "\joe\jill\"

Trim.left("\\\\joe\\jill\\\\", "\") == "joe\\jill\\\\"