I have a string like this:
String line="[x1,x2,x3]";
I want to remove both of the square brackets from the String (replace them with an empty string). I tried it like this:
String x=line.replace("[","").replace("]","");
Is there a more efficient way than this?
String is immutable so you can
String x = line.replace("[","").replace("]","");
another more efficient way is using regex with a pattern for both [ and ]
like
String x = line.replaceAll("[\\]\\[]+", "");
You may use .replaceAll if you want to use a regex to remove both [ and ] at the same time anywhere inside the string (actually, the equivalent of OP two line code snippet):
String x = line.replaceAll("[\\]\\[]+", "");
The [\]\[]+ pattern matches one or more (+) [ or ] characters (these must be escaped inside [....], a character class).
String line="[x1,x2,x3]";
String x = line.replaceAll("[\\[\\]]+", "");
System.out.println(x);
// => x1,x2,x3
If the brackets are always present, and always at the start and end of the string as shown in the example, you can just use substring:
String x2 = line.substring(1, line.length()-1);
Another way:
line = line.substring(1, line.length()-1);
If you want to preserve the square brackets that comes in between, then you can use:
String line="[x1,[x2],x3]";
line = line.replaceAll("^\\[(.*)\\]$", "$1");
Output
x1,[x2],x3
If brackets can only appear at the beginning and at the end of the string, the fastest method (in terms of performance) would be using substring which does not involve regular expressions.
If brackets are guaranteed to be present:
line = line.substring(1, line.length()-1);
If brackets are optional:
int len = line.length();
int trimStart = (len > 0 && line.charAt(0) == '[') ? 1 : 0;
int trimEnd = (len > 0 && line.charAt(len-1) == ']') ? 1 : 0;
if (trimStart > 0 || trimEnd > 0) {
line = line.substring(trimStart, len - trimEnd);
}
null- please clarify.x=line.replaceAll("^\\[|\\]$", "").