Im trying to replace part of a String based on a certain phrase being present within it. Consider the string "Hello my Dg6Us9k. I am alive.".
I want to search for the phase "my" and remove 8 characters to the right, which removes the hash code. This gives the string "Hello. I am alive." How can i do this in Java?
You could achieve this through string.replaceAll function.
string.replaceAll("\\bmy.{8}", "");
Add \\b if necessary. \\b called word boundary which matches between a word character and a non-word character. .{8} matches exactly the following 8 characters.
To remove also the space before my
System.out.println("Hello my Dg6Us9k. I am alive.".replaceAll("\\smy.{8}", ""));
This should do it:
String s = ("Hello my Dg6Us9k. I am alive");
s.replace(s.substring(s.indexOf("my"), s.indexOf("my")+11),"");
That is replacing the string starts at "my" and is 11 char long with nothing.
Use regex like this :
public static void main(String[] args) {
String s = "Hello my Dg6Us9k. I am alive";
String newString=s.replaceFirst("\\smy\\s\\w{7}", "");
System.out.println(newString);
}
O/P :
Hello. I am alive
a-zA-Z_
Java strings are immutable, so you cannot change the string. You have to create a new string. So, find the index i of "my". Then concatenate the substring before (0...i) and after (i+8...).
int i = s.indexOf("my");
if (i == -1) { /* no "my" in there! */ }
string ret = s.substring(0,i);
ret.concat(s.substring(i+2+8));
return ret;
If you want to be flexible about the hash code length, use the folowing regexp:
String foo="Hello my Dg6Us9k. I am alive.";
String bar = foo.replaceFirst("\\smy.*?\\.", ".");
System.out.println(bar);
