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Cisco Certified Network Associate Exam,640-802 CCNA All Answers ~100/100. Daily update
String foo = "VWXYZ";
ReplyDeletefoo = "%" + foo.replaceAll("(.)","$1%");
System.out.println(foo);
Output:
%V%W%X%Y%Z%
You don't need a StringBuilder. The compiler will take care of that simple concatenation prior to the regex for you by using one.
Edit in response to comment below:
replaceAll() uses a Regular Expression (regex).
The regex (.) says "match any character, and give me a reference to it" . is a wildcard for any character, the parenthesis create the backreference. The $1 in the second argument says "Use backreference #1 from the match".
replaceAll() keeps running this expression over the whole string replacing each character with itself followed by a percent sign, building a new String which it then returns to you.
string str="AA";
ReplyDeleteStringBuilder sb = new StringBuilder();
for(int i = 0; i < str.length(); s++)
{
sb.append("%");
sb.append(str.charAt(i));
}
sb.append("%");
This is the simpliest way to do that.
Try something like this:
ReplyDeleteString test = "ABC";
StringBuilder builder = new StringBuilder("");
builder.append("%");
for (char achar : test.toCharArray()) {
builder.append(achar);
builder.append("%");
}
System.out.println(builder.toString());
public static String escape(String s) {
ReplyDeleteStringBuilder buf = new StringBuilder();
boolean wasLetter = false;
for (char c: s.toCharArray()) {
boolean isLetter = Character.isLetter(c);
if (isLetter && !wasLetter) {
buf.append('%');
}
buf.append(c);
if (isLetter) {
buf.append('%');
}
wasLetter = isLetter;
}
return buf.toString();
}
StringBuilder sb = new StringBuilder("AAAAAAA");
ReplyDeletefor(int i = sb.length(); i >= 0; i--)
{
sb.insert(i, '%');
}
You may see this.
ReplyDeleteString s="AAAA";
StringBuilder builder = new StringBuilder();
char[] ch=s.toCharArray();
for(int i=0;i<ch.length;i++)
{
builder.append("%"+ch[i]);
}
builder.append("%");
System.out.println(builder.toString());
Output
%A%A%A%A%