Say i have a string as follows:
property typename ttl=10 cval=30 test myprogram
Now i am trying to get ttl value i.e 10. I am just confused how to use substring() method or if it eases with regex (i dont know how to use it for this case)
It would be helpful if someone can hint me to solve this problem.
Without regex, you can find the position of the value with the help of the String.indexOf(String) and String.indexOf(String, int) methods (in case where property typename is of variable length.)
Then, you can in fact extract the value with String.substring(int, int).
- http://java.sun.com/j2se/1.5.0/docs/api/java/lang/String.html
With regex, you might use an expression that matches the non-whitespace characters that follow "ttl="
You could also use an expression that allows you to extract all the key=value pairs.
- http://java.sun.com/j2se/1.5.0/docs/api/java/util/regex/Pattern.html
- http://java.sun.com/j2se/1.5.0/docs/api/java/util/regex/Matcher.html
Hi,
Whether the particular text is a pattern (i.e., it will remain same as it is expect the values) or one-time string. Please let me know..
Regards,
Loga
> you can read this:
>
> http://mindprod.com/jgloss/substring.html
I know how to use substring. But i am confused how to retrieve the value 10 from that string.
say, my string variable is str,
str.substring(str.indexOf("ttl"), ?) // what have to come in ? place
String str = "property typename ttl=10 cval=30 test myprogram";
System.out.println( str.substring(str.indexOf("10"),str.indexOf("c")).trim() );
display 10.
> String str = "property typename ttl=10 cval=30 test myprogram";
>
> System.out.println(str.substring(str.indexOf("10"),str.indexOf("c")).trim() );
> display 10.
So do:System.out.println("10");
Now, without knowing the value in advance, you could try something like :int startIndex = input.indexOf("ttl=")+4; // index of the char that follows "ttl=" (i.e. the first char of the value)
int endIndex = input.indexOf(" ", startIndex); // index of the whitespace following the value
String value = input.substring(startIndex, endIndex);
s/he only ask to display 10... ^_^
s/he could also try to use StringTokenizer to separate ttl = 10.
> s/he only ask to display 10... ^_^
So my (first) solution still applies...
> s/he could also try to use StringTokenizer to
> separate ttl = 10.
I would probably go with a regex anyway:Matcher m = Pattern.compile("(?<=ttl=)(\\S+)").matcher(input);
if(m.find()) {
System.out.println("'"+m.group()+"'");
}
orMatcher m = Pattern.compile("(\\S+)=(\\S+)").matcher(input);
while(m.find()) {
System.out.println(m.group(1) + " = '" + m.group(2) + "'");
}
Hi,
The following program will extract the value (it will work for both +ve as well as -ve numbers). I used the regex patterns.
public class SplitStringTest {
private static final String STRING_SPLIT = "property typename ttl=10 cval=30 test myprogram";
/**
* @param args
* loganathank
* void
*
*/
public static void main(String[] args) {
String ttlRegExPattern = ".*ttl\\p{Blank}*=\\p{Blank}*-*\\p{Digit}+.*";
if(STRING_SPLIT.matches(ttlRegExPattern)) {
System.out.println("ttl found in the given string");
//ttl is found in the given string
//now extract the ttl.
String ttlExtractRegExPattern = ".*ttl\\p{Blank}*=\\p{Blank}*";
String[] splitString = STRING_SPLIT.split(ttlExtractRegExPattern);
//The first String will be blank
//So directly fetch the second string
String excludeTTL = splitString[1];
//After getting the second string, split the string based on blank spaces
//get the first element, that will give the value
String ttlValueExtractRegExPattern = "\\p{Blank}";
String[] splitString1 = excludeTTL.split(ttlValueExtractRegExPattern);
System.out.println(" The Value of ttl in the given String => " + splitString1[0]);
}
}
}
Please have a look into it.
Regards,
Loga
> Please have a look into it.
I'm not sure I should comment on your approach, but.. nah... nevermind.
> No probs !!! you can comment on my approach.
Basically, I would say it is kind of overkill.
- First you check the format of the input String in order to determine whether your process will work.
- Then, you split the input in order to extract the right-hand part of "ttl=".
- Then you split this part in order to keep the substring until first whitespace.
The Matcher class is aimed at finding occurences of patterns in a character sequence (e.g. a String.)
Here, we are trying to find the value (i.e. a sequence of non-whitespace characters) that follows the sequence "ttl=".
This can be translated into the regex: (?<=ttl=)\\S+
A single call to Matcher.find() will find it if it exists.
the one who post this topic should comment on both your approach... ^_^
> the one who post this topic should comment on both
> your approach... ^_^
The three step approach should be promoted whenever possible. The three step approach is the most simple and easily learned and is the one where everything can be done naturally. Dixit Maine State Candlepin Bowling Association
Any comment ?
I have another doubt. As mentioned in my first post in this thread, from the following string:
property typename ttl=10 cval=30 test myprogram
How to get string "test". I tried following but no use. say the above string is stored in a variable named "line"
line = line.substring(line.lastIndexOf("\\s")+1, line.length());
but i dont get any output. can you please help me.
The parameter of the lastIndexOf method is treated as a litteral, not a regex.
Therefore "\\s" doesn't mean whitespace, but simply backslash followed by 's'.
Note that it returns -1 if there is no occurence of the given substring.
The parameters to the substring methods are :
beginIndex - the beginning index, inclusive.
endIndex - the ending index, exclusive.
Knowing that, you can determine what the following statement does (Hint: not much) :line = line.substring(line.lastIndexOf("\\s")+1, line.length());
You have to determine the criteria that make "test" findable.
Assuming that it is the last but one token of your input String considering whitespace as token separator, you might use the String.split method (which splits the string around matches of the given regex):String[] tokens = input.split("\\s");
String lastButOne = tokens[tokens.length-2];
System.out.println("'" + lastButOne + "'");
Tim - Loga, looks like we finally have to split the string... ;-)
