PARSE NUMBERS FROM STRING
I have a string that I read that contains two numbers as strings for directories. eg. 0-63, a forward slash, and 0 -127 (64 rows, 128 cols). Is there a way to check the string for the possible patterns. I am trying
Pattern p = Pattern.compile("^0-77"+ "\\" + "0-177");
Matcher m = p.matcher(infile);
This does not work. It throws an Illegal octal escape sequence near index 7
^0-77\0-177
^
Would appreciate if anyone knows how to do this. the string I want to return is 0/0 thru 63/127.
[531 byte] By [
seadazea] at [2007-11-27 4:27:30]
If it's "a forward slash" why the "\\" ?
it comes in as a backward slash from the file I am reading. I convert it to a forward slash later.
In a java string, you have to escape a \ by writing \\. The String "\\" gets turned into "\" when compiled. When the regex reads that patter "\", it interprets it as an escape character. You have to escape the \ in the regex also. So use "\\\\" (4 backslashes) to end up with a regex of \\
Pattern p = Pattern.compile("^0-77\\\\0-177");
Here's a regex for your problem.
Pattern p = Pattern.compile("^(([0-9])|([0-5][0-9])|(6[0-3]))/(([0-9])|([1-9][0-9])|(1[01][0-9])|(12[0-7]))$");
Matcher m = null;
String[] testCases = {"0/0", "5/10", "10/25", "59/107", "63/127", "64/133", "bogus/", "105/106"};
for( String s : testCases ) {
m = p.matcher( s );
System.out.println( s + " matches? " + m.matches() );
}
using this string C:\\Data\\out\\19\\116\\cities202803.bin and this regex Pattern.compile("^(([0-9])|([0-5][0-9])|(6[0-3]))\\\\(([0-9])|([1-9][0-9])|(1[0-1][0-9])|(12[0-7]))$");it does not find the 19\116. Any idea why?
It works for me. Did you use the String "19\\116"? Here's the exact code I tested:
Pattern p = Pattern.compile("^(([0-9])|([0-5][0-9])|(6[0-3]))\\\\(([0-9])|([1-9][0-9])|(1[0-1][0-9])|(12[0-7]))$");
Matcher m = null;
String[] testCases = {"0\\0", "5\\10", "10\\25", "59\\107", "63\\127", "64\\133", "bogus\\", "105\\106", "19\\116"};
for( String s : testCases ) {
m = p.matcher( s );
System.out.println( s + " matches? " + m.matches() );
}
Output:
0\0 matches? true
5\10 matches? true
10\25 matches? true
59\107 matches? true
63\127 matches? true
64\133 matches? false
bogus\ matches? false
105\106 matches? false
19\116 matches? true
yes and it worked on that but not on my string......just trying to figure it out.
k = C:\\Data\\out\\19\\116\\cities202803.bin
Matcher mm = pp.matcher(infile);
if (mm.find())
{
start = mm.start();
end = mm.end();
}
String result = k.substring(start, end);
The metacharacters '^' and '$' at the ends of the regex anchor it to the beginning and end of the target text. That's why the regex works with hunter's test data, but not with real, absolute file paths. Here's one way to adapt the regex: Pattern p = Pattern.compile("\\\\(\\d+)\\\\(\\d+)\\\\");
Matcher m = p.matcher(inFile);
if (m.find())
{
// it's good
}
I used (\\d+) as a placeholder; the important changes are the replacement of the anchors with backslashes, and the use of the find() method instead of matches().
Listen to uncle_alice, (he|she) is the (king|queen) of regex around here. ;p
thanks!Tried something like that but just didn't have it right!
Listen to uncle_alice, (he|she) is the (king|queen) of regex around here. ;p^^ ^^^^ ^_^
