I am trying to extract the 00 and 02 from the line below into Strings.
invokestatic:indexbyte1=00 indexbyte2=02
I am using this code, but it's not working correctly:
String parse = "invokestatic:indexbyte1=00 indexbyte2=02";
String first = parse.substring(check.indexOf("=") + 1);
String second= parse.substring(check.lastIndexOf("=") + 1);
This seems to work for the seconds string, but the first strings value is
00 indexbyte2=02
I want to catch just the two digits and not the rest of the string.
If you don't specify the second parameter in substring method it will result in a substring from the starting index to the end of string that's why you get "00 indexbyte2=02" for first.
Specify the last index only to extract two digits when you extract value for first
String first = parse.substring(check.indexOf("=") + 1, check.indexOf("=") + 3);
You can use a regex pattern with groups, like this:
public static void main(String[] args) {
String input = "invokestatic:indexbyte1=00 indexbyte2=02";
Pattern pattern = Pattern.compile(".*indexbyte1=(\\d*) indexbyte2=(\\d*)");
Matcher m = pattern.matcher(input);
if (m.matches()) {
System.out.println(m.group(1));
System.out.println(m.group(2));
}
}
Try this:
String first = parse.substring(check.indexOf("=") + 1, check.indexOf("=") + 3);
check.indexOf("=") + 3 will take the 02 and will be the endindex for the substring. Presently you are not specifying the endindex hence it is taking the indexbyte2=02 as well since substring does not know where to stop hence it parses down till the end.
String parse = "invokestatic:indexbyte1=00 indexbyte2=02";
String first = parse.substring(parse.indexOf("=") + 1,
parse.indexOf("=") + 3);
String second = parse.substring(parse.lastIndexOf("=") + 1);
System.out.println(first + ", " + second);
You could use Pattern, Matcher clases.
Matcher m = Pattern.compile("(?<==)\\d+").matcher(string);
while(m.find())
{
System.out.println(m.group());
}
substring also has an endIndex. See the docs: http://docs.oracle.com/javase/7/docs/api/java/lang/String.html#substring(int,%20int)
If the input has the basic form invokestatic:indexbyte1=00 indexbyte2=02 ... indexbyte99=99 you could use a regex:
Pattern p = Pattern.compile("indexbyte\\d+=([a-fA-F0-9]{2})");
Matcher m = p.matcher(input);
while( m.find() ) {
String idxByte = m.group(1);
//handle the byte here
}
This assumes that the identifier for those bytes is indexbyteN but this can be replaced with another identifier. Further this assumes the bytes are provided in hex, i.e. 2 hex characters (case insensitive here).
Related
Hi I get this String from server :
id_not="autoincrement"; id_obj="-"; id_tr="-"; id_pgo="-"; typ_not=""; tresc="Nie wystawił"; datetime="-"; lon="-"; lat="-";
I need to create a new String e.x String word and send a value which I get from String tresc="Nie wystawił"
Like #Jan suggest in comment you can use regex for example :
String str = "id_not=\"autoincrement\"; id_obj=\"-\"; id_tr=\"-\"; id_pgo=\"-\"; typ_not=\"\"; tresc=\"Nie wystawił\"; datetime=\"-\"; lon=\"-\"; lat=\"-\";";
Pattern p = Pattern.compile("tresc(.*?);");
Matcher m = p.matcher(str);
if (m.find()) {
System.out.println(m.group());
}
Output
tresc="Nie wystawił";
If you want to get only the value of tresc you can use :
Pattern p = Pattern.compile("tresc=\"(.*?)\";");
Matcher m = p.matcher(str);
if (m.find()) {
System.out.println(m.group(1));
}
Output
Nie wystawił
Something along the lines of
Pattern p = Pattern.compile("tresc=\"([^\"]+)\");
Matcher m = p.matcher(stringFromServer);
if(m.find()) {
String whatYouWereLookingfor = m.group(1);
}
should to the trick. JSON parsing might be much better in the long run if you need additional values
Your question is unclear but i think you get a string from server and from that string you want the string/value for tresc. You can first search for tresc in the string you get. like:
serverString.substring(serverString.indexOf("tresc") + x , serverString.length());
Here replace x with 'how much further you want to pick characters.
Read on substring and delimiters
As values are separated by semicolon so annother solution could be:
int delimiter = serverstring.indexOf(";");
//in string thus giving you the index of where it is in the string
// Now delimiter can be -1, if lets say the string had no ";" at all in it i.e. no ";" is not found.
//check and account for it.
if (delimiter != -1)
String subString= serverstring.substring(5 , iend);
Here 5 means tresc is on number five in string, so it will five you tresc part.
You can then use it anyway you want.
I have the following string;
String s = "Hellow world,how are you?\"The other day, where where you?\"";
And I want to replace the , but only the one that is inside the quotation mark \"The other day, where where you?\".
Is it possible with regex?
String s = "Hellow world,how are you?\"The other day, where where you?\"";
Pattern pattern = Pattern.compile("\"(.*?)\"");
Matcher matcher = pattern.matcher(s);
while (matcher.find()) {
s = s.substring(0, matcher.start()) + matcher.group().replace(',','X') +
s.substring(matcher.end(), s.length());
}
If there are more then two quotes this splits the text into in quote/out of quote and only processes inside quotes. However if there are odd number of quotes (unmatched quotes), the last quote is ignored.
If you are sure this is always the last "," you can do that
String s = "Hellow world,how are you?\"The other day, where where you?\"";
int index = s.lastIndexOf(",");
if( index >= 0 )
s = new StringBuilder(s).replace(index , index + 1,"X").toString();
System.out.println(s);
Hope it helps.
I have a long string let's say
I like this #computer and I want to buy it from #XXXMall.
I know the regular expression pattern is
Pattern tagMatcher = Pattern.compile("[#]+[A-Za-z0-9-_]+\\b");
Now i want to get all the hashtags in an array. How can i use this expression to get array of all hash tags from string something like
ArrayList hashtags = getArray(pattern, str)
You can write like?
private static List<String> getArray(Pattern tagMatcher, String str) {
Matcher m = tagMatcher.matcher(str);
List<String> l = new ArrayList<String>();
while(m.find()) {
String s = m.group(); //will give you "#computer"
s = s.substring(1); // will give you just "computer"
l.add(s);
}
return l;
}
Also you can use \\w- instead of A-Za-z0-9-_ making the regex [#]+[\\w]+\\b
This link would surely be helpful for achieving what you want.
It says:
The find() method searches for occurrences of the regular expressions
in the text passed to the Pattern.matcher(text) method, when the
Matcher was created. If multiple matches can be found in the text, the
find() method will find the first, and then for each subsequent call
to find() it will move to the next match.
The methods start() and end() will give the indexes into the text
where the found match starts and ends.
Example:
String text =
"This is the text which is to be searched " +
"for occurrences of the word 'is'.";
String patternString = "is";
Pattern pattern = Pattern.compile(patternString);
Matcher matcher = pattern.matcher(text);
int count = 0;
while(matcher.find()) {
count++;
System.out.println("found: " + count + " : "
+ matcher.start() + " - " + matcher.end());
}
You got the hint now.
Here is one way, using Matcher
Pattern tagMatcher = Pattern.compile("#+[-\\w]+\\b");
Matcher m = tagMatcher.matcher(stringToMatch);
ArrayList<String> hashtags = new ArrayList<>();
while (m.find()) {
hashtags.add(m.group());
}
I took the liberty of simplifying your regex. # does not need to be in a character class. [A-Za-z0-9_] is the same as \w, so [A-Za-z0-9-_] is the same as [-\w]
You can use :
String val="I like this #computer and I want to buy it from #XXXMall.";
String REGEX = "(?<=#)[A-Za-z0-9-_]+";
List<String> list = new ArrayList<String>();
Pattern pattern = Pattern.compile(REGEX);
Matcher matcher = pattern.matcher(val);
while(matcher.find()){
list.add(matcher.group());
}
(?<=#) Positive Lookbehind - Assert that the character # literally be matched.
you can use the following code for getting the names
String saa = "#{akka}nikhil#{kumar}aaaaa";
Pattern regex = Pattern.compile("#\\{(.*?)\\}");
Matcher m = regex.matcher(saa);
while(m.find()) {
String s = m.group(1);
System.out.println(s);
}
It will print
akka
kumar
This is the string that I have:
KLAS 282356Z 32010KT 10SM FEW090 10/M13 A2997 RMK AO2 SLP145 T01001128 10100 20072 51007
This is a weather report. I need to extract the following numbers from the report: 10/M13. It is temperature and dewpoint, where M means minus. So, the place in the String may differ and the temperature may be presented as M10/M13 or 10/13 or M10/13.
I have done the following code:
public String getTemperature (String metarIn){
Pattern regex = Pattern.compile(".*(\\d+)\\D+(\\d+)");
Matcher matcher = regex.matcher(metarIn);
if (matcher.matches() && matcher.groupCount() == 1) {
temperature = matcher.group(1);
System.out.println(temperature);
}
return temperature;
}
Obviously, the regex is wrong, since the method always returns null. I have tried tens of variations but to no avail. Thanks a lot if someone can help!
This will extract the String you seek, and it's only one line of code:
String tempAndDP = input.replaceAll(".*(?<![M\\d])(M?\\d+/M?\\d+).*", "$1");
Here's some test code:
public static void main(String[] args) throws Exception {
String input = "KLAS 282356Z 32010KT 10SM FEW090 M01/M13 A2997 RMK AO2 SLP145 T01001128 10100 20072 51007";
String tempAndDP = input.replaceAll(".*(?<![M\\d])(M?\\d+/M?\\d+).*", "$1");
System.out.println(tempAndDP);
}
Output:
M01/M13
The regex should look like:
M?\d+/M?\d+
For Java this will look like:
"M?\\d+/M?\\d+"
You might want to add a check for white space on the front and end:
"\\sM?\\d+/M?\\d+\\s"
But this will depend on where you think you are going to find the pattern, as it will not be matched if it is at the end of the string, so instead we should use:
"(^|\\s)M?\\d+/M?\\d+($|\\s)"
This specifies that if there isn't any whitespace at the end or front we must match the end of the string or the start of the string instead.
Example code used to test:
Pattern p = Pattern.compile("(^|\\s)M?\\d+/M?\\d+($|\\s)");
String test = "gibberish M130/13 here";
Matcher m = p.matcher(test);
if (m.find())
System.out.println(m.group().trim());
This returns: M130/13
Try:
Pattern regex = Pattern.compile(".*\\sM?(\\d+)/M?(\\d+)\\s.*");
Matcher matcher = regex.matcher(metarIn);
if (matcher.matches() && matcher.groupCount() == 2) {
temperature = matcher.group(1);
System.out.println(temperature);
}
Alternative for regex.
Some times a regex is not the only solution. It seems that in you case, you must get the 6th block of text. Each block is separated by a space character. So, what you need to do is count the blocks.
Considering that each block of text does NOT HAVE fixed length
Example:
String s = "KLAS 282356Z 32010KT 10SM FEW090 10/M13 A2997 RMK AO2 SLP145 T01001128 10100 20072 51007";
int spaces = 5;
int begin = 0;
while(spaces-- > 0){
begin = s.indexOf(' ', begin)+1;
}
int end = s.indexOf(' ', begin+1);
String result = s.substring(begin, end);
System.out.println(result);
Considering that each block of text does HAVE fixed length
String s = "KLAS 282356Z 32010KT 10SM FEW090 10/M13 A2997 RMK AO2 SLP145 T01001128 10100 20072 51007";
String result = s.substring(33, s.indexOf(' ', 33));
System.out.println(result);
Prettier alternative, as pointed by Adrian:
String result = rawString.split(" ")[5];
Note that split acctualy receives a regex pattern as parameter
I want to parse a line from a CSV(comma separated) file, something like this:
Bosh,Mark,mark#gmail.com,"3, Institute","83, 1, 2",1,21
I have to parse the file, and instead of the commas between the apostrophes I wanna have ';', like this:
Bosh,Mark,mark#gmail.com,"3; Institute","83; 1; 2",1,21
I use the following Java code but it doesn't parse it well:
Pattern regex = Pattern.compile("(\"[^\\]]*\")");
Matcher matcher = regex.matcher(line);
if (matcher.find()) {
String replacedMatch = matcher.group();
String gr1 = matcher.group(1);
gr1.trim();
replacedMatch = replacedMatch.replace(",", ";");
line = line.replace(matcher.group(), replacedMatch);
}
the output is:
Bosh,Mark,mark#gmail.com,"3; Institute";"83; 1; 2",1,21
anyone have any idea how to fix this?
This is my solution to replace , inside quote to ;. It assumes that if " were to appear in a quoted string, then it is escaped by another ". This property ensures that counting from start to the current character, if the number of quotes " is odd, then that character is inside a quoted string.
// Test string, with the tricky case """", which resolves to
// a length 1 string of single quote "
String line = "Bosh,\"\"\"\",mark#gmail.com,\"3, Institute\",\"83, 1, 2\",1,21";
Pattern pattern = Pattern.compile("\"[^\"]*\"");
Matcher matcher = pattern.matcher(line);
int start = 0;
StringBuilder output = new StringBuilder();
while (matcher.find()) {
// System.out.println(m.group() + "\n " + m.start() + " " + m.end());
output
.append(line.substring(start, matcher.start())) // Append unrelated contents
.append(matcher.group().replaceAll(",", ";")); // Append replaced string
start = matcher.end();
}
output.append(line.substring(start)); // Append the rest of unrelated contents
// System.out.println(output);
Although I cannot find any case that will fail the method of replace the matched group like you did in line = line.replace(matcher.group(), replacedMatch);, I feel safer to rebuild the string from scratch.
Here's a way:
import java.util.regex.*;
class Main {
public static void main(String[] args) {
String in = "Bosh,Mark,mark#gmail.com,\"3, \"\" Institute\",\"83, 1, 2\",1,21";
String regex = "[^,\"\r\n]+|\"(\"\"|[^\"])*\"";
Matcher matcher = Pattern.compile(regex).matcher(in);
StringBuilder out = new StringBuilder();
while(matcher.find()) {
out.append(matcher.group().replace(',', ';')).append(',');
}
out.deleteCharAt(out.length() - 1);
System.out.println(in + "\n" + out);
}
}
which will print:
Bosh,Mark,mark#gmail.com,"3, "" Institute","83, 1, 2",1,21
Bosh,Mark,mark#gmail.com,"3; "" Institute","83; 1; 2",1,21
Tested on Ideone: http://ideone.com/fCgh7
Here is the what you need
String line = "Bosh,Mark,mark#gmail.com,\"3, Institute\",\"83, 1, 2\",1,21";
Pattern regex = Pattern.compile("(\"[^\"]*\")");
Matcher matcher = regex.matcher(line);
while(matcher.find()){
String replacedMatch = matcher.group();
String gr1 = matcher.group(1);
gr1.trim();
replacedMatch = replacedMatch.replace(",", ";");
line = line.replace(matcher.group(), replacedMatch);
}
line will have value you needed.
Have you tried to make the RegExp lazy?
Another idea: inside the [] you should use a " too. If you do that, you should have the expected output with global flag set.
Your regex is faulty. Why would you want to make sure there are no ] within the "..." expression? You'd rather make the regex reluctant (default is eager, which means it catches as much as it can).
"(\"[^\\]]*\")"
should be
"(\"[^\"]*\")"
But nhadtdh is right, you should use a proper CSV library to parse it and replace , to ; in the values the parser returns.
I'm sure you'll find a parser when googling "Java CSV parser".
Shouldn't your regex be ("[^"]*") instead? In other words, your first line should be:
Pattern regex = Pattern.compile("(\"[^\"]*\")");
Of course, this is assuming you can't have quotes in the quoted values of your input line.