I have text like:
לשלום קוראים לי משהmy test is עלות 39.40, כל מיני data 1.1.2015 ויש גם data 123456 מידע
This text have Hebrew and English characters, I need to eliminate all except the 6 digit number (may be 5, this num: 123456).
Can you help me with regular expression for this?
Tried:
String patternS = "[אבגדהוזחטיכךלמםנןסעפףצץקרשתa-fA-F0-9]{5,10}.*";
Pattern pattern = Pattern.compile(patternString);
With no success
To match everything except the number use:
\d+(?:[^\d]\d+)+|[\p{L}\p{M}\p{Z}\p{P}\p{S}\p{C}]+
String resultString = subjectString.replaceAll("\\d+(?:[^\\d]\\d+)+|[\\p{L}\\p{M}\\p{Z}\\p{P}\\p{S}\\p{C}]+", "");
This will give you every 6 didgit combination in your string.
(\d{6,6})
We can't give you a more detailled regex since we do now know the pattern of those strings.
In case there is always the "data " prefix you can also use this to make the pattern more accurate:
data (\d{6,6})
Try something like this:
String patternS = "(\d{5,6})";
Pattern pattern = Pattern.compile(patternS);
Matcher m = pattern.matcher(yourText);
int number = Integer.parseInt(m.group(1));
where yourText is the Hebrew/English text you want to match.
This would work for this specific example.
String s = " לשלום קוראים לי מש my test is עלות 39.40, כל מיני data 1.1.2015 ויש גם data 123456 מידע1234";
System.out.println(s.replaceAll(".*\\b(\\d{5,6})\\b.*", "$1"));
Related
What I'm trying to do:
if a String is
String a = "Love is rare"; Or
String a = "l o v e is rare"; Or
String a = "LO VE is rare";
Then,
I want the word "love" to be replaced by "hate" ignoring the space and case
My code:
a=a.toLowerCase().replace("love","hate").replace("\\s+","");
But the problem is that it removes all the spaces and change all the words to lower case. I just want to check if there is a word love by ignoring all the spaces and cases and if the word is there then replace it with some other word.
Please help!
Thank you.
Try it like this using an optional whitespace ? and the case insensitive flag. If you want to match zero or more whitespaces, you could use a whitespace and *
l ?o ?v ?e
String regex = "l ?o ?v ?e";
String[] lines = {"Love is rare", "l o v e is rare", "LO VE is rare"};
String subst = "hate";
Pattern pattern = Pattern.compile(regex, Pattern.CASE_INSENSITIVE);
for (String line: lines) {
Matcher matcher = pattern.matcher(line);
String result = matcher.replaceAll(subst);
System.out.println(result);
}
That would give you:
hate is rare
hate is rare
hate is rare
Demo Java
Thanks for suggesting the term regex, Studied thoroughly and come up with this one line function which will replace "love" by "hate" ignoring the spaces and cases:
String a = "Lo ve is rare.";
Code:
a = a.replaceAll("(?i)l(\\s*)o(\\s*)v(\\s*)e","hate");
TextView.setText(a);
This would give:
hate is rare.
So I have the following string:
String text = "\t\t\torder #168\n\t\t\tpaid\n\t\t\tview 4 items\n\t\t\tpicked up\n\t\t\tcomplete pickup\n\t\t\t2 stops";
How do I parse this string so that I always get the 2 in front of stops? I have tried the following, but it always returns 2 stops.
String substr = "complete pickup";
String numberOfStops = text.substring(text.indexOf(substr) + substr.length());
numberOfStops = numberOfStops.replaceAll("^\\s+","").replaceAll("\\s+$","");
The short way:
numberOfStops = numberOfStops.replaceAll("^\\s+","").replaceAll("\\s+$","").replace("stops","");
The flexible way is using Regex, and Pattern and Match classes. Let me know if you need it
I try to get only this part "9916-4203" in "Region Code:9916-4203 " in android. How can I do this?
I tried below code, I used substring method but it doesn't work:
firstNumber = Integer.parseInt(message.substring(11, 19));
If you know that string contains "Region Code:" couldn't you do a replace?
message = message.replace("Region Code:", "");
Assumed that you have only one phone number in your String, the following will remove any non-digit characters and parse the resulting number:
public static int getNumber(String num){
String tmp = "";
for(int i=0;i<num.length();i++){
if(Character.isDigit(num.charAt(i)))
tmp += num.charAt(i);
}
return Integer.parseInt(tmp);
}
Output in your case: 99164203
And as already mentioned, you won't be able to parse any String to Integer in case there are any non-digit characters
Im going to guess that what you want to extract is the full region code text minus the title. So maybe using regex would be a good simple fit for you?
String myString = "Region Code:9916-4203";
String match = "";
String pattern = "\:(.*)";
Pattern regEx = Pattern.compile(pattern);
Matcher m = regEx.matcher(myString);
// Find instance of pattern matches
Matcher m = regEx.matcher(myString);
if (m.find()) {
match = m.group(0);
}
Variable match will contain "9916-4203"
This should work for you.
Java code sourced from http://android-elements.blogspot.in/2011/04/regular-expressions-in-android.html
In Java the substring() method works with the first parameter being inclusive and the second parameter being exclusive. Meaning "Hello".substring(0, 2); will result in the string He.
In addition to excluding the parsing of something that isn't a number like #Opiatefuchs mentioned, your substring method should instead be message.substring(12, 21).
I am currently working on an android project. I would like to know if a string contains the '\r' or '\n' as the last character. How to do the pattern match?
Try this.
String string = "stackoverflow";
lastchar = string.substring(string.length() - 1);
It will give you the result "w".
try
String patterntomatch ="^[_A-Za-z0-9-]*(\r\n)$";
Pattern pattern=Pattern.compile(patterntomatch);
Matcher matcher=pattern.matcher(matchfromedittext);
boolean matcher.matches();
String last = your_string.substring(Math.max(your_string.length() - 2, 0));
//It will give you the last 2 characters of the string. If the string is null
//or has less than 2 characters, then it gives you the original string.
if(last.equals("\r") || last.equals("\n")){
//String's last two characters are either \n or \r. Now do something.
}
In my Android Application when I am reading the particular data from NFC chip it's giving garbage values as show follows which is printed on Log
����������������
I used following line to remove garbage value
str.replaceAll("[^\\p{ASCII}]", "")
but it is not working.
Please provide me solution.
That is because � is not an ASCII character. It is a unicode character with (int) � returning 65533.
And your code str.replaceAll("[^\\p{ASCII}]", "") works perfectly fine.
scala> val str ="����������������"
str: String = ����������������
scala> str.replaceAll("[^\\p{ASCII}]", "")
res8: String = ""
You need to show more code and explain what exactly you are trying to do.
Better to retrieve the data in the form of UTF-8 format then it helps. try it out.
or convert the string to UTF-8 format
i.e, String _data=new String(str.getBytes(),"UTF-8");
it returns the data in UTF-8 format
One solution using this method .replaceAll("[^\\x00-\\x7F]", "")
String str = "jorgesys���������������� was here!";
str = str.replaceAll("[^\\x00-\\x7F]", ""));
so the result of str is:
jorgesys was here!