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Extract specific string by regular expression in Android [duplicate]

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Select part of line in regular expression
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I have this strings: https://regex101.com/r/7Er0Ch/6
I want put all my http://esupb.tabriz.ir:808x/srvSC.svc into array list.So in order to i used matcher like blow:
String regx= "#\\d+#";
Pattern pattern = Pattern.compile(regx);
Matcher matcher = pattern.matcher(url);
String[] metadata = new String[4];
while (matcher.find()) {
metadata[0] = matcher.group(1);
metadata[1] = matcher.group(2);
metadata[2] = matcher.group(3);
metadata[3] = matcher.group(4);
}
but I got not appropriate result. What is my mistake?

From requirement your regex will be
"(#\d+#)(http[^#]*svc)(#\d+#)"
group(0): (#\\d+#)(http[^#]*svc)(#\\d+#)
group(1): (#\\d+#)
group(2): (http[^#]*svc)
group(3): (#\\d+#)
Change your code to
List<String> urls = new ArrayList<>();
String url =
"#1#http://test.com:8080/srv.svc#1# " +
"#2#http://test.com:8081/srv.svc#2# " +
"#3#http://test.com:8082/srv.svc#3# " +
"#4#http://test.com:8083/srv.svc#4# " +
"#5#http://test.com:8084/srv.svc#5# ";
String regx = "(#\\d+#)(http[^#]*svc)(#\\d+#)";
Pattern pattern = Pattern.compile(regx);
Matcher matcher = pattern.matcher(url);
int from = 0;
while (matcher.find(from)) {
urls.add(matcher.group(2));
from = matcher.start() + 1;
}

You regex #\\d+#matches # followed by matching one or more times a digit and then another # .It does not use capturing groups.
For your example data you could remove that match from the string giving you the desired result leaving out matching any pattern for the string that is left. It could also match inside the string instead of only at the start and the end.
To match your example string(s) like http://esupb.tabriz.ir:808x/srvSC.svc you might use your regex to match the start and the end, and capture in a group what is in between.
^#\d+#(https?://test.ir:808\d/srvSC\.svc)#\d+#$
In Java
^#\\d+#(https?://test.ir:808\\d/srvSC\\.svc)#\\d+#$
Regex demo
Demo Java
Explanation
^ Assert the start of the string
#\d+# Match #, one or more times a digit and another #
( Start capturing group
https?://test.ir:808\d Match the start of the url with an optional s s? and a digit after 808. Use \d+ to match one or more digits.
/srvSC\.svc Match /srvSC.svc
#\d+# Match #, one or more times a digit and another #
) Close caputring group
$ Assert the end of the string

What regular expression not work in android

Hi i want a regular expression for a-z A-Z 0_9آ-ی( Persian)
String regex ="^(?:([a-z A-Z 0-9])||([آ-ی]))+$";
i do not want allow special characters like !##$%^&*()?><+- .
and my commllet code is :
public int check_first_last_name(String name){
String regex ="^(?:([a-zA-Z0-9])||([آ-ی]))+$";
Matcher matcher = Pattern.compile( regex ).matcher(name);
if (!matcher.find( )){
return 2;
}
return 1;
}

this regex accept every words in English ,Persian and Arabic if you want allow space change end of regex to "...0-9 "
if(name.matches("^([\\u0600-\\u06FF\\uFB8A\\u067E\\u0686\\u06AFa-zA-Z0-9])+$")){
//do what you want
}
please check in code you must have 2 backslash before every unicode not 4!
and if you want accept all tabs newline and space you can use this regex:
"^([\\u0600-\\u06FF\\uFB8A\\u067E\\u0686\\u06AFa-zA-Z0-9]+\\s?)+$"

Is there any method to add some string to what found by regex in Android Studio?(ctrl+shift+R)

Basically, what I am trying to do is add double quote to the heads and tails of the numbers
String a = 1;
String b = 2;
String c = 3;
to
String a = "1";
String b = "2";
String c = "3";
So, I use [1-9] to find all numbers. Then, all of a sudden, it comes to me that I don't know how to get the values which regex found, like don't know what to set between double quotes.
Hence, I am wondering if it's possible.

You should use \d+ instead of [1-9] or at the very least [0-9]+ to include the 0
The reason why you need the + is because your regex would not find 10 or any digits that has more than 1 digit. You can reference the groups that you have found by using $1 (first group) $2 (second group) and so on. So you could do "$1" as your substitution and (\d+) as your search although you might want to use a better regex ie:
=\s*(\d)+;
replace to
= "$1";
See https://regex101.com/r/SaT6nK/1

how to get text with using substring

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).

Regular expression with hebrew

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"));