This question already has answers here:
Java Regex Capturing Groups
(4 answers)
Closed 6 years ago.
I need to get the string between by_ and _on.
So far I have this, but don't understand how to truncate the actual "string delimiters":
by_(.*)_on
Sample input:
Files_by_wesasegeaazedude_on_January_26.jpg
Current Match:
by_wesasegeaazedude_on
Needed Match:
wesasegeaazedude
Your expression is good*. All you need to do is extracting the content of the first capturing group:
Pattern regex = Pattern.compile("by_(.*)_on");
String str = "Files_by_wesasegeaazedude_on_January_26.jpg";
Matcher m = regex.matcher(str);
if (m.find()) {
String res = m.group(1);
}
Demo.
* Well, almost good. If you expect inputs with multiple file names on the same line, you may want to consider using reluctant qualifier, i.e. by_(.*?)_on
I would do this without regular expressions.
int start = str.indexOf("by_");
int end = str.indexOf("_on", start + 1); // or lastIndexOf("_on"), for greedy match.
assert start > 0 && end > start;
String part = str.substring(start + 3, end);
You can simply use positive lookarounds:
String regex = "(?<=by_).*(?=_on)";
What this regex does is:
match anything: .*
that is preceded by by_: (?<=by_)
and followed by _on: (?=_on)
Related
This question already has answers here:
How to extract numbers from a string and get an array of ints?
(13 answers)
Closed 1 year ago.
For example, if I have these strings, is there any way I can get 123 of all these strings, or 777 or 888?
https://www.example.com/any/123/ and
https://www.example.com/any/777/123/ and
https://www.example.com/any/777/123/888
What I mean is how to match the first or second or the third last number in the string.
You can use capture groups to solve this as
val strList = listOf("https://www.example.com/any/777/123/888", "https://www.example.com/any/123/", "https://www.example.com/any/777/123/")
val intList = mutableListOf<Int>()
val regex = Regex("/?(\\d+)")
strList.forEach { str ->
regex.findAll(str).forEach {
intList.add(it.groupValues[1].toInt())
}
}
Assuming the digits all follow a slash and nothing intervenes,
(?<=/)\d+(?=/\d+){0}$ parses the last number
(?<=/)\d+(?=/\d+){1}$ parses the second to last number
(?<=/)\d+(?=/\d+){2}$ parses the third to last,
etc.
With Java, You can make use of the Pattern and Matcher class from the java.util.regex package.
e.g for your case above, you want to match integers - use \d Predefined character class to match digits.
String str = "https://www.example.com/any/777/123/";
Pattern pattern = Pattern.compile("\\d+");
Matcher matcher = pattern.matcher(str);
for(; matcher.find(); System.out.println(matcher.group()));
In the above you loop through the String finding matches, and printing each subsequent found match.
This question already has answers here:
Java Find word in a String
(5 answers)
Find word in random string
(3 answers)
How to find index of whole word in string in java
(1 answer)
Regex to find a specific word in a string in java
(3 answers)
Closed 5 years ago.
I have a message that is of the format:
FixedWord1 variable1 FixedWord2 on FixedWord3 variable2/variable3, variable4 = variable5
I need to extract only variable3 from the above message.
Here is what I tried:
String example = "FixedWord1 variable1 FixedWord2 on FixedWord3 variable2/variable3, variable4 = variable5";
I know that the length of FixedWord3 is 6. So,
example.substring(example.lastIndexOf("FixedWord3") + 6 , example.lastIndexOf(",")); //To get {variable2}/{variable3}
And then,
String requiredString[] = example.split("/", 2); //requiredString[1] would contain {variable3} even if it contains /
Can you suggest a more efficient solution to this problem?
EDIT:
This regex should do the trick.
Pattern pattern = Pattern.compile(".+(Device).+[/]([A-Z].+)[,][ ].+");
Matcher matcher = pattern.matcher(yourstring);
if(matcher.matches())
System.out.println(matcher.group(2));
Assumption to make this work:
Variable2 has no slash '/' followed by upper case letter
Variable3 has no comma ',' followed by space ' '
Since you know that variable2 cannot contain a "/" and you know the length of FixedWord3 then how about this?
String deviceName = example.substring(example.lastIndexOf("Device") + 6, example.lastIndexOf(","));
String lastPart = deviceName.substring(deviceName.indexOf("/") + 1);
System.out.println(deviceName);
System.out.println(lastPart);
Prints:
SJ-ME3600X-185/GigabitEthernet0/4
GigabitEthernet0/4
Regex for the help.
One possible approach is catching the match that's after "{variable2}":
{variable2}\/{([^}]+)}
Then you can use Matcher and Pattern and maybe other tools to make it work in Java.
See here for explanation and live demo.
Using Regex patterns are the efficient way to extract the word from a message in java.
String s = "FixedWord1 {variable1} FixedWord2 on FixedWord3 {variable2}/{variable3}, {variable4} = {variable5}";
Pattern p = Pattern.compile("/(\\{([^}]*)\\})");
Matcher m = p.matcher(s);
while (m.find()) {
System.out.println(m.group(1));
}
Output {variable3}
This question already has answers here:
Regex to replace all string literals in a Java file
(4 answers)
Closed 8 years ago.
The sample source code to match is
String string="welcome";
String k="a\"welcome";
I am using "(\"[^(\")]*\")" regex in java.
But this extracts
0:"welcome"
0:"a\"
Expected output is
0:"welcome"
0:"a\"welcome"
What change should i make in regex to get the expected output ?
Java source :
private static String pattern1="(\"[^(\")]*\")";
public void getStrings(){
Pattern r = Pattern.compile(pattern1);
Matcher m = r.matcher("String string=\"welcome\";\n" +
"String k=\"a\\\"welcome\";");
while(m.find()){
System.out.println("0:"+m.group(0));
}
}
Just use lookahead and lookbehind in your regex,,
(?<==)(".*?")(?=;)
Get the value from group index 1.
DEMO
Pattern r = Pattern.compile("(?<==)(\".*?\")(?=;)");
Matcher m = r.matcher("String string=\"welcome\";\n" +
"String k=\"a\\\"welcome\";");
while(m.find()){
System.out.println("0:"+m.group(1));
}
Output:
0:"welcome"
0:"a\"welcome"
OR
Use the greediness of *,
Pattern r = Pattern.compile("(\".*\")");
OR
It skips the double quotes which are preceded by a backslash,
Pattern r = Pattern.compile("(\\\".*?(?<=[^\\\\])\\\")");
Why do you even bother with variable assignment. You know that everything within "" is a string.
"(.+)"\s*; should do it just fine.
This question already has an answer here:
SCJP6 regex issue
(1 answer)
Closed 7 years ago.
Sample code
Pattern p = Pattern.compile("\\d?");
Matcher m = p.matcher("ab34ef");
boolean b = false;
while (m.find())
{
System.out.print(m.start());// + m.group());
}
Answer: 012456
But string total length is 6. So How m.start will give 6 in the output, as index starts
from 0.
\d? matches zero or one character, so it starts beyond the last character of the string as well, as a zero-width match.
Note that your output is not in fact attained by \d?, but by \d*. You should change either one or the other to make the question self-consistent.
\d? matches zero or one digit, which matches every digit, but also matches every character boundary.
Try matching at least one digit:
Pattern p = Pattern.compile("\\d+");
This question already has answers here:
Closed 10 years ago.
Possible Duplicate:
Regular expression for excluding special characters
How would I check if a String contains a symbol? let's say I have this String
"SUGAR :::: SUGAR"
I would want to check if that string contains = the following Symbols
":,?,!##$%^&*()";
I tried this
Pattern p = Pattern.compile("[?,!,#,$,%,^,&,*,(,)]");
Matcher m = p.matcher("?");
boolean b = m.matches();
System.out.println(b);
But what if the text contains multiple occurrences of that symbol
Try this:
Pattern pat = Pattern.compile("[:?!##$%^&*()]");
String str = "SUGAR :::: SUGAR";
Matcher m = pat.matcher(str);
if (m.find()) {
System.out.println("string contains pattern");
}
The above will check if any part of the string contains at least one occurrence of any of the symbols in the pattern (no need to separate them with ,).
guava may be?
CharMatcher matcher = CharMatcher.anyOf(":,?,!##$%^&*()");
boolean result = matcher.matchesAnyOf("SUGAR :: SUGAR");
System.out.println(result);
Just simply match against a set containing these characters: [:,?!##$%^&*()]
If you want to check against the symbol -, position it to the end of the set: [...-]
There's no need to separate elements in a set by comma.
If you want to check for ANY non alphanumeric use \W instead: [\W_] (because \W does not match _)