I have an input string that will follow the pattern /user/<id>?name=<name>, where <id> is alphanumeric but must start with a letter, and <name> is a letter-only string that can have multiple spaces. Some examples of matches would be:
/user/ad?name=a a
/user/one111?name=one ONE oNe
/user/hello?name=world
I came up with the following regex:
String regex = "/user/[a-zA-Z]+\\w*\\?name=[a-zA-Z\\s]+";
All of the above examples match the regex, but it only looks at the first word in <name>. Shouldn't the sequence \s allow me to have white spaces?
The code that I made to test what it is doing is:
String regex = "/user/[a-zA-Z]+\\w*\\?name=[a-zA-Z\\s]+";
// Check to see that input matches pattern
if(Pattern.matches(regex, str) == true){
str = str.replaceFirst("/user/", "");
str = str.replaceFirst("name=", "");
String[] tokens = str.split("\\?");
System.out.println("size = " + tokens.length);
System.out.println("tokens[0] = " + tokens[0]);
System.out.println("tokens[1] = " + tokens[1]);
} else
System.out.println("Didn't match.");
So for example, one test might look like:
/user/myID123?name=firstName LastName
size = 2
tokens[0] = myID123
tokens[1] = firstName
whereas the desired output would be
tokens[1] = firstName LastName
How can I change my regex to do this?
Not sure what you think is the problem in your code. tokens[1] will indeed contain firstName LastName in your example.
Here's an ideone.com demo showing this.
However, have you considered using capturing groups for the id and the name.
If you write it like
String regex = "/user/(\\w+)\\?name=([a-zA-Z\\s]+)";
Matcher m = Pattern.compile(regex).matcher(input);
you can get hold of myID123 and firstName LastName through m.group(1) and m.group(2)
I don't find any fault in your code but you may capture group like this:
String str = "/user/myID123?name=firstName LastName ";
String regex = "/user/([a-zA-Z]+\\w*)\\?name=([a-zA-Z\\s]+)";
Pattern p = Pattern.compile(regex);
Matcher m = p.matcher(str);
if(m.find()) {
System.out.println(m.group(1) + ", " + m.group(2));
}
The problem is that * is greedy by default (it matches the whole string), so you need to modify your regex by adding a ? (making it reluctant):
List<String> str = Arrays.asList("/user/ad?name=a a", "/user/one111?name=one ONE oNe", "/user/hello?name=world");
String regex = "/user/([a-zA-Z]+\\w*?)\\?name=([a-zA-Z\\s]+)";
for (String s : str) {
Matcher matcher = Pattern.compile(regex).matcher(s);
if (matcher.matches()) {
System.out.println("user: " + matcher.group(1));
System.out.println("name: " + matcher.group(2));
}
}
Output:
user: ad
name: a a
user: one111
name: one ONE oNe
user: hello
name: world
Related
I have string like this: "Welcome Vitalii Mckay "
I need to cut from this string my name and surname, it should left in new string: "Mckay, Vitalii".
But it should be good not just for my name, it should works for other names with different length, for example:
"Welcome John Smith " -> "Smith, John"
or
"Welcome Andrea J. " -> "J., Andrea".
String name = "Welcome Vitalii Mckay";
String[] parts = name.split("\\ ");
name = parts[2] + ", " + parts[1];
Based on #Vishal's answer and OP's comment on #Max's answer, I believe this will work:
String name = " Welcome Vitalii Mckay "; // with spaces in the beginning and in the end
String[] parts = name.trim().split(" "); // you don't really need the \\
name = parts[2] + ", " + parts[1];
Just make sure you trim your String input.
Could you just use a delimiter?
i.e. use a delimiter to separate the three strings, and only print out the two needed values (Surname [2]/Firstname [1])
String s = "Welcome Vitalii Mckay";
String[] split = s.split("\\s+");
System.out.println(split[2] + ", " + split[1]);
// "Welcome"
// followed by the not of space one or more times
// then a space
// followed by anything one or more times
Pattern pattern = Pattern.compile("Welcome ([^ ]+) (.+)");
Matcher matcher = pattern.matcher("Welcome Vitalii Mckay");
if (!matcher.matches()) throw new Exception();
String firstName = matcher.group(1); // groups are captured between ()
String lastName = matcher.group(2); // groups are captured between ()
I have a string some thing like this:
If message contains sensitive info like: {Password:123456, tmpPwd : tesgjadgj, TEMP_PASSWORD: kfnda}
My pattern should look for the particular words Password or tmpPwd or TEMP_PASSWORD.
How can I create a pattern for this kind of search?
I think you are looking for the values after these words. You need to set capturing groups to extract those values, e.g.
String content = "If message contains sensitive info like: {Password:123456, tmpPwd : tesgjadgj, TEMP_PASSWORD: kfnda} ";
Pattern p = Pattern.compile("\\{Password\\s*:\\s*([^,]+)\\s*,\\s*tmpPwd\\s*:\\s*([^,]+)\\s*,\\s*TEMP_PASSWORD:\\s*([^,]+)\\s*\\}");
Matcher m = p.matcher(content);
while (m.find()) {
System.out.println(m.group(1) + ", " + m.group(2) + ", " + m.group(3));
}
See IDEONE demo
This will output 123456, tesgjadgj, kfnda.
To just find out if there are any of the substrings, use contains method:
System.out.println(content.contains("Password") ||
content.contains("tmpPwd") ||
content.contains("TEMP_PASSWORD"));
See another demo
And if you want a regex-solution for the keywords, here it is:
String str = "If message contains sensitive info like: {Password:123456, tmpPwd : tesgjadgj, TEMP_PASSWORD: kfnda} ";
Pattern ptrn = Pattern.compile("Password|tmpPwd|TEMP_PASSWORD");
Matcher m = ptrn.matcher(str);
while (m.find()) {
System.out.println("Match found: " + m.group(0));
}
See Demo 3
Finally I am using it like as per my requirement .
private final static String censoredWords =
"(?i)PASSWORD|pwd";
The (?i) makes it case-insensitive
I am trying to do a replacement using regex. The relevant piece of code is as follows:
String msg =" <ClientVerificationResult>\n " +
" <VerificationIDCheck>Y</VerificationIDCheck>\n" +
" </ClientVerificationResult>\n";
String regex = "(<VerificationIDCheck>)([Y|N])(</VerificationIDCheck>)";
String replacedMsg= msg.replaceAll(regex, "$2".matches("Y") ? "$1YES$3" : "$1NO$3") ;
System.out.println(replacedMsg);
The output of this is
<ClientVerificationResult>
<VerificationIDCheck>NO</VerificationIDCheck>
</ClientVerificationResult>
When it should be
<ClientVerificationResult>
<VerificationIDCheck>YES</VerificationIDCheck>
</ClientVerificationResult>
I guess the problem is that "$2".matches("Y") is returning false. I have tried doing "$2".equals("Y"); and weird combinations inside matches() like "[Y]" or "([Y])", but still nothing.
If I print "$2" the output is Y. Any hints on what am I doing wrong?
You cannot use Java code as the replacement argument for replaceAll which is supposed to be a string only. Better use Pattern and Matcher APIs and evaluate matcher.group(2) for your replacement logic.
Suggested Code:
String msg =" <ClientVerificationResult>\n " +
" <VerificationIDCheck>Y</VerificationIDCheck>\n" +
" </ClientVerificationResult>\n";
String regex = "(<VerificationIDCheck>)([YN])(</VerificationIDCheck>)";
Pattern p = Pattern.compile(regex);
Matcher m = p.matcher( msg );
StringBuffer sb = new StringBuffer();
while (m.find()) {
String repl = m.group(2).matches("Y") ? "YES" : "NO";
m.appendReplacement(sb, m.group(1) + repl + m.group(3));
}
m.appendTail(sb);
System.out.println(sb); // replaced string
You are checking the literal string "$2" to see if it matches "Y". This will never happen.
I have this regex:
String regexPattern = "[0-9A-Za-z]+(st|nd|rd|th)" + " " + "floor";
I want to test it against:
String lineString = "8th floor, Prince's Building, 12 Chater Road";
so I do:
boolean isMatching = lineString.matches(regexPattern);
and it return false. Why?
I thought it had something to do with whitespaces in Java, so I removed the whitespace in the regexPattern variable so it reads
regexPattern = "[0-9A-Za-z]+(st|nd|rd|th)floor";
and matched it with a string without white space:
String lineString = "8thfloor,Prince'sBuilding,12ChaterRoad"
it still returns false. Why? Any help very much appreciated.
String.matches() only returns true if the entire string matches the pattern.
Try adding .* to the beginning and end of your regex.
Example:
String regex = ".*[0-9A-Za-z]+(st|nd|rd|th)" + " " + "floor.*";
This is not the best approach, however...
Here's a better alternative:
String input = "8th floor, Prince's Building, 12 Chater Road";
String regex = "[0-9A-Za-z]+(st|nd|rd|th)" + " " + "floor";
Pattern p = Pattern.compile(regex);
boolean isMatch = p.matcher(input).find();
If you want to extract the floor number, do this:
String input = "8th floor, Prince's Building, 12 Chater Road";
String regex = "([0-9A-Za-z])+(st|nd|rd|th)" + " " + "floor";
Pattern p = Pattern.compile(regex);
Matcher m = p.matcher(input);
if (m.find()) {
String num = m.group(1);
String suffix = m.group(2);
System.out.println("Welcome to the " + num + suffix + " floor!");
// prints 'Welcome to the 8th floor!'
}
Check out the Pattern API for a boatload of info about Java regular expressions.
Edited, per comments ...
The [0-9A-Za-z]+ part is greedily matching until the end of th.
Try [0-9] instead.
I'm not new to Java, but have not dealt with Regex and Patterns before. What I'm looking to do is take a string like
"Class: " + data1 + "\nFrom: " + data2 + " To: " + data3 + "\nOccures: " + data4 + " In: " + data5 + " " + data6;
and pull out only data_1 to data_n.
I appreciate any help.
Use this regex:
Pattern pattern = Pattern.compile("Class: (.+?)\nFrom: (.+?) To: (.+?)\nOccures: (.+?) In: (.+?) (.+?)");
Matcher matcher = pattern.matcher(yourInputString);
if (matcher.find())
{
String data1 = matcher.group(1);
String data2 = matcher.group(2);
String data3 = matcher.group(3);
String data4 = matcher.group(4);
String data5 = matcher.group(5);
String data6 = matcher.group(6);
} else
{
// String didn't match the specified format
}
Explanation:
.+? will match any character for undefined times, but non-greedy.
(), using brackets will create a group. A group is given an index starting by 1 (since group 0 is the entire match)
So, (.+?) will creates groups of any character.
And what the matcher does, is searching for the whole pattern somewhere in the input string. But since you specified the format, we know exactly how your entire string is going to look like. The only thing you have to do is copy the format and replace the data you want to extract with "something" (.+?), which you give an index by creating a group of it.
Afterwards, the matcher will try to find the pattern (done by matcher.find()) and you ask them what the content is of the groups 1 up to 6.
how about using split() with ":", then from the splitted String[] get string[2i+1] ? (i from 0)