Regex Validation only works with one single Character - java

I want to check if a userInput is free from special characters, here is my code:
public class ValidateHelper {
public boolean userInputContainsNoSpecialCharacters(String input){
Pattern p = Pattern.compile("[a-zA-Z_0-9 ]");
Matcher m = p.matcher(input);
boolean b = m.matches();
if (b)
return true;
else
return false;
}
}
This works if I type one character in a textfield -> "a" in the textfield -> the method returns true
"ab" in the textfield -> method returns false.
can somebody help please?
beste regards Daniel


Pattern p = Pattern.compile("[a-zA-Z_0-9 ]+");


Change "[a-zA-Z_0-9 ]" to "[a-zA-Z_0-9 ]+"
The + matches "one or more" of that group.


That is because you are using the character class []. If you would like to capture a limited amount, any amount or a range of characters you need to modify it.
[a-zA-Z_0-9 ]+ //1 or more characters
[a-zA-Z_0-9 ]{1,5} //1 - 5 characters


You should use the following regex:
[A-Za-z0-9]+


change your pattern from [a-zA-Z_0-9 ] to ^[a-zA-Z_0-9 ]*$ (for 0 or more valid characters) or ^[a-zA-Z_0-9 ]+$ if you want to ensure they enter a value
A + indicates 1 or more repetitions.
A * indicates 0 or more repetitions.
The ^ and $ denote the start end of the line respectively.

Related

REGEX extract two double number separated from hypen

I have strings like:
some foo text
some foo
1-2
1.00-2.00
3.21-1.23
2.12-2.12
I have to check if the string format contains two numbers separated by hyphen.
How can I do it?
Thanks

Regex for float is: ^[1-9]\d*\.\d+$ if decimals are optional : ^[1-9]\d*(?:\.\d+)?$
Repeat it twice with hyphen in between:
`^[1-9]\d*(?:\.\d+)?-[1-9]\d*(?:\.\d+)?$`

You can use the regex:
^\d+(\.\d+)?-\d+(\.\d+)?$
Explanation can be found here.
Using java you can create a method that checks whether your desired pattern exists or not:
public static boolean returnMatch(String input) {
Pattern p1 = Pattern.compile("^\\d+(\\.\\d+)?-\\d+(\\.\\d+)?$");
Matcher m1 = p1.matcher(input);
return m1.find() ? true : false;
}
Now call it using:
System.out.println(returnMatch("some foo text")); // false
System.out.println(returnMatch("1.00-2.00")); // true
System.out.println(returnMatch("2.12-2.12")); // true
System.out.println(returnMatch("10-20")); // true

Use a simple Regex:
(\d+(?:\.\d+)?)-(\d+(?:\.\d+)?)
This solution assumes there is always a decimal part present (at least one digit). Demo at Regex101.
\d is a digit
\d+ is at least one digit
\. matches a dot (.) literally
() is a capturing group
(?:\.\d+)? is a non-capturing group which optionally matches the decimal part
Don't forget the proper escaping in Java String regex = "(\\d+(?:\\.\\d+)?)-(\\d+(?:\\.\\d+)?)";
In case one or more spaced or blank characters appear between the dash and numbers, use:
(\d+(?:\.\d+)?)\s*-\s*(\d+(?:\.\d+)?)

Matcher cannot recognize the second group of regular expression in java

I've got a problem when I'm using Matcher for finding a symbol from the group of regular expressions, it cannot recognize the second group .Maybe the code below make it clear :
public void set(String n){
String pat = "(\\d+)[!##$%^&*()_+-=}]";
Pattern r;
r = Pattern.compile(pat);
System.out.println(r);
Matcher m;
m = r.matcher(n);
if (m.find()) {
JOptionPane.showMessageDialog(null,
"Not a correct form", "ERROR_NAME_MATCH", 0);
}else{
name = n;
}
}
After running the code the first group is recognizable but the second one [!##$%^&*()_+-=}] is not.I'm totally sure that the expression is true I've checked it with 'RegexBuddy'. There must be a problem with concatenating two or more groups in one line.
Thank you for your help.

Your regex - (\d+)[!##$%^&*()_+=}-] - matches a sequence of 1+ digits followed with a symbol from the specified set.
You want to test a string and return true if a single character from the specified set is present in the string.
So, just move \d to the character class and certainly move the - to the end of this class:
String pat = "[\\d!##$%^&*()_+=}-]";
^^^
If you need to match a digit or special char, use
String pat = "\\d|[!##$%^&*()_+=}-]";
If you need both irrespective of the order:
String pat = "^(?=\\D*\\d)(?=[^!##$%^&*()_+=}-]*[!##$%^&*()_+=}-])";

Writing regex for string containing no only numbers

I need to write a regex containing not only digits [0-9]. How can I do that without explicitly specifying all possible charaters in a group. Is it possible to do through lookahead/lookbehind? Examples:
034987694 - doesn't match
23984576s9879 - match
rtfsdbhkjdfg - match
=-0io[-09uhidkbf - match
9347659837564983467 - doesn't match

^(?!\\d+$).*$
This should do it for you.See demo.
https://regex101.com/r/fM9lY3/1
The negative will lookahead will check if the string doesnt have integers from start to end.You need $ to make sure the check is till end or else it will just check at the start.

If you just need to detect whether the string is not numbers-only, then you can simply test for /\D/ - "succeed if there is a non-digit anywhere".

Why not check if it only contains digits, if not it matches
String[] strings = {"034987694", "23984576s9879",
"rtfsdbhkjdfg",
"=-0io[-09uhidkbf",
"9347659837564983467"};
for (String s : strings) {
System.out.printf("%s = %s%n", s, !s.matches("\\d*"));
}
output
034987694 = false
23984576s9879 = true
rtfsdbhkjdfg = true
=-0io[-09uhidkbf = true
9347659837564983467 = false

You may try the below,
string.matches(".*\\D.*");
This expects atleast 1 non-digit character.

Regex to get first number in string with other characters

I'm new to regular expressions, and was wondering how I could get only the first number in a string like 100 2011-10-20 14:28:55. In this case, I'd want it to return 100, but the number could also be shorter or longer.
I was thinking about something like [0-9]+, but it takes every single number separately (100,2001,10,...)
Thank you.

/^[^\d]*(\d+)/
This will start at the beginning, skip any non-digits, and match the first sequence of digits it finds
EDIT:
this Regex will match the first group of numbers, but, as pointed out in other answers, parseInt is a better solution if you know the number is at the beginning of the string

Try this to match for first number in string (which can be not at the beginning of the string):
String s = "2011-10-20 525 14:28:55 10";
Pattern p = Pattern.compile("(^|\\s)([0-9]+)($|\\s)");
Matcher m = p.matcher(s);
if (m.find()) {
System.out.println(m.group(2));
}

Just
([0-9]+) .*
If you always have the space after the first number, this will work

Assuming there's always a space between the first two numbers, then
preg_match('/^(\d+)/', $number_string, $matches);
$number = $matches[1]; // 100
But for something like this, you'd be better off using simple string operations:
$space_pos = strpos($number_string, ' ');
$number = substr($number_string, 0, $space_pos);
Regexs are computationally expensive, and should be avoided if possible.

the below code would do the trick.
Integer num = Integer.parseInt("100 2011-10-20 14:28:55");

[0-9] means the numbers 0-9 can be used the + means 1 or more times. if you use [0-9]{3} will get you 3 numbers

Try ^(?'num'[0-9]+).*$ which forces it to start at the beginning, read a number, store it to 'num' and consume the remainder without binding.

This string extension works perfectly, even when string not starts with number.
return 1234 in each case - "1234asdfwewf", "%sdfsr1234" "## # 1234"
public static string GetFirstNumber(this string source)
{
if (string.IsNullOrEmpty(source) == false)
{
// take non digits from string start
string notNumber = new string(source.TakeWhile(c => Char.IsDigit(c) == false).ToArray());
if (string.IsNullOrEmpty(notNumber) == false)
{
//replace non digit chars from string start
source = source.Replace(notNumber, string.Empty);
}
//take digits from string start
source = new string(source.TakeWhile(char.IsDigit).ToArray());
}
return source;
}

NOTE: In Java, when you define the patterns as string literals, do not forget to use double backslashes to define a regex escaping backslash (\. = "\\.").
To get the number that appears at the start or beginning of a string you may consider using
^[0-9]*\.?[0-9]+ # Float or integer, leading digit may be missing (e.g, .35)
^-?[0-9]*\.?[0-9]+ # Optional - before number (e.g. -.55, -100)
^[-+]?[0-9]*\.?[0-9]+ # Optional + or - before number (e.g. -3.5, +30)
See this regex demo.
If you want to also match numbers with scientific notation at the start of the string, use
^[0-9]*\.?[0-9]+([eE][+-]?[0-9]+)? # Just number
^-?[0-9]*\.?[0-9]+([eE][+-]?[0-9]+)? # Number with an optional -
^[-+]?[0-9]*\.?[0-9]+([eE][+-]?[0-9]+)? # Number with an optional - or +
See this regex demo.
To make sure there is no other digit on the right, add a \b word boundary, or a (?!\d)
or (?!\.?\d) negative lookahead that will fail the match if there is any digit (or . and a digit) on the right.

public static void main(String []args){
Scanner s=new Scanner(System.in);
String str=s.nextLine();
Pattern p=Pattern.compile("[0-9]+");
Matcher m=p.matcher(str);
while(m.find()){
System.out.println(m.group()+" ");
}

\d+
\d stands for any decimal while + extends it to any other decimal coming directly after, until there is a non number character like a space or letter

How to find if a special character is present in a string in Java

I want to check whether a string contains # or not.
Then if it contains #, I want to find the content after #.
For example,
test#1 — This should return me 1.
test*1 — This should not return anything.
test#123Test — This should return 123Test.
Please let me know. Thanx in advance.

I'd use simple string operations rather than a regular expression:
int index = text.indexOf('#');
return index == -1 ? "" : text.substring(index + 1);
(I'm assuming "should not return anything" means "return empty string" here - you could change it to return null if you want.)

// Compile a regular expression: A hash followed by any number of characters
Pattern p = Pattern.compile("#(.*)");
// Match input data
Matcher m = p.matcher("test#1");
// Check if there is a match
if (m.find()) {
// Get the first matching group (in parentheses)
System.out.println(m.group(1));
}

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