I have this kind of String ('123','12345678') and i would validate it throw a regex.
I have write this code, but i'm not shure it work.
String field = "('123','12345678')";
String regex = "^('\\d{3}','\\d{8}')$";
public void valideField(String field, String regex){
{
if(!field.matches(regex)){
System.out.println("Not validated!");
}
}
Is correct regex or not? You have any suggestions or help?
You need to escape parentheses (using backslashes) as they represent capturing groups in regular expressions:
String regex = "^\\('\\d{3}','\\d{8}'\\)$";
You have to escape the single quotes and parenthesis too with \\' and \\(
The correct regex would be:
String regex = "\\(\\'\\d{3}\\',\\'\\d{8}\\'\\)";
Related
I have been taking a look at the regular expressions and how to use it in Java for the problem I have to solve. I have to insert a \ before every ". This is what I have:
public class TestExpressions {
public static void main (String args[]) {
String test = "$('a:contains(\"CRUCERO\")')";
test = test.replaceAll("(\")","$1%");
System.out.println(test);
}
}
The ouput is:
$('a:contains("%CRUCERO"%)')
What I want is:
$('a:contains(\"CRUCERO\")')
I have changed % for \\ but have an error StringIndexOutofBounds don't know why. If someone can help me I would appreciate it, thank you in advance.
I have to insert a \ before every "
You can try with replace which automatically escapes all regex metacharacters and doesn't use any special characters in replacement part so you can simply use String literals you want to be put in matched part.
So lets just replace " with \" literal. You can write it as
test = test.replace("\"", "\\\"");
If you want to insert backspace before quote then use:
test = test.replaceAll("(\")","\\\\$1"); // $('a:contains(\"CRUCERO\")')
Or if you want to avoid already escaped quote then use negative lookbehind:
String test = "$('a:contains(\\\"CRUCERO\")')";
test = test.replaceAll("((?<!\\\\)\")","\\\\$1"); // $('a:contains(\"CRUCERO\")')
String result = subject.replaceAll("(?i)\"CRUCERO\"", "\\\"CRUCERO\\\"");
EXPLANATION:
Match the character string “"CRUCERO"” literally (case insensitive) «"CRUCERO"»
Ignore unescaped backslash «\»
Insert the character string “"CRUCERO” literally «"CRUCERO»
Ignore unescaped backslash «\»
Insert the character “"” literally «"»
If your goal is escape text for Java strings, then instead of regular expressions, consider using
String escaped = org.apache.commons.lang.StringEscapeUtils.
escapeJava("$('a:contains(\"CRUCERO\")')");
System.out.println(escaped);
Output:
$('a:contains(\"CRUCERO\")')
JavaDoc: http://commons.apache.org/proper/commons-lang/javadocs/api-2.6/org/apache/commons/lang/StringEscapeUtils.html#escapeJava(java.lang.String)
I'm trying to match any string consisting of:
Any alphanumeric string of 1+ chars; then
Two periods (".."); then
Any alphanumeric string of 1+ chars
For example:
mydatabase..mytable
anotherDatabase23..table28
etc.
Given the following function:
public boolean isValidDBTableName(String candidate) {
if(candidate.matches("[a-zA-Z0-9]+..[a-zA-Z0-9]+"))
return true;
else
return false;
}
Passing this function the value "mydb..tablename" causes it to return false. Why? Thanks in advance!
As NeplatnyUdaj has pointed out in comment, your current regex should return true for the input "mydb..tablename".
However, your regex has the problem of over-matching, where it returns true for invalid names such as nodotname.
You need to escape ., since in Java regex, it will match any character except for line separators:
"[a-zA-Z0-9]+\\.\\.[a-zA-Z0-9]+"
In regex, you can escape meta-characters (character with special meaning) with \. To specify \ in string literal, you need to escape it again.
You must escape the period in regexes. As a \ must also be escaped, this gives
"[a-zA-Z0-9]+\\.\\.[a-zA-Z0-9]+"
I just tried your regex in Eclipse and it worked. Or at least did not fail. Try stripping whitespace characters.
#Test
public void test()
{
String testString = "mydb..tablename";
Assert.assertTrue("no match", testString.matches("[a-zA-Z0-9]+..[a-zA-Z0-9]+"));
Assert.assertFalse("falsematch", "a.b".matches("[a-zA-Z0-9]+..[a-zA-Z0-9]+"));
}
Is there any method in Java or any open source library for escaping (not quoting) a special character (meta-character), in order to use it as a regular expression?
This would be very handy in dynamically building a regular expression, without having to manually escape each individual character.
For example, consider a simple regex like \d+\.\d+ that matches numbers with a decimal point like 1.2, as well as the following code:
String digit = "d";
String point = ".";
String regex1 = "\\d+\\.\\d+";
String regex2 = Pattern.quote(digit + "+" + point + digit + "+");
Pattern numbers1 = Pattern.compile(regex1);
Pattern numbers2 = Pattern.compile(regex2);
System.out.println("Regex 1: " + regex1);
if (numbers1.matcher("1.2").matches()) {
System.out.println("\tMatch");
} else {
System.out.println("\tNo match");
}
System.out.println("Regex 2: " + regex2);
if (numbers2.matcher("1.2").matches()) {
System.out.println("\tMatch");
} else {
System.out.println("\tNo match");
}
Not surprisingly, the output produced by the above code is:
Regex 1: \d+\.\d+
Match
Regex 2: \Qd+.d+\E
No match
That is, regex1 matches 1.2 but regex2 (which is "dynamically" built) does not (instead, it matches the literal string d+.d+).
So, is there a method that would automatically escape each regex meta-character?
If there were, let's say, a static escape() method in java.util.regex.Pattern, the output of
Pattern.escape('.')
would be the string "\.", but
Pattern.escape(',')
should just produce ",", since it is not a meta-character. Similarly,
Pattern.escape('d')
could produce "\d", since 'd' is used to denote digits (although escaping may not make sense in this case, as 'd' could mean literal 'd', which wouldn't be misunderstood by the regex interpeter to be something else, as would be the case with '.').
Is there any method in Java or any open source library for escaping (not quoting) a special character (meta-character), in order to use it as a regular expression?
If you are looking for a way to create constants that you can use in your regex patterns, then just prepending them with "\\" should work but there is no nice Pattern.escape('.') function to help with this.
So if you are trying to match "\\d" (the string \d instead of a decimal character) then you would do:
// this will match on \d as opposed to a decimal character
String matchBackslashD = "\\\\d";
// as opposed to
String matchDecimalDigit = "\\d";
The 4 slashes in the Java string turn into 2 slashes in the regex pattern. 2 backslashes in a regex pattern matches the backslash itself. Prepending any special character with backslash turns it into a normal character instead of a special one.
matchPeriod = "\\.";
matchPlus = "\\+";
matchParens = "\\(\\)";
...
In your post you use the Pattern.quote(string) method. This method wraps your pattern between "\\Q" and "\\E" so you can match a string even if it happens to have a special regex character in it (+, ., \\d, etc.)
I wrote this pattern:
Pattern SPECIAL_REGEX_CHARS = Pattern.compile("[{}()\\[\\].+*?^$\\\\|]");
And use it in this method:
String escapeSpecialRegexChars(String str) {
return SPECIAL_REGEX_CHARS.matcher(str).replaceAll("\\\\$0");
}
Then you can use it like this, for example:
Pattern toSafePattern(String text)
{
return Pattern.compile(".*" + escapeSpecialRegexChars(text) + ".*");
}
We needed to do that because, after escaping, we add some regex expressions. If not, you can simply use \Q and \E:
Pattern toSafePattern(String text)
{
return Pattern.compile(".*\\Q" + text + "\\E.*")
}
The only way the regex matcher knows you are looking for a digit and not the letter d is to escape the letter (\d). To type the regex escape character in java, you need to escape it (so \ becomes \\). So, there's no way around typing double backslashes for special regex chars.
The Pattern.quote(String s) sort of does what you want. However it leaves a little left to be desired; it doesn't actually escape the individual characters, just wraps the string with \Q...\E.
There is not a method that does exactly what you are looking for, but the good news is that it is actually fairly simple to escape all of the special characters in a Java regular expression:
regex.replaceAll("[\\W]", "\\\\$0")
Why does this work? Well, the documentation for Pattern specifically says that its permissible to escape non-alphabetic characters that don't necessarily have to be escaped:
It is an error to use a backslash prior to any alphabetic character that does not denote an escaped construct; these are reserved for future extensions to the regular-expression language. A backslash may be used prior to a non-alphabetic character regardless of whether that character is part of an unescaped construct.
For example, ; is not a special character in a regular expression. However, if you escape it, Pattern will still interpret \; as ;. Here are a few more examples:
> becomes \> which is equivalent to >
[ becomes \[ which is the escaped form of [
8 is still 8.
\) becomes \\\) which is the escaped forms of \ and ( concatenated.
Note: The key is is the definition of "non-alphabetic", which in the documentation really means "non-word" characters, or characters outside the character set [a-zA-Z_0-9].
Use this Utility function escapeQuotes() in order to escape strings in between Groups and Sets of a RegualrExpression.
List of Regex Literals to escape <([{\^-=$!|]})?*+.>
public class RegexUtils {
static String escapeChars = "\\.?![]{}()<>*+-=^$|";
public static String escapeQuotes(String str) {
if(str != null && str.length() > 0) {
return str.replaceAll("[\\W]", "\\\\$0"); // \W designates non-word characters
}
return "";
}
}
From the Pattern class the backslash character ('\') serves to introduce escaped constructs. The string literal "\(hello\)" is illegal and leads to a compile-time error; in order to match the string (hello) the string literal "\\(hello\\)" must be used.
Example: String to be matched (hello) and the regex with a group is (\(hello\)). Form here you only need to escape matched string as shown below. Test Regex online
public static void main(String[] args) {
String matched = "(hello)", regexExpGrup = "(" + escapeQuotes(matched) + ")";
System.out.println("Regex : "+ regexExpGrup); // (\(hello\))
}
Agree with Gray, as you may need your pattern to have both litrals (\[, \]) and meta-characters ([, ]). so with some utility you should be able to escape all character first and then you can add meta-characters you want to add on same pattern.
use
pattern.compile("\"");
String s= p.toString()+"yourcontent"+p.toString();
will give result as yourcontent as is
I have a String, which I want to split into parts using delimeter }},{". I have tried using:
String delims="['}},{\"']+";
String field[]=new String[50];
field=subResult.split(delims);
But it is not working :-( do you know, what expression in delims should I use?
Thanks for your replies
A { is a regex meta-character which marks the beginning of a character class. To match a literal { you need to escape it by preceding it with a \\ as:
String delims="}},\\{";
String field[] = subResult.split(delims);
You need not escape the } in your regex as the regex engine infers that it is a literal } as it is not preceded by a opening {. That said there is no harm in escaping it.
See it
If the delimiter is simply }},{ then subResult.split("\\}\\},\\{") should work
String fooo = "asdf}},{bar}},{baz";
System.out.println(Arrays.toString(fooo.split("\\}\\},\\{")));
You should be escaping it.
String.split("\\}\\},\\{");
You could be making it more complex than you need.
String text = "{{aaa}},{\"hello\"}";
String[] field=text.split("\\}\\},\\{\"");
System.out.println(Arrays.toString(field));
Use:
Pattern p = Pattern.compile("[}},{\"]");
// Split input with the pattern
String[] result = p.split(MyTextString);
This seems like a well known title, but I am really facing a problem in this.
Here is what I have and what I've done so far.
I have validate input string, these chars are not allowed :
&%$###!~
So I coded it like this:
String REGEX = "^[&%$###!~]";
String username= "jhgjhgjh.#";
Pattern pattern = Pattern.compile(REGEX);
Matcher matcher = pattern.matcher(username);
if (matcher.matches()) {
System.out.println("matched");
}
Change your first line of code like this
String REGEX = "[^&%$##!~]*";
And it should work fine. ^ outside the character class denotes start of line. ^ inside a character class [] means a negation of the characters inside the character class. And, if you don't want to match empty usernames, then use this regex
String REGEX = "[^&%$##!~]+";
i think you want this:
[^&%$###!~]*
To match a valid input:
String REGEX = "[^&%$##!~]*";
To match an invalid input:
String REGEX = ".*[&%$##!~]+.*";