I have one question related with regular expression. In my case, I have to make sure that
first letter is alphabet, second onwards it can be any alphanumeric + some special characters.
Regards,
Anto
Try something like this:
^[a-zA-Z][a-zA-Z0-9.,$;]+$
Explanation:
^ Start of line/string.
[a-zA-Z] Character is in a-z or A-Z.
[a-zA-Z0-9.,$;] Alphanumeric or `.` or `,` or `$` or `;`.
+ One or more of the previous token (change to * for zero or more).
$ End of line/string.
The special characters I have chosen are just an example. Add your own special characters as appropriate for your needs. Note that a few characters need escaping inside a character class otherwise they have a special meaning in the regular expression.
I am assuming that by "alphabet" you mean A-Z. Note that in some other countries there are also other characters that are considered letters.
More information
Character Classes
Repetition
Anchors
Try this :
/^[a-zA-Z]/
where
^ -> Starts with
[a-zA-Z] -> characters to match
I think the simplest answer is to pick and match only the first character with regex.
String str = "s12353467457458";
if ((""+str.charAt(0)).matches("^[a-zA-Z]")){
System.out.println("Valid");
}
Related
I want a regular expression that accept only alphabets,hyphen,apostrophe,underscore.
I tried
/^[ A-Za-z-_']*$/
but its not working. Please help.
Your regex is wrong. Try this:
/^[0-9A-Za-z_#'-]+$/
OR
/^[\w#'-]+$/
Hyphen needs to be at first or last position inside a character class to avoid escaping. Also if empty string isn't allowed then use + (1 or more) instead of * (0 or more)
Explanation:
^ assert position at start of the string
[\w#'-]+ match a single character present in the list below
Quantifier: Between one and unlimited times, as many times as possible
\w match any word character [a-zA-Z0-9_]
#'- a single character in the list #'- literally
$ assert position at end of the string
Move the hyphen at the end or the beginig of the character class or escape it:
^[ A-Za-z_'-]*$
or
^[- A-Za-z_']*$
or
^[ A-Za-z\-_']*$
If you want all letters:
^[ \pL_'-]*$
or
When using a hyphen in a character class, be sure to place it at the end of the character class as a best practice.
The reason for this is because the hyphen is used to signify a range of characters in the character class, and when it is at the end of the class, it will not create any ranges.
My best bet would be :
/[A-Za-z-\'_#0-9]+/g
You can use the following (in Java):
String acceptHyphenApostropheUnderscoreRegEx = "^(\\p{Alpha}*+((['_-]+)\\p{Alpha})?)*+$";
If you want to have spaces and # also (as some have given above) try:
String acceptHyphenApostropheUnderscoreRegEx = "^(\\p{Alpha}*+((\\s|['#_-]+)\\p{Alpha})?)*+$";
I have some large number of strings which starts like DD_filename.
How can I extract the characters before _ using regular expression.
I tried learning using from here and in that it is given a.b will retrieve characters starting from a and ending on b
I tried similarly ^._ but it is not working for me.
^._ will only match one character before _. Try this pattern:
^.*?(?=_)
Starting from the beginning of the string, capture all non-underscore characters:
"^[^_]*"
The first ^ (caret) character means that the match starts from the beginning of the string. The brackets allow you to define a set of possible characters (character class). The second ^ character means "not". So the character class is "not underscore". The star means "zero or more". So in plain English: "match from the start of the string zero or more non underscore characters".
You can try something like
.*?(?=_)
. matches any character and *? is a reluctant quantifier. (?=_) is a positive lookahead to ensure our match is followed by an _.
If you want to only extract characters that occur at the beginning of a string you can add the ^ anchor: ^.*?(?=_). ^ matches the position before the first character in the string.
Just capture all characters that are not an underscore:
"[^_]*"
Regular Expression to get all characters before "-"
Check out #stema's answer. He gives four ways to do this, but the first is probably the best.
Match result = Regex.Match(text, #"^.*?(?=-)");
Console.WriteLine(result);
I am having problems creating a regex validator that checks to make sure the input has uppercase or lowercase alphabetical characters, spaces, periods, underscores, and dashes only. Couldn't find this example online via searches. For example:
These are ok:
Dr. Marshall
sam smith
.george con-stanza .great
peter.
josh_stinson
smith _.gorne
Anything containing other characters is not okay. That is numbers, or any other symbols.
The regex you're looking for is ^[A-Za-z.\s_-]+$
^ asserts that the regular expression must match at the beginning of the subject
[] is a character class - any character that matches inside this expression is allowed
A-Z allows a range of uppercase characters
a-z allows a range of lowercase characters
. matches a period
rather than a range of characters
\s matches whitespace (spaces and tabs)
_ matches an underscore
- matches a dash (hyphen); we have it as the last character in the character class so it doesn't get interpreted as being part of a character range. We could also escape it (\-) instead and put it anywhere in the character class, but that's less clear
+ asserts that the preceding expression (in our case, the character class) must match one or more times
$ Finally, this asserts that we're now at the end of the subject
When you're testing regular expressions, you'll likely find a tool like regexpal helpful. This allows you to see your regular expression match (or fail to match) your sample data in real time as you write it.
Check out the basics of regular expressions in a tutorial. All it requires is two anchors and a repeated character class:
^[a-zA-Z ._-]*$
If you use the case-insensitive modifier, you can shorten this to
^[a-z ._-]*$
Note that the space is significant (it is just a character like any other).
I have this regex: /[^a-zA-Z0-9_-]/
What I want to add to above is:
first character can be only a-zA-Z
How I could make this regular expression?
Try something like this:
^[a-zA-Z][a-zA-Z0-9.,$;]+$
Explanation:
^ Start of line/string.
[a-zA-Z] Character is in a-z or A-Z.
[a-zA-Z0-9.,$;] Alphanumeric or `.` or `,` or `$` or `;`.
+ One or more of the previous token (change to * for zero or more).
$ End of line/string.
I think this would also work
^[a-zA-Z].*
If you wanted to test just the first character as being alphabetical and the rest of the string can be anything.
I need a regular expression to check a string should contain only letters and space.No other character other than letter [A-Z] and space are allowed.
Please help.
The complete regex looks like this
^[A-Z ]+$
You can simply create a character class and put the characters in that you want to allow:
[A-Z ]
if you want to allow also lower case letters then use
[A-Za-z ]
or use the i (IgnoreCase) option
So your character class matches 1 character. you want to repeat it to match more than one character.
+ would be at least one character, where
* would additionally match 0 characters
As last step you need to ensure that the complete string is matched, you can do this using anchors.
^ matches the beginning of the string
$ matches the end of the string (or a newline if you use the m (multiline) option
A character class should be sufficient
[A-Z ]+
i.e. one or more of letters between A-Z and space
Check that the string matches the following:
^[a-zA-Z ]*$
Regex character classes can be negated by putting a ^ symbol at the begining of them.
Your example could be negated like this: [^A-Z]. Add a space to allow the full range of characters you want to check for and you have [^A-Z ].
Now you have a validator that meets your criteria: If that regex returns true then your validation fails.
Since you didn't specify the programming language you're working in, I can't help you much further than that.
This will match what you need:
^[A-Z\s]+$
try matching with this regex
^[A-Za-z\s]+$
this should do the trick