What is the regex that matches these examples(6 characters, first is a letter, others are numbers):
u78945 - valid
s56123 - valid
456a12 - invalid
78561d - invalid
1234567 - invalid
i don't know if regular expressions are the same for every programming language. I need it for Regular Expression Validator control using VB ASP.NET.
Use this pattern:
^[a-z][0-9]{5}$
This will match any Latin letter (lower-case unless using case-insensitive matching) followed by 5 decimal digits.
Note: You could use \d instead of [0-9], but read this for an explanation about why they are different.
[a-zA-Z]\d{5}
If you are searching explicitly from the beginning of the line use ^
^[a-zA-Z]\d{5}
and append $ for the end of the line.
^[a(?i)-z(?i)]\d{5}$
The (?i) code enables the expression to accept any letter without case-sensitivity. The \d{5} looks for a sequence of numbers whose length is exactly 5.
Related
This may be a very simple question but I have not much experience with regex expressions. This page is a good source of regex expressions but could not figure out how to include them into my following code:
data %>% filter(grepl("^A01H1", icl))
Question
I would like to extract the values in one column of my data frame starting with this A01H1 up to 2 more digits, for example A01H100, A01H140, A01H110. I could not find a solution despite my few attempts:
Attempts
I looked at this question from which I used ^A01H1[0-9].{2} to select up tot two more digits.
I tried with adding any character ^A01H1[0-9][0-9][x-y] to stop after two digits.
Any help would be much appreciated :)
You can use "^A01H1\\d{1,2}$".
The first part ("^A01H1"), you figured out yourself, so what are we doing in the second part ("\\d{1,2}$")?
\d includes all digits and is equivalent to [0-9], since we are working in R you need to escape \ and thus we use \\d
{1,2} indicates we want to have 1 or 2 matches of \\d
$ specifies the end of the string, so nothing should come afterwards and this prevents to match more than 2 digits
It looks as if you want to match a part of a string that starts with A01H1, then contains 1 or 2 digits and then is not followed with any digit.
You may use
^A01H1\d{1,2}(?!\d)
See the regex demo. If there can be no text after two digits at all, replace (?!\d) with $.
Details
^ - start of strinmg
A01H1 - literal string
\d{1,2} - one to two digits
(?!\d) - no digit allowed immediately to the right
$ - end of string
In R, you could use it like
grepl("^A01H1\\d{1,2}(?!\\d)", icl, perl=TRUE)
Or, with the string end anchor,
grepl("^A01H1\\d{1,2}$", icl)
Note the perl=TRUE is only necessary when using PCRE specific syntax like (?!\d), a negative lookahead.
I need to identify matching course number that have xx.3xxxxxx.
These are some examples of the course numbers.
26.3730004
27.0210000
26.3730009
26.7114001
23.9610071
26.0A34430
23.3670005
26.0B05430
I tried many patterns one example I used is the pattern below. It did not get any match.
"[^0-9]{2}\Q.\E3[^0-9]+$"
I tried using grep and grepl. I actually need the code to return indexes.
This code shows my attempt to tag the rows that have matches.
Teacher$virtual[
which(
grepl("[^0-9]{2}\\Q.\\E3[^0-9]+$",Teacher$CourseNumber))]
<- "1"
I need to remove any row from my dataframe that have the course number with that pattern. XX.3XXXXXX
But, my code did not find any match. Can you please help me?
You should use
grepl("^[0-9]{2}\\.3", Teacher$CourseNumber)
See the regex graph:
Details:
^ - start of a string
[0-9]{2} - two digits
\\. - a dot (note that a regex escape is a literal backslash, but inside a string literal, "...", a single backslash is used to form string escape sequences, hence the backslash must be double to obtain a literal backslash char necessary for a regex escape)
3 - a 3 char.
NOTE: If you want to use in-pattern quoting with \Q and \E (in between which all chars are treated literally) you need to use PCRE regex, add perl=TRUE and use
grepl("^[0-9]{2}\\Q.\\E3", Teacher$CourseNumber, perl=TRUE)
Now, the dot is treated as a literal dot, not a . metacharacter that matches any char but a line break char (in a PCRE regex, . does not match line break chars by default).
Here, this simple expression would likely cover that:
^[0-9]{2}\.[3].+$
which has a [3] boundary right after the .. It would probably work without start and end anchors:
[0-9]{2}\.[3].+
Demo
We can add or reduce the boundaries, if it'd be necessary.
I need to use an asp:RegularExpressionValidator control in my asp.net site and I want to restrict the characters to digits and semi-colon only.
So it needs to be able to take any amount of digits and ;
Valid values:
123
123;456
123;456;789;....
What is the regex for that?
Try this simple regex: [0-9;]+ or [\d;]+
\d match a digit [0-9]
; the literal character ;
+ means that it match between one character and unlimited times, as many times as possible,
If you want to guarantee that at least numbers are present in your expression you could do this too:
#npinti has a valid point better will be: ^[\d;]+$
where ^ indicates the begging of your expression and $ the end of it.
Online Demo
How would I go about building a regex that allows only digits, with no spaces, and an optional "+" at the beginning?
try this
^\+?\d+$
^ anchors it to the start of the string, $ to the end
\+? is the optional +
\d is a digit and the following + is the quantifier that says at least one (digit).
A useful resource to learn regular expressions is the tutorial of regular-expressions.info
And Regexr is a very useful resource to test regular expressions, see this regex here online
This one should work: ^\+?\d+$
You need to match a +,maybe, followed by digits. The + is a special character, so you need to escape it. To match a telephone number on its own (nothing else in the string) do ^\+?\d+$, to match it in a larger string omit the ^ and $ for just \+?\d+. You can obviously also change \d+ to \d{7} if you know how many digits there should be.
I'm using the following:
(^\+?[0-9]{10,15})$
The + in the beginning is optional as indicated above, with added length restrictions (being minimum 10 digits & maximum 15)
i have got a code to edit some function. There is a text box in that web application. It using a regular expression validator control to validate the text box. the validation expression is
ValidationExpression="[\w]{3,15}"
it accept all letters,numbers and underscores. but it do not accept special characters like \,/ * . i want to change the above regular expression to accept / .
i hope someone can explain what the above regular expression means and how to change that expression to accept / without affecting current regular expression
i am using asp.net and c#
string ValidationExpression= "[\w/]{3,15}"
[...] match a single character presents in the list between brackets
[...]{3,15} match between 3 and 15 characters presents between brackets
\w match a word character (letter, digit, underscore...)
/ match the character /
So [\w/]{3,15} match a word character or '/' between 3 and 15 times.
You current regular expression can be deconstructed as follows :
[] brackets represents regular expression group. Regex engine will try to match all the characters or group of characters given inside [] with the input string.
\w - Allow all the alpha numberic characters which includes upper case and lower case alphabets and 0 to 9 numbers and and underscore (This does not include other special characters like / or # ,etc ).
{3,15} means minimum 3 and maximum 15 alphanumeric characters must be provided in order to successfully match the string.
To add other charters, you need to add them explicitly. If you want to add / your regex should be like [\w/]{3,15}.
You can learn everything about regex here.