I need a regular expression that check my string contains one alphabet+many digits.Eg, A546456 or B456 or N455,asp.net
I have tried
^[a-z|A-Z|]+[a-z|A-Z|0-9]+[0-9]*
and
[a-zA-Z0-9]+
A regular expression to validate if a string starts with a single letter A-Z in any case and has 1 or more digits and nothing else is: ^[A-Za-z]\d+$
Explanation:
^ ... beginning of string (or beginning of line in other context).
[A-Za-z] ... a character class definition for a single character (no multiplier appended) matching a letter in range A-Z in any case. \w cannot be used as it matches also an underscore and letters from other languages as for example German umlauts.
\d ... a digit which is equal the expression [0-9].
+ ... previous expression (any digit) 1 or more times.
$ ... end of string (or end of line in other context).
Be careful with a regular expression containing a backslash as used here for \d put into a string as the backslash character is in many programming/scripting languages also the escape character inside strings and must be therefore often escaped with one more backslash, i.e. "^[A-Za-z]\\d+$"
Related
I want to check if a character can be safely converted to a numeric by using a regex.
However, I don't see my error. Example:
stringr::str_detect("4.", pattern = "-{0,1}[0-9]+(.[0-9]+){0,1}")
This produces a TRUE. My intention was to specifiy that whenever a . follows the first sequence of numbers, there must be at least one other number, therefore (.[0-9]+){0,1}.
What's wrong here?
Note:
(.[0-9]+){0,1} is an optional pattern because {0,1} (=?) makes the .[0-9]+ pattern sequence match one or zero times. So, yes, one or more digits ([0-9]+) must follow any char other than line break chars (matched with an unescaped .), but this pattern is optional, and thus you cannot require anything with it.
. is unescaped, so it matches any char other than line break chars. Escape it to match a literal dot
Your regex is not anchored, and can match partial substrings in a longer string. Use ^ and $ to make the pattern match the whole string.
So, consider using
stringr::str_detect("4.", pattern = "^-?[0-9]+(?:\\.[0-9]+)?$")
where
^ - start of string
-? - an optional - char
[0-9]+ - one or more digits
(?:\.[0-9]+)? - a non-capturing group matching an optional sequence of a . and then one or more digits
$ - end of string.
I'm creating a small example to be put into mutate(). Not sure why this doesn't work.
> str_extract("rs1234-<b>C</b>","^rs*\\d$")
[1] NA
I'd be great if you can point to my misunderstanding of the language instead of merely providing a solution. I expect to get "rs1234".
The ^rs*\d$ regex matches
^ - start of string
rs* - r and zero or more occurrences of s char
\d - a digit
$ - end of string.
So, your pattern matches strings like rsssss1, r3, etc.
You need
str_extract("rs1234-<b>C</b>", "^rs\\d+")
where ^rs\d+ matches rs at the start of string and then one or more digits. See this regex demo.
But if I just want the substring in between "rs" and the last number. What should I do?
You would use rs.*\d:
str_extract("rs1234-<b>C</b>", "rs.*\\d")
where rs.*\d matches rs, then any zero or more chars other than line break chars as many as possible and then a digit.
NOTE: If you need to match line endings, too, you need to prepend the last pattern with (?s) inline DOTALL modifier.
See this regex demo.
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 write a regular expression for RegularExpressionValidator ASP.NET Web Controls.
The regular expression should ALLOW all alphabetic characters but not numbers or special characters (example: |!"£$%&/().
Any idea how to do it?
^[A-Za-z]+$
validates a string of length 1 or greater, consisting only of ASCII letters.
^[^\W\d_]+$
does the same for international letters, too.
Explanation:
[^ # match any character that is NOT a
\W # non-alphanumeric character (letters, digits, underscore)
\d # digit
_ # or underscore
] # end of character class
Effectively, you get \w minus (\d and _).
Or, you could use the fact that ASP.NET supports Unicode properties:
^\p{L}+$
validates a string of Unicode letters of length 1 or more.
Including spaces:
"^[a-zA-Z ]*$"
Excluding Spaces:
"^[a-zA-Z]*$"
To make it non-optional, change the * to a +
You can use the regex:
^[a-zA-Z]+$
Explanation:
^ : Start anchor
[..] : Char class
+ : one or more repetations
$ : End anchor
I have a question about a regex. Given this part of a regex:
(.[^\\.]+)
The part [^\.]+ Does this mean get everything until the first dot? So with this text:
Hello my name is Martijn. I live in Holland.
I get 2 results: both sentences. But when I leave the + sign, I get 2 two characters: he, ll, o<space>, my, etc. Why is that?
Your regex .[^\\.]+ means:
Match any character
Match any character until you get slash or a dot ".". Note that [^\\.] means NOT slash or NOT dot, which means either a dot or a slash is not a match. It will keep on matching characters until it founds a dot or slash because of the "+" at the end. It is called a greedy quantifier because of that.
When you input (quotes not included): "Hello my name is Martijn. I live in Holland."
The matches are:
Hello my name is Martijn
. I live in Holland
Note that the dot is not included in the first match since it stops at n in Martijn and the second match starts with the dot.
When you remove the +: (.[^\\.])
It just means:
Match any character
Match any character except a dot or a slash.
Because a dot outside a character class (ie, not between []) means (almost) any character.
So, .[^\\.] means match (almost) any character followed by something which is not a dot nor a backslash (dots don't need to be escaped in a character class to mean just a dot, but backslashes do),
This, in your example, is h (any character) e (not a dot nor a backslash) and so on and so forth.
Whereas with a + (one or more of not a dot nor a backslash) you will match all characters which are not dots until a dot.
The regex means:
any one character followed by more than zero characters that are not a backslash or a period.