What is an unicode code for grave accent mark used to specify regex literal in Frege?
The character is called Acute Accent and the unicode for that is 00B4. In ubuntu, you can type that using Ctrl+Shift+u and then type 00B4 then space. However you don't really have to use that if your regex literal is more than one character in which case you can just use apostrophes.
Quoting the doc:
Regular expression literals have type Regex and are written:
´\b(foo|bar)\b´ -- string enclosed in grave accents
'\w+' -- string with length > 1 enclosed in apostrophes
The notation with the apostrophes has been introduced because many have a hard time entering a grave accent mark on their terminal. However, it is not possible to write a regular expressions with length 1 this way, because then the literal gets interpreted as Char literal. (One can write something like '(?:X)' for a Regex that matches a single 'X').
Related
I'm trying replace CT with COURT regardless of where it appears in a string (using Snowflake SQL). I would expect this to work:
select
regexp_replace('36 HERITAGE CT', '\bCT\b', 'COURT'),
regexp_replace('36 HERITAGE CT #204', '\bCT\b', 'COURT')
But the output is always 36 HERITAGE CT no matter what I do.
Anyone know what I'm doing wrong?
As per documentation, \b is supported.
Use
regexp_replace('36 HERITAGE CT', '\\bCT\\b', 'COURT')
See Note:
In single-quoted string constants, you must escape the backslash character in the backslash-sequence. For example, to specify \d, use \\d. For details, see Specifying Regular Expressions in Single-Quoted String Constants (in this topic).
You do not need to escape backslashes if you are delimiting the string with pairs of dollar signs ($$) (rather than single quotes).
I have username in this pattern
ref_2_34_aaa_dos
ref_2_34_bbb_dos
How can I use regexp_like for this?
SELECT username FROM all_users WHERE regexp_like(username, '^ref_2_34_[:alpha:]_dos$')
does not work. Nor can I use ESCAPE '\' with regexp_like. it would give a syntax error.
You need to put the POSIX character class [:alpha:] into a bracket expression (i.e. [...]) and apply a + quantifier to it:
regexp_like(username, '^ref_2_34_[[:alpha:]]+_dos$')
The + quantifier means there can be 1 or more letters between the last but one and last underscores.
If your string may have no user name at that location (it is empty), and you want to get those entries too, you would need to replace the plus with the * quantifier that matches zero or more occurrences of the quantified subpattern.
Since comments require some more clarifications, here is some bracket expression and POSIX character class reference.
Bracket expressions
Matches any single character in the list within the brackets. The following operators are allowed within the list, but other metacharacters included are treated as literals:
Range operator: -
POSIX character class: [: :]
POSIX collation element: [. .]
POSIX character equivalence class: [= =]
A dash (-) is a literal when it occurs first or last in the list, or as an ending range point in a range expression, as in [#--]. A right bracket (]) is treated as a literal if it occurs first in the list.
POSIX Character Class (can be a part of the bracket expression):
[:class:] - Matches any character belonging to the specified POSIX character class. You can use this operator to search for characters with specific formatting such as uppercase characters, or you can search for special characters such as digits or punctuation characters. The full set of POSIX character classes is supported.
...
The expression [[:upper:]]+ searches for one or more consecutive uppercase characters.
Bracket expressions can be considered a kind of a "container" construct for multiple atoms, that, as a whole regex unit, matches some class of characters you defined. If you need to match a <, or >, or letters, you may combine them into 1 bracket expression [<>[:alpha:]]. To match zero or more of <, > or letters, add a * quantifier after ]: [<>[:alpha:]]*.
Or, to imitate a trailing word boundary, one might use [^_[:alnum:]] (say, in a ($|[^_[:alnum:]]) pattern) that matches any character but a _, digits and letters ([:alnum:] matches alphanumerical symbols).
I'm writing a regular expression in Objective-C.
The escape sequence \w is illegal and emits a warning, so the regular expression /\w/ must be written as #"\\w"; the escape sequence \? is valid, apparently, and doesn't emit a warning, so the regular expression /\?/ must be written as #"\\?" (i.e., the backslash must be escaped).
Question marks aren't invisible like \t or \n, so why is \? a valid escape sequence?
Edit: To clarify, I'm not asking about the quantifier, I'm asking about a string escape sequence. That is, this doesn't emit a warning:
NSString *valid = #"\?";
By contrast, this does emit a warning ("Unknown escape sequence '\w'"):
NSString *invalid = #"\w";
It specifies a literal question mark. It is needed because of a little-known feature called trigraphs, where you can write a three-character sequence starting with question marks to substitute another character. If you have trigraphs enabled, in order to write "??" in a string, you need to write it as "?\?" in order to prevent the preprocessor from trying to read it as the beginning of a trigraph.
(If you're wondering "Why would anybody introduce a feature like this?": Some keyboards or character sets didn't include commonly used symbols like {. so they introduced trigraphs so you could write ??< instead.)
? in regex is a quantifier, it means 0 or 1 occurences. When appended to the + or * quantifiers, it makes it "lazy".
For example, applying the regex o? to the string foo? would match o.
However, the regex o\? in foo? would match o?, because it is searching for a literal question mark in the string, instead of an arbitrary quantifier.
Applying the regex o*? to foo? would match oo.
More info on quantifiers here.
examples:
"""Romeo and Juliet"""
'another string that is quoted with '' single quotes'
the problem is that the string can have characters used for db escaping even in it's beginning and end, so regex should look if given char is used in sequence of odd length that is 1,3,5... at the end of matched string
Try this regex: '(?:[^']|'')*' for single quotes. The same for double quotes, i.e. full regex:
'(?:[^']|'')*'|"(?:[^"]|"")*"
In string hello 'my ''beautiful''' 'world'! """Romeo and Juliet""" it will find:
'my ''beautiful'''
'world'
"""Romeo and Juliet"""
Where your string includes (or might include) characters that cause problems in your sql, you should always escape.
If you're using PHP with a mysql database, use mysql_real_escape_string($text); (docs)
For other databases and languages, you'll have to check for your specific purposes, but there's likely to be an existing method.
I need a complete list of characters that should be escaped in sql string parameters to prevent exceptions. I assume that I need to replace all the offending characters with the escaped version before I pass it to my ObjectDataSource filter parameter.
No, the ObjectDataSource will handle all the escaping for you. Any parametrized query will also require no escaping.
As others have pointed out, in 99% of the cases where someone thinks they need to ask this question, they are doing it wrong. Parameterization is the way to go. If you really need to escape yourself, try to find out if your DB access library offers a function for this (for example, MySQL has mysql_real_escape_string).
SQL Books online:
Search for String Literals:
String Literals
A string literal consists of zero or more characters surrounded by quotation marks. If a string contains quotation marks, these must be escaped in order for the expression to parse. Any two-byte character except \x0000 is permitted in a string, because the \x0000 character is the null terminator of a string.
Strings can include other characters that require an escape sequence. The following table lists escape sequences for string literals.
\a
Alert
\b
Backspace
\f
Form feed
\n
New line
\r
Carriage return
\t
Horizontal tab
\v
Vertical tab
\"
Quotation mark
\
Backslash
\xhhhh
Unicode character in hexadecimal notation
Here's a way I used to get rid of apostrophes. You could do the same thing with other offending characters that you run into. (example in VB.Net)
Dim companyFilter = Trim(Me.ddCompany.SelectedValue)
If (Me.ddCompany.SelectedIndex > 0) Then
filterString += String.Format("LegalName like '{0}'", companyFilter.Replace("'", "''"))
End If
Me.objectDataSource.FilterExpression = filterString
Me.displayGrid.DataBind()