How to use VS Find/Replace to replace:
this: $('a[name="lnkFind"]').on('click', function
with this: $(document).on("click", "a[name='lnkFind']", function
I'm not sure which characters need to be escaped - single or double quotes or both? None of the patters I've tried seem to find a match.
You'll need to escape many of these characters.
Find/Replace will complain about the un-escaped ( and ), even the bare ( at the end because it's missing a matching ). Also the square brackets, which are used for character sets, and finally the $.
So this should work as the pattern:
\$\('a\[name="lnkFind"\]'\).on\('click', function
You should look at a list of special characters in Regular Expressions.
$, ., [, ] should all be escaped.
http://www.fon.hum.uva.nl/praat/manual/Regular_expressions_1__Special_characters.html
Except in special cases (such as vim regex), in general you can escape any and all special characters in regex to get their literal form, i.e. escaping a special character that doesn't need to be escaped, won't do any harm.
That said, here's the minimum that needs to be escaped:
\$\('a\[name="lnkFind"]')\.on\('click', function
I don't think you'll need to escape anything in the replacement, because only a $ or \ followed by a number will be interpreted.
Related
I am struggling with writing regex expression in Snowflake.
SELECT
'DEM7BZB01-123' AS SKU,
RLIKE('DEM7BZB01-123','^DEM.*\d\d$') AS regex
I would like to find all strings that starts with "DEM" and ends with two digits. Unfortunately the expression that I am using returns FALSE.
I was checking this expression in two regex generators and it worked.
In snowflake the backslash character \ is an escape character.
Reference: Escape Characters and Caveats
So you need to use 2 backslashes in a regex to express 1.
SELECT
'DEM7BZB01-123' AS SKU,
RLIKE('DEM7BZB01-123', '^DEM.*\\d\\d$') AS regex
Or you could write the regex pattern in such a way that the backslash isn't used.
For example, the pattern ^DEM.*[0-9]{2}$ matches the same as the pattern ^DEM.*\d\d$.
You need to escape your backslashes in your SQL before it can be parsed as a regex string. (sometimes it gets a bit silly with the number of backslashes needed)
Your example should look like this
RLIKE('DEM7BZB01-123','^DEM.*\\d\\d$') AS regex
RLIKE (which is an alias in Snowflake for the SQL Standard REGEXP_LIKE function) implicitly adds ^ and $ to your search pattern...
The function implicitly anchors a pattern at both ends (i.e. '' automatically becomes '^$', and 'ABC' automatically becomes '^ABC$').
so you can remove them, and that then allows you to use $$ quoting
In single-quoted string constants, you must escape the backslash character in the backslash-sequence. For example, to specify \d, use \d. For details, see Specifying Regular Expressions in Single-Quoted String Constants (in this topic).
You do not need to escape backslashes if you are delimiting the string with pairs of dollar signs ($$) (rather than single quotes).
so you can simply use the regex DEM.*\d\d to find all strings that starts with DEM and ends with two digits without extra escaping as follows
SELECT
'DEM7BZB01-123' AS SKU
, RLIKE('DEM7BZB01-123', $$DEM.*\d\d$$) AS regex
which gives
SKU |REGEX|
-------------+-----+
DEM7BZB01-123|true |
With regular expression I would like to get all characters between round brackets, but \( and \) characters should be also included in the result.
Examples:
input: fo(ob)a)r
output: ob
input: foo(bar\(qwerty\))baz
output: bar\(qwerty\)
This is what I used for finding text between brackets:
(?<=\()([^\s\(\)]+)(?=\)), but I can't make exceptions for brackets preceded by \.
You could do something like this :
.*(?<!\\)\((.*?)(?<!\\)\)
Basically, it matches as many characters as possible until it sees an open parenthesis without a backslash (using a negative lookbehind), then groups the next matching characters until a closing parenthesis (still without a backslash).
Note that this regex may not work properly if you escape the backslashes.
Example : https://regex101.com/r/BqVKZp/1
This regex works for both your examples, without any lookaheads and lookbehinds:
\((.+[^\\])\)
A U flag is needed.
Im trying to find files that are looking like this:
access_log-20160101
access_log-20160304
...
with perl regex i came up with something like this:
/^access_log-\d{8}$/
But im not sure about the "_" and the "-". are these metacharacter?
What is the expression for this?
i read that "_" in regex is something like \w, but how do i use them in my exypression?
/^access\wlog-\d{8}$/ ?
Underscore (_) is not a metacharacter and does not need to be quoted (though it won't change anything if you quote it).
Hyphen (-) IS a metacharacter that defines the range between two symbols inside a bracketed character class. However, in this particular position, it will be interpreted verbatim and doesn't need quoting since it is not inside [] with a symbol on both sides.
You can use your regexp as is; hyphens (-) might need quoting if your format changes in future.
Your regex pattern is exactly right
Neither underscore _ nor hyphen - need to be escaped. Outside a square-bracketed character class, the twelve Perl regex metacharacters are
Brackets ( ) [ {
Quantifiers * + ?
Anchors ^ $
Alternator |
Wild character .
The escape itself \
and only these must be escaped
If the pattern of your file names doesn't vary from what you have shown then the pattern that you are using
^access_log-\d{8}$
is correct, unless you need to validate the date string
Within a character class like [A-F] you must escape the hyphen if you want it to be interpreted literally. As it stands, that class is the equivalent to [ABCDEF]. If you mean just the three characters A, - or F then [A\-F] will do what you want, but it is usual to put the hyphen at the start or end of the class list to make it unambiguous. [-AF] and [AF-] are the same as [A\-F] and rather more readable
Can't work this one out, this matches a single star:
// Escaped multiply
Text = Text.replace(new RegExp("\\*", "g"), '[MULTIPLY]');
But I need it to match \*, I've tried:
\\*
\\\\*
\\\\\*
Can't work it out, thanks for any help!
You were close, \\\\\\* would have done it.
Better use verbatim strings, that makes it easier:
RegExp(#"\\\*", "g")
\\ matches a literal backslash (\\\\ in a normal string), \* matches an asterisk (\\* in a normal string).
Remember that there are two 'levels' of escaping.
First, you are escaping your strings for the C# compiler, and you are also escaping your strings for the Regex engine.
If you want to match "\*" literally, then you need to escape both of these characters for the regex engine, since otherwise they mean something different. We escape these with backslashes, so you will have "\\\*".
Then, we have to escape the backslashes in order to write them as a literal string. This means replacing each backslash with two backslashes: "\\\\\\*".
Instead of this last part, we could use a "verbatim string", where no escapes are applied. In this case, you only need the result from the first escaping: #"\\\*".
Your syntax is completely wrong. It looks more like Javascript than C#.
This works fine:
string Text = "asdf*sadf";
Text = Regex.Replace(Text, "\\*", "[MULTIPLY]");
Console.WriteLine(Text);
Output:
asdf[MULTIPLY]sadf
To match \* you would use the pattern "\\\\\\*".
I have the following regular expression for eliminating spaces, tabs, and new lines: [^ \n\t]
However, I want to expand this for certain additional characters, such as > and <.
I tried [^ \n\t<>], which works well for now, but I want the expression to not match if the < or > is preceded by a \.
I tried [^ \n\t[^\\]<[^\\]>], but this did not work.
Can any one of the sequences below occur in your input?
\\>
\\\>
\\\\>
\blank
\tab
\newline
...
If so, how do you propose to treat them?
If not, then zero-width look-behind assertions will do the trick, provided that your regular expression engine supports it. This will be the case in any engine that supports Perl-style regular expressions (including Perl's, PHP, etc.):
(?<!\\)[ \n\t<>]
The above will match any un-escaped space, newline, tab or angled braces. More generically (using \s to denote any space characters, including \r):
(?<!\\)\s
Alternatively, using complementary notation without the need for a zero-width look-behind assertion (but arguably less efficiently):
(?:[^ \n\t<>]|\\[<>])
You may also use a variation of the latter to handle the \\>, \\\>, \\\\> etc. cases as well up to some finite number of preceding backslashes, such as:
(?:[^ \n\t<>]|(?:^|[^<>])[\\]{1,3,5,7,9}[<>])
According to the grep man page:
A bracket expression is a list of
characters enclosed by [ and ]. It
matches any single character in that
list; if the first character of the
list is the caret ^ then it matches
any character not in the list.
This means that you can't match a sequence of characters such as \< or \> only single characters.
Unless you have a version of grep built with Perl regex support then you can use lookarounds like one of the other posters mentioned. Not all versions of grep have this support though.
Maybe you can use egrep and put your pattern string inside quotes. This should obliterate the need for escaping.