I know this line escapes a javascript string by adding \ before special characters. But can anyone explain how it does that? And also. Can it cause any problems in further string manipulations.
str = str.replace(/[-\\/\\^$*+?.()|[\]{}]/g, '\\$&');
The regular expression replaces anything it matches with backslash and the matched substring. It matches any single characters of the set between [ and ]. That is:
- simple dash
\\ single backslash (preceded by another backslash, because it server as escape character)
/ slash
\\ single backslash again (it is obsolete here)
^, $, *, +, ?, ., (, ), |, [ single characters
\] closing bracket is preceded by escape character, otherwise it would close set of characters to match
{, } single characters
The replacement string \\$& simply says:
\\ add single backslash
$& add the whole substring matched by the expression
Related
var str = "Hello\\\World\\\";
var newStr = str.replace("\\\", "");
alert(newStr); // I want this to alert: HelloWorld
The number of slashes is always 3, not more not less. How can I replace them? The code above doesn't work at all. I've played around a bit with the global flag, escaping the slashes etc but can't figure it out.
Firstly, you need to escape each slash with another backslash, as mentioned by #Bathsheba.
Additionally, you want your replacement regex to be global:
var str = "Hello\\\\\\World\\\\\\";
var newStr = str.replace(/\\\\\\/g, "");
alert(newStr); // I want this to alert: HelloWorld
If you want three slashes in a row in a string literal then you need to escape each one in turn:
var str = "Hello\\\\\\World\\\\\\";
var newStr = str.replace("\\\\\\", "");
In your current string, \\\W would be one slash and an error as \W is not a valid sequence. (Some more examples: \\ is a single slash, \t a tab, \" a quotation character).
Try this regex \\\\\\ for replace
\\ indicates \
There are 12 characters with special meanings: the backslash \, the caret ^, the dollar sign $, the period or dot ., the vertical bar or pipe symbol |, the question mark ?, the asterisk or star *, the plus sign +, the opening parenthesis (, the closing parenthesis ), and the opening square bracket [, the opening curly brace {, These special characters are often called "metacharacters".
If you want to use any of these characters as a literal in a regex, you need to escape them with a backslash.
I want to have a regex for text field in ExtJs(maskRe) which matches all java code pattern
I've used
maskRe:/^[A-Za-z0-9 _=//~'"|{}();*:?+,.]*$/
I also want to include [,], but it seems /[, /], //[, //] is not working..
Any inputs please
The problem is you need to escape your forward slash. Change // to \/:
/^[A-Za-z0-9 _=\/~'"|{}();*:?+,.]*$/
However this regular expression does not match any Java code. Java code can contain almost any Unicode character. int møøse = 42; is valid Java.
To strip special characters from its magic powers you have to escape them, by putting backslash \ in front of character. I.e. to match [ you type \[.
And since backslash acts as special character as well, to match it literally, you escape it the same way: \\.
And since you used / as patter delimiter, you need to escape its occurrences within pattern:
/^[A-Za-z0-9 _=\/~'"|{}();*:?+,.]*$/
The way to escape regex meta-characters is using a backslash (\), not a forwards slash (/).
[,] should be \[,\]
// should be \/
I have found the following regular expression
new RegExp("(^|\\s)hello(\\s|$)");
I refer http://www.javascriptkit.com/jsref/escapesequence.shtml for regular expressions..
But i cannot see \s escape sequence there..I know \s indicate whitespace character...
But what does the preceding \ do ..Which character is escaped?
I found similar regular expression in the Treewalker code in the following document http://ejohn.org/blog/getelementsbyclassname-speed-comparison/
The double \\ is to escape the backslash inside the string. In other word, \\ will be interpreted as \ for the regular expression.
The extra \ in this case is to escape the \ in the \s. Because we are inside a string declaration, you have to double up the \ to escape it. Once the string is processed and saved, it is reduced down to (^|\s)hello(\s|$)
The character immediately following the first \ is escaped. Normally \s escapes the s to mean "whitespace". In your example, the character which is escaped is \.
What you have is an expression which builds a regex (presumably to pass elsewhere) of (^|\s)hello(\s|$) — the word "hello" preceded either by whitespace or the start of the string, and followed by whitespace or the end of the string.
Essentially what the reg ex is doing, is looking for the opening and closing items of text surrounding the word hello and literally interpreting the '\s' as string content at the same time.
In laymans terms it's looking for a string that exactly matches:
|\shello\s|
As others have said the double \ is to escape the single \ so that instead of the reg ex engine looking for white-space it actually looks for '\s' as a string.
The ^ means start of line, the $ means end of line and the 2 | are interpreted as actual chars to look for
Lastly your start and end markers are bracketed () which means they will be extracted and placed in matches, which for you using C# means you can get at them by using:
myRegex.Matches.Group[1].Value
myRegex.Matches.Group[2].Value
1 being the beginning grouping, and 2 being the end.
I need to match these characters. This quote is from an API documentation (external to our company):
Valid characters: 0-9 A-Z a-z & # - . , ( ) / : ; ' # "
I used this Regex to match characters:
^[0-9a-z&#-\.,()/:;'""#]*$
However, this wrongly matches characters like %, $, and many other characters. What's wrong?
You can test this regular expression online using http://regexhero.net/tester/, and this regular expression is meant to work in both .NET and JavaScript.
You are not escaping the dash -, which is a reserved character. If you add replace the dash with \- then the regex no longer matches those characters between # and \
Move the literal - to the front of the character set:
^[-0-9a-z&#\.,()/:;'""#]*$
otherwise it is taken as specifying a range like when you use it in 0-9.
- sign, when not escaped, has special meaning in square brackets. #-\. is transformed into #-. (BTW, backslash before dot is not necessary in square brackets), which means "any character between # (ASCII 0x23) and . (ASCII 0x2E). The correct notation is
^[0-9a-z&#\-.,()/:;'"#]*$
The special characters in a character class are the closing bracket (]), the backslash (\), the caret (^) and the hyphen (-).
As such, you should either escape them with a backslash (\), or put them in a position where there is no ambiguity and they do not need escaping. In the case of a hyphen, this would be the first or last position.
You also do not need to escape the dot (.).
Your regex thus becomes:
^[-0-9a-z&#.,()/:;'"#]*$
As a side note, there are many available regex evaluators which provide code hinting. This way, you can simply hover your mouse over your regular expression and it can be explained in English words.
One such free one is RegExr.
Typing your original regex in it and hovering over the hyphen shows:
Matches characters in the range '#-\'
Try that
^[0-9a-zA-Z\&\#\-\.\,\(\)\/\:\;\'\"\#]*$
var MM = '\' + obj[0]['MM '] + '/';
I get two errors while using this code...
missing; before statement and
unterminated string literal
The character \ is "special" because it's used to allow the use of all printable characters in strings. In your case '\' is not a string composed by the only character \, but the beginning of a string starting with the single quote character '.
For exampe if you want the string Hello Andrea "6502" Griffini you can use single quotes
string1 = 'Hello Andrea "6502" Griffini';
and if you want single quotes in the string you can do the opposite
string2 = "Hello Andrea '6502' Griffini";
But what if you want both kind of quotes in the same string? This is where the escape \ character comes handy:
string3 = "'llo Andrea \"6502\" Griffini";
Basically \ before a quote or double quote in a string tells javascript that the following character is just a regular character, with no special meaning attached to it.
Note that the very same character is also used in regular expressions... for example if you want to look for an open bracket [ you must prefix it with a backslash because [ in a regular expression has a special meaning.
The escape is also used to do the opposite... in a string if you put a backslash in front of a normal character you are telling javascript that that character is indeed special... for example
alert("This is\na test");
In the above line the \n sequence means a newline code, so the message displayed will be on two lines ("This is" and "a test").
You may now wonder... what if I need a backslash character in my string? Just double it in that case. In your code for example just use '\\'.
Here is a table for the possible meanings of backslash in strings
\" just a regular double-quote character, it doesn't end the string
\' just a regular single-quote character, it doesn't end the string
\b a backspace character (ASCII code 0x08)
\t a tab character (ASCII code 0x09)
\n a newline character (ASCII code 0x0A)
\v a vertical tab character (ASCII code 0x0B)
\f a form feed character (ASCII code 0x0C)
\r a carriage return character (ASCII code 0x0D)
\033 the character with ASCII code 033 octal = 27 ("ESC" in this case)
\x41 the character with ASCII code 0x41 = 65 ("A" in this case)
\u05D0 the unicode character 0x05D0 (Aleph from the Hebrew charset)
\\ just regular backslash character, not an escape prefix
\ is an escape character. You'll have to double it to literally mean a backslash character, otherwise it'll augment the following character (In this case the next single quote)
You need to properly escape the backslash:
var lastMenstrualPeriod = '\\' + obj[0]['LastMenstrualPeriod'] + '/';
Being escape character, the JS "compiler" is expecting another character to follow, for example \n is newline constant, \t is tab etc.. so \\ is one single backslash in a string.
It is also mentioned in Douglas Crockford book.
You are forgetting to escape '\'
Do this:
var lastMenstrualPeriod = '\\' + obj[0]['LastMenstrualPeriod'] + '/';