I am trying to build a regexp from static text plus a variable in javascript. Obviously I am missing something very basic, see comments in code below. Help is very much appreciated:
var test_string = "goodweather";
// One regexp we just set:
var regexp1 = /goodweather/;
// The other regexp we built from a variable + static text:
var regexp_part = "good";
var regexp2 = "\/" + regexp_part + "weather\/";
// These alerts now show the 2 regexp are completely identical:
alert (regexp1);
alert (regexp2);
// But one works, the other doesn't ??
if (test_string.match(regexp1))
alert ("This is displayed.");
if (test_string.match(regexp2))
alert ("This is not displayed.");
First, the answer to the question:
The other answers are nearly correct, but fail to consider what happens when the text to be matched contains a literal backslash, (i.e. when: regexp_part contains a literal backslash). For example, what happens when regexp_part equals: "C:\Windows"? In this case the suggested methods do not work as expected (The resulting regex becomes: /C:\Windows/ where the \W is erroneously interpreted as a non-word character class). The correct solution is to first escape any backslashes in regexp_part (the needed regex is actually: /C:\\Windows/).
To illustrate the correct way of handling this, here is a function which takes a passed phrase and creates a regex with the phrase wrapped in \b word boundaries:
// Given a phrase, create a RegExp object with word boundaries.
function makeRegExp(phrase) {
// First escape any backslashes in the phrase string.
// i.e. replace each backslash with two backslashes.
phrase = phrase.replace(/\\/g, "\\\\");
// Wrap the escaped phrase with \b word boundaries.
var re_str = "\\b"+ phrase +"\\b";
// Create a new regex object with "g" and "i" flags set.
var re = new RegExp(re_str, "gi");
return re;
}
// Here is a condensed version of same function.
function makeRegExpShort(phrase) {
return new RegExp("\\b"+ phrase.replace(/\\/g, "\\\\") +"\\b", "gi");
}
To understand this in more depth, follows is a discussion...
In-depth discussion, or "What's up with all these backslashes!?"
JavaScript has two ways to create a RegExp object:
/pattern/flags - You can specify a RegExp Literal expression directly, where the pattern is delimited using a pair of forward slashes followed by any combination of the three pattern modifier flags: i.e. 'g' global, 'i' ignore-case, or 'm' multi-line. This type of regex cannot be created dynamically.
new RegExp("pattern", "flags") - You can create a RegExp object by calling the RegExp() constructor function and pass the pattern as a string (without forward slash delimiters) as the first parameter and the optional pattern modifier flags (also as a string) as the second (optional) parameter. This type of regex can be created dynamically.
The following example demonstrates creating a simple RegExp object using both of these two methods. Lets say we wish to match the word "apple". The regex pattern we need is simply: apple. Additionally, we wish to set all three modifier flags.
Example 1: Simple pattern having no special characters: apple
// A RegExp literal to match "apple" with all three flags set:
var re1 = /apple/gim;
// Create the same object using RegExp() constructor:
var re2 = new RegExp("apple", "gim");
Simple enough. However, there are significant differences between these two methods with regard to the handling of escaped characters. The regex literal syntax is quite handy because you only need to escape forward slashes - all other characters are passed directly to the regex engine unaltered. However, when using the RegExp constructor method, you pass the pattern as a string, and there are two levels of escaping to be considered; first is the interpretation of the string and the second is the interpretation of the regex engine. Several examples will illustrate these differences.
First lets consider a pattern which contains a single literal forward slash. Let's say we wish to match the text sequence: "and/or" in a case-insensitive manner. The needed pattern is: and/or.
Example 2: Pattern having one forward slash: and/or
// A RegExp literal to match "and/or":
var re3 = /and\/or/i;
// Create the same object using RegExp() :
var re4 = new RegExp("and/or", "i");
Note that with the regex literal syntax, the forward slash must be escaped (preceded with a single backslash) because with a regex literal, the forward slash has special meaning (it is a special metacharacter which is used to delimit the pattern). On the other hand, with the RegExp constructor syntax (which uses a string to store the pattern), the forward slash does NOT have any special meaning and does NOT need to be escaped.
Next lets consider a pattern which includes a special: \b word boundary regex metasequence. Say we wish to create a regex to match the word "apple" as a whole word only (so that it won't match "pineapple"). The pattern (as seen by the regex engine) needs to be: \bapple\b:
Example 3: Pattern having \b word boundaries: \bapple\b
// A RegExp literal to match the whole word "apple":
var re5 = /\bapple\b/;
// Create the same object using RegExp() constructor:
var re6 = new RegExp("\\bapple\\b");
In this case the backslash must be escaped when using the RegExp constructor method, because the pattern is stored in a string, and to get a literal backslash into a string, it must be escaped with another backslash. However, with a regex literal, there is no need to escape the backslash. (Remember that with a regex literal, the only special metacharacter is the forward slash.)
Backslash SOUP!
Things get even more interesting when we need to match a literal backslash. Let's say we want to match the text sequence: "C:\Program Files\JGsoft\RegexBuddy3\RegexBuddy.exe". The pattern to be processed by the regex engine needs to be: C:\\Program Files\\JGsoft\\RegexBuddy3\\RegexBuddy\.exe. (Note that the regex pattern to match a single backslash is \\ i.e. each must be escaped.) Here is how you create the needed RegExp object using the two JavaScript syntaxes
Example 4: Pattern to match literal back slashes:
// A RegExp literal to match the ultimate Windows regex debugger app:
var re7 = /C:\\Program Files\\JGsoft\\RegexBuddy3\\RegexBuddy\.exe/;
// Create the same object using RegExp() constructor:
var re8 = new RegExp(
"C:\\\\Program Files\\\\JGsoft\\\\RegexBuddy3\\\\RegexBuddy\\.exe");
This is why the /regex literal/ syntax is generally preferred over the new RegExp("pattern", "flags") method - it completely avoids the backslash soup that can frequently arise. However, when you need to dynamically create a regex, as the OP needs to here, you are forced to use the new RegExp() syntax and deal with the backslash soup. (Its really not that bad once you get your head wrapped 'round it.)
RegexBuddy to the rescue!
RegexBuddy is a Windows app that can help with this backslash soup problem - it understands the regex syntaxes and escaping requirements of many languages and will automatically add and remove backslashes as required when pasting to and from the application. Inside the application you compose and debug the regex in native regex format. Once the regex works correctly, you export it using one of the many "copy as..." options to get the needed syntax. Very handy!
You should use the RegExp constructor to accomplish this:
var regexp2 = new RegExp(regexp_part + "weather");
Here's a related question that might help.
The forward slashes are just Javascript syntax to enclose regular expresions in. If you use normal string as regex, you shouldn't include them as they will be matched against. Therefore you should just build the regex like that:
var regexp2 = regexp_part + "weather";
I would use :
var regexp2 = new RegExp(regexp_part+"weather");
Like you have done that does :
var regexp2 = "/goodweather/";
And after there is :
test_string.match("/goodweather/")
Wich use match with a string and not with the regex like you wanted :
test_string.match(/goodweather/)
While this solution may be overkill for this specific question, if you want to build RegExps programmatically, compose-regexp can come in handy.
This specific problem would be solved by using
import {sequence} from 'compose-regexp'
const weatherify = x => sequence(x, /weather/)
Strings are escaped, so
weatherify('.')
returns
/\.weather/
But it can also accept RegExps
weatherify(/./u)
returns
/.weather/u
compose-regexp supports the whole range of RegExps features, and let one build RegExps from sub-parts, which helps with code reuse and testability.
Related
The regex allows chars that are: alphanumeric, space, '-', '_', '&', '()' and '/'
this is the expression
[\s\/\)\(\w&-]
I have tested this in various online testers and know it works, I just can't get it to work correctly in code. I get sysntax errors with anything I try.. any suggestions?
var programProductRegex = new RegExp([\s\/\)\(\w&-]);
You can use the regular expression syntax:
var programProductRegex = /[\s\/\)\(\w&-]/;
You use forward slashes to delimit the regex pattern.
If you use the RegExp object constructor you need to pass in a string. Because backslashes are special escape characters inside JavaScript strings and they're also escape characters in regular expressions, you need to use two backslashes to do a regex escape inside a string. The equivalent code using a string would then be:
var programProductRegex = new RegExp("[\\s\\/\\)\\(\\w&-]");
All the backslashes that were in the original regular expression need to be escaped in the string to be correctly interpreted as backslashes.
Of course the first option is better. The constructor is helpful when you obtain a string from somewhere and want to make a regular expression out of it.
var programProductRegex = new RegExp(userInput);
If you are using a String and want to escape characters like (, you need to write \\( (meaning writing backslash, then the opening parenthesis => escaping it).
If you are using the RegExp object, you only need one backslash for each character (like \()
Enclose your regex with delimiters:
var programProductRegex = /[\s\/)(\w&-]/;
This question already has answers here:
JavaScript Regex, where to use escape characters?
(3 answers)
Closed 8 years ago.
I am having a small issue with placing a RegExp pattern inside a string, I have 2 patterns which are both really the same. The first doesn't work I presume due to the \d - is it being seen as an escape character?
var pattern = '^.{1,5}-\d{1,5}$'; // Doesn't work
var pattern = '^[a-zA-Z]{1,5}-[0-9]{1,5}$'; // Works
Is there anyway around this ? apart from replacing the \d with [0-9]?
Here is the extra code I am using
var regex = new RegExp(pattern);
var result = regex.test(value);
Thanks in advance
As found in the documentation you have several different ways to create a RegExp
Regular expression literal,
var regex = /^.{1,5}-\d{1,5}$/;
Constructor function of the RegExp object
var regex = new RegExp("^.{1,5}-\\d{1,5}$");
since it is a string, you need to escape any \
Same for \w and other backslashed chars
The second version is mostly used if you have variables you need to add to the regexp
If you want the way your are writing the regex to work, you can double escape the d:
var pattern = '^.{1,5}-\\d{1,5}$'; // Should work
var regex = new RegExp(pattern);
Otherwise, you can use the regex directly using the delimiters /:
var pattern = /^.{1,5}-\d{1,5}$/;
In the first instance, you are storing the pattern in a string, and the actual characters that are being passed to the variables are: ^.{1,5}-\d{1,5}$ because \d has no meaning in a string, but \\d is a backslash and a literal d. You can try putting a backslash in a string:
console.log('\'); // Won't run
console.log(' \ '); // Returns a space
console.log('\n'); // Returns a newline character
So that if you mean a literal backslash, you have to escape it.
Using:
var pattern = '^.{1,5}-\\d{1,5}$'; // Should work
var regex = new RegExp(pattern);
should be faster though, if you are using the regex several times, because here, you are compiling the regex so that you can use it multiple times.
The other way will require compiling the regex each time it is called for.
I have a custom regular expression which I use to detect whole numbers, fractions and floats.
var regEx = new RegExp("^((^[1-9]|(0\.)|(\.))([0-9]+)?((\s|\.)[0-9]+(/[0-9])?)?)$");
var quantity = 'd';
var matched = quantity.match(regEx);
alert(matched);
(The code is also found here: http://jsfiddle.net/aNb3L/ .)
The problem is that for a single letter it matches, and I can't figure out why. But for more letters it fails(which is good).
Disclaimer: I am new to regular expressions, although in http://gskinner.com/RegExr/ it doesn't match a single letter
It's easier to use straight regular expression syntax:
var regEx = /^((^[1-9]|(0\.)|(\.))([0-9]+)?((\s|\.)[0-9]+(\/[0-9])?)?)$/;
When you use the RegExp constructor, you have to double-up on the backslashes. As it is, your code only has single backslashes, so the \. subexpressions are being treated as . — and that's how single non-digit characters are slipping through.
Thus yours would also work this way:
var regEx = new RegExp("^((^[1-9]|(0\\.)|(\\.))([0-9]+)?((\\s|\\.)[0-9]+(/[0-9])?)?)$");
This happens because the string syntax also uses backslash as a quoting mechanism. When your regular expression is first parsed as a string constant, those backslashes are stripped out if you don't double them. When the string is then passed to the regular expression parser, they're gone.
The only time you really need to use the RegExp constructor is when you're building up the regular expression dynamically or when it's delivered to your code via JSON or something.
Well, for a whole number this would be your regex:
/^(0|[1-9]\d*)$/
Then you have to account for the possibility of a float:
/^(0|[1-9]\d*)(.\d+)?$/
Then you have to account for the possibility of a fraction:
/^(0|[1-9]\d*)((.\d+)|(\/[1-9]\d*)?$/
To me this regex is much easier to read than your original, but it's up to you of course.
What am I doing wrong as both strings below are returning false when tested below?
var pattern = "^[\s\da-zA-ZåäöÅÄÖ_]+$"
var reg = new RegExp(pattern);
console.log(reg.test("This should be invalid as it is full with invalid chars. #!¤%&/()=?"));
console.log(reg.test("This is an valid string, despite that Swedish chars such as ÅÄÖ are used"));
You need to double-up on the backslashes in the pattern.
var pattern = "^[\\s\\da-zA-ZåäöÅÄÖ_]+$"
The problem is that when you build regular expression objects that way, there are two passes made over the string: one to interpret it as a string, and then a second to interpret it as a regular expression. Both of those micro-syntaxes use \ to mean something, so by doubling them you get a single backslash out of the string constant parse.
If your pattern is really a constant, and not something that you construct dynamically from separate parts, then you can just use the native syntax for regular expressions:
var pattern = /^[\s\da-zA-ZåäöÅÄÖ_]+$/;
Only one backslash is necessary because the pattern is only parsed once, as a regular expression.
Why isn't my code around the test function in Object RegExp working the same way as the regex pattern. Am I missing something or Am I using the wrong escape regex
<html>
<body>
<script type="text/javascript">
var str = "info#test.com";
//This isn't working
var regStr = "^([\w-]+(?:\.[\w-]+)*)#((?:[\w-]+\.)*\w[\w-]{0,66})\.([a-z]{2,6}(?:\.[a-z]{2})?)$"; //This string can be any regex get from XSLT
//Escape function get from: http://stackoverflow.com/a/6969486/193850
regStr = regStr.replace(/[\-\[\]\/\{\}\(\)\*\+\?\.\\\^\$\|]/g, "\\$&");
console.log(regStr); // \^\(\[w\-\]\+\(\?:\.\[w\-\]\+\)\*\)#\(\(\?:\[w\-\]\+\.\)\*w\[w\-\]\{0,66\}\)\.\(\[a\-z\]\{2,6\}\(\?:\.\[a\-z\]\{2\}\)\?\)\$
var re = new RegExp(regStr , "i");
console.log(re.test(str)); //false
var filter=/^([\w-]+(?:\.[\w-]+)*)#((?:[\w-]+\.)*\w[\w-]{0,66})\.([a-z]{2,6}(?:\.[a-z]{2})?)$/i
console.log(filter.test(str)); //true
</script>
</body>
</html>
You have to double your backslashes when you write a regular expression as a string.
Why? The string literal syntax also observes its own backslash-quoting convention, for things like quote characters, newlines, etc. Therefore, when JavaScript parses your string constant that contains the regular expression, the backslashes will disappear. Thus, you need to quote them with another backslash so that when you pass the string to the RegExp constructor it sees the regular expression you actually intended.
See, there's a confusion. The function you've used is a nice way of preprocessing your string-to-be regexes so you don't have to worry about escaping regex metacharacters - i.e., symbols that will control the regex behaviour, and not just taken literally.
But the point is that your string has already been parsed before it was taken by this escaping function: \w and \. sequences became just w and . respectively, the preceding slash was lost.
For characters not listed in Table 2.1, a preceding backslash is
ignored, but this usage is deprecated and should be avoided.
The escaper function, actually, did restore the slash before the ., but w wasn't special for it in any kind. ) Therefore the string that went into RegExp constructor had [w] instead of [\w].
It's actually quite easy to check: just console.log(regStr) after the replacement operation.