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.
Related
I am using this regex in JavaScript which is to evaluate if a given string matches some german phone number patterns.
var reg = new RegExp("(?:\+\d+)?\s*(?:\(\d+\)\s*(?:[/–-]\s*)?)?\d+(?:\s*(?:[\s/–-]\s*)?\d+)*");
When using it I get this error:
SyntaxError: invalid quantifier
...eg = new RegExp("(?:\+\d+)\s*(?:\(\d+\)\s*(?:[/–-]\s*))\d+(?:\s*(?:[\s/–-]\s*)\d...
I'm trying hard to learn reading regular expressions, but can not understand them full yet. I did not write this expression by myself and I am struggling to understand it.
Why I am getting this error?
Because you're using a string litteral, you need to escape each backslash:
(?:\\+\\d+)?\\s*(?:\\(\\d+\\)\\s*(?:[/–-]\\s*)?)?\\d+(?:\\s*(?:[\\s/–-]\\s*)?\\d+)*
The other solution would be to use the regex litteral:
var reg = /(?:\+\d+)?\s*(?:\(\d+\)\s*(?:[\/–-]\s*)?)?\d+(?:\s*(?:[\s\/–-]\s*)?\d+)*/;
new RegExp requires a string, and since backslashes already have meaning inside strings, they need to be escaped again.
In your case, though, you're using a static pattern, so you'd be better off with a literal:
var reg = /(?:\+\d+)?\s*(?:\(\d+\)\s*(?:[\/–-]\s*)?)?\d+(?:\s*(?:[\s\/–-]\s*)?\d+)*/;
Just be aware that you need to escape / here ;)
As an additional tip, you can simplify the need to escape things to some extent by doing stuff like [+] for a literal +. I think it looks nicer then \+, but that's just my opinion.
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.
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.
I'd like to compare 2 strings with each other, but I got a little problem with the Brackets.
The String I want to seek looks like this:
CAPPL:LOCAL.L_hk[1].vorlauftemp_soll
Quoting those to bracket is seemingly useless.
I tried it with this code
var regex = new RegExp("CAPPL:LOCAL.L_hk\[1\].vorlauftemp_soll","gi");
var value = "CAPPL:LOCAL.L_hk[1].vorlauftemp_soll";
regex.test(value);
Somebody who can help me??
It is useless because you're using string. You need to escape the backslashes as well:
var regex = new RegExp("CAPPL:LOCAL.L_hk\\[1\\].vorlauftemp_soll","gi");
Or use a regex literal:
var regex = /CAPPL:LOCAL.L_hk\[1\].vorlauftemp_soll/gi
Unknown escape characters are ignored in JavaScript, so "\[" results in the same string as "[".
In value, you have (1) instead of [1]. So if you expect the regular expression to match and it doesn't, it because of that.
Another problem is that you're using "" in your expression. In order to write regular expression in JavaScript, use /.../g instead of "...".
You may also want to escape the dot in your expression. . means "any character that is not a line break". You, on the other hand, wants the dot to be matched literally: \..
You are generating a regular expression (in which [ is a special character that can be escaped with \) using a string (in which \ is a special character).
var regex = /CAPPL:LOCAL.L_hk\[1\].vorlauftemp_soll/gi;
I am trying use javascript regular expressions to do some matching and I found a really unusual behavior that I was hoping someone could explain.
The string I was trying to match was: " 0 (IR) " and the code block was
finalRegEx = new RegExp("[0-9]");
match = finalRegEx.exec(str);
except that when I put "\d" instead of "[0-9]" it didn't find a match. I'm really confused by this.
If you use RegExp with "\d" to build the regular expression, the "\d" will result in just "d". Either use two back slashes to escape the slash like "\\d" or simply use the regular expression literals /…/ instead:
match = /\d/.exec(str)
You need to escape it because you're using the constructor, otherwise it matches d literally:
new RegExp('\\d').test('1')
new RegExp should only be used for dynamic matching. Otherwise use a literal:
var foo = /\d/;
foo.test(1)
You probably need to escape the backslash: finalRegEx = new RegExp("\\d");