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Differences between Javascript regexp literal and constructor
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Closed 7 years ago.
I have to put a given variable into a regular expression. When I do it with hard coded data it works. Here is my code for that
/^(?=.*[a-z])+(?=.*[A-Z])+(?=.*[0-9##$-/:-?{-~!"^_`\[\]])+((?!ccarn).)*$/
This should ( and does ) look for a word (password in this case) that is case sensitive, has at least one capitol and one lowercase letter, and one number or symbol. It cannot, however, contain the word "ccarn" in it. Again when I put this in as my regex all works out. When I try to turn it into a string that gets passed in, it doesn't work. Here is my code for that
var regex = new RegExp('/^(?=.*[a-z])+(?=.*[A-Z])+(?=.*[0-9##$-/:-?{-~!"^_`\[\]])+((?!' + $scope.username + ').)*$/');
I feel like I may just be missing something in translation/transition, but can't seem to get it right. TIA
When you use the new RegExp() constructor to construct a regex from a string, you shouldn't include the leading and trailing / within the string. The /.../ form is only to be used when specifying a regex literal, which isn't what you're doing here.
When you do, say, var r = new RegExp('/foo/'), the regex you're actually getting is equivalent to doing var r = /\/foo\//, which clearly isn't what you want. So your constructor should actually look like this:
var regex = new RegExp('^(?=.*[a-z])+(?=.*[A-Z])+(?=.*[0-9##$-/:-?{-~!"^_`\[\]])+((?!' + $scope.username + ').)*$');
// ↑↑ ↑↑
// no "/" at the locations pointed to above
You probably also need to double your backslashes (since backslashes are escape characters in strings, but not in regex literals). So, [0-9##$-/:-?{-~!"^_`\[\]] needs to become [0-9##$-/:-?{-~!"^_`\\[\\]].
If you look closely the '/' character gets delimited when you give it inside the quotes so essentially the
var regex = new RegExp('/^(?=.*[a-z])+(?=.*[A-Z])+(?=.*[0-9##$-/:-?{-~!"^_`\[\]])+((?!' + $scope.username + ').)*$/');
The regular expression would be like this
/\/^(?=.*[a-z])+(?=.*[A-Z])+(?=.*[0-9##$-\/:-?{-~!"^_`[]])+((?!ccarn).)*$\//
The right way to go is to remove the '/' character from the RegEx and it should work
var regex = new RegExp('/^(?=.*[a-z])+(?=.*[A-Z])+(?=.*[0-9##$-/:-?{-~!"^_`\[\]])+((?!' + $scope.username + ').)*$/');
The output for the above would be
/^(?=.*[a-z])+(?=.*[A-Z])+(?=.*[0-9##$-\/:-?{-~!"^_`[]])+((?!ccarn).)*$/
which is exactly what you need ?
Hope it helps
Please before doing anything else, read a regex tutorial!
Mistakes:
A lookahead is a zero width assertion ( in other words, it's only a test and doesn't match anything ), putting a quantifier for a zero width assertion doesn't make any sense: (?=.*[a-z])+ (it is like repeating something empty, zero or more times. Note that the regex engine will protest if you write something like this.)
When you use the oop syntax to define a pattern (ie:var pattern = new RegExp("...), you don't need to add delimiters. You need to put double backslashes instead simple backslashes.
Related
var serialNumber = $('#SerialNumber').val();
var serialNumberPattern = new RegExp('^[\s\da-zA-z\-.]+$');
if (!serialNumberPattern.test(serialNumber)) {
}
Above is the code I am using to validate a serial number which has alphanumeric characters, dots (.), dashes (-), and slashes (/) in it but somehow it's not working. Where am I going wrong? Please help.
When you're passing regex to RegExp constructor which uses " as regex delimiter, you have to escape all the backslashes one more time. Or otherwise it would be treated as an escape sequence.
var serialNumberPattern = new RegExp("^[\\s\\da-zA-Z.-]+$");
alphanumeric,dot(.),Dash(-),Slash(/) in it.
var serialNumberPattern = new RegExp("^[\\da-zA-Z./-]+$");
Just use /^[\s\da-zA-Z\-.\/]+$/, it's simple and works just fine.
You should only use the RegExp constructor when parts of the expression use a variable. This is not true in your case and just adds additional confusion.
document.write(/^[\s\da-zA-Z\-.\/]+$/.test('23 43-89'))
In Javascript, I want to match a string pattern that goes something like:
\left((some expression here to be matched)\right)
It's a little more complicated than that, so I have to define my pattern using the RegExp constructor. The simplified version is:
var pattern_string = '\\left\\((' + EXPRESSIONpattern + ')\\\\right\\)' ;
var pattern_regexp = new RegExp(pattern_string, 'g');
I realize that \r is the carriage return character in a string, hence having the four back slashes in the pattern_string before the r. Upon implementing, the only way I could get it to work was to also treat \l as a special string character and use:
var pattern_string = '\\\\left\\((' + EXPRESSIONpattern + '\\\\right\\)' ;
var pattern_regexp = new RegExp(pattern_string, 'g');
Why do I need to double escape the l? I can't find a reference that says \l is a special string sequence. What does \l mean in a string? Can someone point me to a reference that includes all characters that need to be double escaped in a string? It would help greatly with debugging to know for sure when I need to double escape.
Thanks in advance for your help,
T
I've seen plenty of regex examples that will not allow any special characters. I need one that requires at least one special character.
I'm looking at a C# regex
var regexItem = new Regex("^[a-zA-Z0-9 ]*$");
Can this be converted to use with javascript? Do I need to escape any of the characters?
Based an example I have built this so far:
var regex = "^[a-zA-Z0-9 ]*$";
//Must have one special character
if (regex.exec(resetPassword)) {
isValid = false;
$('#vsResetPassword').append('Password must contain at least 1 special character.');
}
Can someone please identify my error, or guide me down a more efficient path? The error I'm currently getting is that regex has no 'exec' method
Your problem is that "^[a-zA-Z0-9 ]*$" is a string, and you need a regex:
var regex = /^[a-zA-Z0-9 ]*$/; // one way
var regex = new RegExp("^[a-zA-Z0-9 ]*$"); // another way
[more information]
Other than that, your code looks fine.
In javascript, regexs are formatted like this:
/^[a-zA-Z0-9 ]*$/
Note that there are no quotation marks and instead you use forward slashes at the beginning and end.
In javascript, you can create a regular expression object two ways.
1) You can use the constructor method with the RegExp object (note the different spelling than what you were using):
var regexItem = new RegExp("^[a-zA-Z0-9 ]*$");
2) You can use the literal syntax built into the language:
var regexItem = /^[a-zA-Z0-9 ]*$/;
The advantage of the second is that you only have to escape a forward slash, you don't have to worry about quotes. The advantage of the first is that you can programmatically construct a string from various parts and then pass it to the RegExp constructor.
Further, the optional flags for the regular expression are passed like this in the two forms:
var regexItem = new RegExp("^[A-Z0-9 ]*$", "i");
var regexItem = /^[A-Z0-9 ]*$/i;
In javascript, it seems to be a more common convention to the user /regex/ method that is built into the parser unless you are dynamically constructing a string or the flags.
I've a string done like this: "http://something.org/dom/My_happy_dog_%28is%29cool!"
How can I remove all the initial domain, the multiple underscore and the percentage stuff?
For now I'm just doing some multiple replace, like
str = str.replace("http://something.org/dom/","");
str = str.replace("_%28"," ");
and go on, but it's really ugly.. any help?
Thanks!
EDIT:
the exact input would be "My happy dog is cool!" so I would like to get rid of the initial address and remove the underscores and percentage and put the spaces in the right place!
The problem is that trying to put a regex on Chrome "something goes wrong". Is it a problem of Chrome or my regex?
I'd suggest:
var str = "http://something.org/dom/My_happy_dog_%28is%29cool!";
str.substring(str.lastIndexOf('/')+1).replace(/(_)|(%\d{2,})/g,' ');
JS Fiddle demo.
The reason I took this approach is that RegEx is fairly expensive, and is often tricky to fine tune to the point where edge-cases become less troublesome; so I opted to use simple string manipulation to reduce the RegEx work.
Effectively the above creates a substring of the given str variable, from the index point of the lastIndexOf('/') (which does exactly what you'd expect) and adding 1 to that so the substring is from the point after the / not before it.
The regex: (_) matches the underscores, the | just serves as an or operator and the (%\d{2,}) serves to match digit characters that occur twice in succession and follow a % sign.
The parentheses surrounding each part of the regex around the |, serve to identify matching groups, which are used to identify what parts should be replaced by the ' ' (single-space) string in the second of the arguments passed to replace().
References:
lastIndexOf().
replace().
substring().
You can use unescape to decode the percentages:
str = unescape("http://something.org/dom/My_happy_dog_%28is%29cool!")
str = str.replace("http://something.org/dom/","");
Maybe you could use a regular expression to pull out what you need, rather than getting rid of what you don't want. What is it you are trying to keep?
You can also chain them together as in:
str.replace("http://something.org/dom/", "").replace("something else", "");
You haven't defined the problem very exactly. To get rid of all stretches of characters ending in %<digit><digit> you'd say
var re = /.*%\d\d/g;
var str = str.replace(re, "");
ok, if you want to replace all that stuff I think that you would need something like this:
/(http:\/\/.*\.[a-z]{3}\/.*\/)|(\%[a-z0-9][a-z0-9])|_/g
test
var string = "http://something.org/dom/My_happy_dog_%28is%29cool!";
string = string.replace(/(http:\/\/.*\.[a-z]{3}\/.*\/)|(\%[a-z0-9][a-z0-9])|_/g,"");
I'm trying to write a regex for use in javascript.
var script = "function onclick() {loadArea('areaog_og_group_og_consumedservice', '\x26roleOrd\x3d1');}";
var match = new RegExp("'[^']*(\\.[^']*)*'").exec(script);
I would like split to contain two elements:
match[0] == "'areaog_og_group_og_consumedservice'";
match[1] == "'\x26roleOrd\x3d1'";
This regex matches correctly when testing it at gskinner.com/RegExr/ but it does not work in my Javascript. This issue can be replicated by testing ir here http://www.regextester.com/.
I need the solution to work with Internet Explorer 6 and above.
Can any regex guru's help?
Judging by your regex, it looks like you're trying to match a single-quoted string that may contain escaped quotes. The correct form of that regex is:
'[^'\\]*(?:\\.[^'\\]*)*'
(If you don't need to allow for escaped quotes, /'[^']*'/ is all you need.) You also have to set the g flag if you want to get both strings. Here's the regex in its regex-literal form:
/'[^'\\]*(?:\\.[^'\\]*)*'/g
If you use the RegExp constructor instead of a regex literal, you have to double-escape the backslashes: once for the string literal and once for the regex. You also have to pass the flags (g, i, m) as a separate parameter:
var rgx = new RegExp("'[^'\\\\]*(?:\\\\.[^'\\\\]*)*'", "g");
while (result = rgx.exec(script))
print(result[0]);
The regex you're looking for is .*?('[^']*')\s*,\s*('[^']*'). The catch here is that, as usual, match[0] is the entire matched text (this is very normal) so it's not particularly useful to you. match[1] and match[2] are the two matches you're looking for.
var script = "function onclick() {loadArea('areaog_og_group_og_consumedservice', '\x26roleOrd\x3d1');}";
var parameters = /.*?('[^']*')\s*,\s*('[^']*')/.exec(script);
alert("you've done: loadArea("+parameters[1]+", "+parameters[2]+");");
The only issue I have with this is that it's somewhat inflexible. You might want to spend a little time to match function calls with 2 or 3 parameters?
EDIT
In response to you're request, here is the regex to match 1,2,3,...,n parameters. If you notice, I used a non-capturing group (the (?: ) part) to find many instances of the comma followed by the second parameter.
/.*?('[^']*')(?:\s*,\s*('[^']*'))*/
Maybe this:
'([^']*)'\s*,\s*'([^']*)'