I will try to make my question short yet understandable, I have a simple RegEx I use in javascript to check for characters that aren't alphanumeric (AKA Symbols). It would be "/[$-/:-?{-~!"^_`[]]/"
In javascript, doing
if(/[$-/:-?{-~!"^_`\[\]]/.test( string ))
just works, if any of those characters are in the string, it will give true, else, it will give false. I tried to do the same in PHP, the following way
if(preg_match('/[$-/:-?{-~!"^_`\[\]]/', $string ))
other regexes work when done this way, but this particular one simply will give false no matter what when ran in PHP.
Is there any reason to this? Am I doing something wrong? Does PHP comprehend regexes in a different way? What should I change to make it work?
Thanks for your time.
Since php uses PCRE, you will get a pattern error using delimiter / as seen here http://regex101.com/r/3ILGgE/1
So, it should be escaped correctly.
Using / as the delimiter, the string is
'/[$-\/:-?{-~!"^_`\[\]]/'
Using ~ as the delimiter, the string is
'~[$-/:-?{-\~!"^_`\[\]]~'
Also, be aware you have a couple of range's in the class $-/ and :-? and {-~
that will include the characters between the from/to range characters as well
and does not include the range character - itself as it is an operator.
Related
I'm trying to implement a username form validation in javascript where the username
can't start with numbers
can't have whitespaces
can't have any symbols but only One dot or One underscore or One dash
example of a valid username: the_user-one.123
example of invalid username: 1----- user
i've been trying to implement this for awhile but i couldn't figure out how to have only one of each allowed symbol:-
const usernameValidation = /(?=^[\w.-]+$)^\D/g
console.log(usernameValidation.test('1username')) //false
console.log(usernameValidation.test('username-One')) //true
How about using a negative lookahead at the start:
^(?!\d|.*?([_.-]).*\1)[\w.-]+$
This will check if the string
neither starts with digit
nor contains two [_.-] by use of capture and backreference
See this demo at regex101 (more explanation on the right side)
Preface: Due to my severe carelessness, I assumed the context was usage of the HTML pattern attribute instead of JavaScript input validation. I leave this answer here for posterity in case anyone really wants to do this with regex.
Although regex does have functionality to represent a pattern occuring consecutively within a certain number of times (via {<lower-bound>,<upper-bound>}), I'm not aware of regex having "elegant" functionality to enforce a set of patterns each occuring within a range of number of times but in any order and with other patterns possibly in between.
Some workarounds I can think of:
Make a regex that allows for one of each permutation of ordering of special characters (note: newlines added for readability):
^(?:
(?:(?:(?:[A-Za-z][A-Za-z0-9]*\.?)|\.)[A-Za-z0-9]*-?[A-Za-z0-9]*_?)|
(?:(?:(?:[A-Za-z][A-Za-z0-9]*\.?)|\.)[A-Za-z0-9]*_?[A-Za-z0-9]*-?)|
(?:(?:(?:[A-Za-z][A-Za-z0-9]*-?)|-)[A-Za-z0-9]*\.?[A-Za-z0-9]*_?)|
(?:(?:(?:[A-Za-z][A-Za-z0-9]*-?)|-)[A-Za-z0-9]*_?[A-Za-z0-9]*\.?)|
(?:(?:(?:[A-Za-z][A-Za-z0-9]*_?)|_)[A-Za-z0-9]*\.?[A-Za-z0-9]*-?)|
(?:(?:(?:[A-Za-z][A-Za-z0-9]*_?)|_)[A-Za-z0-9]*-?[A-Za-z0-9]*\.?)
)[A-Za-z0-9]*$
Note that the above regex can be simplified if you don't want usernames to start with special characters either.
Friendly reminder to also make sure you use the HTML attributes to enforce a minimum and maximum input character length where appropriate.
If you feel that regex isn't well suited to your use-case, know that you can do custom validation logic using javascript, which gives you much more control and can be much more readable compared to regex, but may require more lines of code to implement. Seeing the regex above, I would personally seriously consider the custom javascript route.
Note: I find https://regex101.com/ very helpful in learning, writing, and testing regex. Make sure to set the "flavour" to "JavaScript" in your case.
I have to admit that Bobble bubble's solution is the better fit. Here ia a comparison of the different cases:
console.log("Comparison between mine and Bobble Bubble's solution:\n\nusername mine,BobbleBubble");
["valid-usrId1","1nvalidUsrId","An0therVal1d-One","inva-lid.userId","anot-her.one","test.-case"].forEach(u=>console.log(u.padEnd(20," "),chck(u)));
function chck(s){
return [!!s.match(/^[a-zA-Z][a-zA-Z0-9._-]*$/) && ( s.match(/[._-]/g) || []).length<2, // mine
!!s.match(/^(?!\d|.*?([_.-]).*\1)[\w.-]+$/)].join(","); // Bobble bulle
}
The differences can be seen in the last three test cases.
I have the following string
class=use><em>use</em>
that when searched using us I want to transform into
class=use><em><b>us</b>e</em>
I've tried looking at relating answers but I can't quite get it working the way I want it to. I'm especially interested in this answer's callback approach.
Help appreciated
This is a good exercise for writing regular expressions, and here's a possible solution.
"useclass=use><em>use</em>".replace(/([^=]|^)(us)/g, "$1<b>$2</b>");
// returns "<b>us</b>eclass=use><em><b>us</b>e</em>"
([^=]|^) ensures that the prefix of any matched us is either not an equal sign, or it's the start of the string.
As #jamiec pointed out in the comments, if you are using this to parse/modify HTML, just stop right now. It's mathematically impossible to parse a CFG with a regular grammar (even with enhanced JS regexps you will have a bad time trying to achieve that.)
If you can make any assumptions about the structure of your document, you may be better off using an approach that operates on DOM elements directly rather than parsing the whole document with a regex.
Parsing HTML with a regex has certain problems that can be painful to deal with.
var element = document.querySelector('em');
element.innerHTML = element.innerHTML.replace('us', '<b>us</b>');
<div class=use><em>use</em>
</div>
I would first look for any character other than the equals sign [^=] and separate it by parentheses so that I can use it again in my replacement. Then another set of parentheses around the two characters us ought to do it:
var re = /([^=]|^)(us)/
That will give you two capture groups to work with (inside the parentheses), which you can represent with $1 and $2 in your replacement string.
str.replace( /([^=|^])(us)/, '$1<b>$2</b>' );
I'm trying to improve my understanding of Regex, but this one has me quite mystified.
I started with some text defined as:
var txt = "{\"columns\":[{\"text\":\"A\",\"value\":80},{\"text\":\"B\",\"renderer\":\"gbpFormat\",\"value\":80},{\"text\":\"C\",\"value\":80}]}";
and do a replace as follows:
txt.replace(/\"renderer\"\:(.*)(?:,)/g,"\"renderer\"\:gbpFormat\,");
which results in:
"{"columns":[{"text":"A","value":80},{"text":"B","renderer":gbpFormat,"value":80}]}"
What I expected was for the renderer attribute value to have it's quotes removed; which has happened, but also the C column is completely missing! I'd really love for someone to explain how my Regex has removed column C?
As an extra bonus, if you could explain how to remove the quotes around any value for renderer (i.e. so I don't have to hard-code the value gbpFormat in the regex) that'd be fantastic.
You are using a greedy operator while you need a lazy one. Change this:
"renderer":(.*)(?:,)
^---- add here the '?' to make it lazy
To
"renderer":(.*?)(?:,)
Working demo
Your code should be:
txt.replace(/\"renderer\"\:(.*?)(?:,)/g,"\"renderer\"\:gbpFormat\,");
If you are learning regex, take a look at this documentation to know more about greedyness. A nice extract to understand this is:
Watch Out for The Greediness!
Suppose you want to use a regex to match an HTML tag. You know that
the input will be a valid HTML file, so the regular expression does
not need to exclude any invalid use of sharp brackets. If it sits
between sharp brackets, it is an HTML tag.
Most people new to regular expressions will attempt to use <.+>. They
will be surprised when they test it on a string like This is a
first test. You might expect the regex to match and when
continuing after that match, .
But it does not. The regex will match first. Obviously not
what we wanted. The reason is that the plus is greedy. That is, the
plus causes the regex engine to repeat the preceding token as often as
possible. Only if that causes the entire regex to fail, will the regex
engine backtrack. That is, it will go back to the plus, make it give
up the last iteration, and proceed with the remainder of the regex.
Like the plus, the star and the repetition using curly braces are
greedy.
Try like this:
txt = txt.replace(/"renderer":"(.*?)"/g,'"renderer":$1');
The issue in the expression you were using was this part:
(.*)(?:,)
By default, the * quantifier is greedy by default, which means that it gobbles up as much as it can, so it will run up to the last comma in your string. The easiest solution would be to turn that in to a non-greedy quantifier, by adding a question mark after the asterisk and change that part of your expression to look like this
(.*?)(?:,)
For the solution I proposed at the top of this answer, I also removed the part matching the comma, because I think it's easier just to match everything between quotes. As for your bonus question, to replace the matched value instead of having to hardcode gbpFormat, I used a backreference ($1), which will insert the first matched group into the replacement string.
Don't manipulate JSON with regexp. It's too likely that you will break it, as you have found, and more importantly there's no need to.
In addition, once you have changed
'{"columns": [..."renderer": "gbpFormat", ...]}'
into
'{"columns": [..."renderer": gbpFormat, ...]}' // remove quotes from gbpFormat
then this is no longer valid JSON. (JSON requires that property values be numbers, quoted strings, objects, or arrays.) So you will not be able to parse it, or send it anywhere and have it interpreted correctly.
Therefore you should parse it to start with, then manipulate the resulting actual JS object:
var object = JSON.parse(txt);
object.columns.forEach(function(column) {
column.renderer = ghpFormat;
});
If you want to replace any quoted value of the renderer property with the value itself, then you could try
column.renderer = window[column.renderer];
Assuming that the value is available in the global namespace.
This question falls into the category of "I need a regexp, or I wrote one and it's not working, and I'm not really sure why it has to be a regexp, but I heard they can do all kinds of things, so that's just what I imagined I must need." People use regexps to try to do far too many complex matching, splitting, scanning, replacement, and validation tasks, including on complex languages such as HTML, or in this case JSON. There is almost always a better way.
The only time I can imagine wanting to manipulate JSON with regexps is if the JSON is broken somehow, perhaps due to a bug in server code, and it needs to be fixed up in order to be parseable.
Could someone please tell me why this RegEx fails?
http://jsfiddle.net/SrKPG/
^(\+[0-9]+ )[1-9]{2,} [0-9]{2,}(\-[0-9]+|)$
The funny thing is - when I test it at http://jsregex.com/ it works.
But in my code it fails.
The reason you're failing to match is because your second sequence of numbers does not accept zeroes:
^([+][0-9]+ )[1-9]{2,} [0-9]{2,}(\-[0-9]+|)$
+43 660 1234556
It fails because you write it as a string, without escaping the \.
You could write
var regex = "^(\\+[0-9]+ )[1-9]{2,} [0-9]{2,}(\\-[0-9]+|)$";
But, instead of using a string and the RegExp constructor, you should directly use a regex literal :
text.match(/^(\+[0-9]+ )[1-9]{2,} [0-9]{2,}(\-[0-9]+|)$/g);
You were also refusing 0 in the middle, which doesn't comply with your test string. It seems that what you want is
text.match(/^(\+[0-9]+ )[0-9]{2,} [0-9]{2,}(\-[0-9]+|)$/g);
Yours
"^(\+[0-9]+ )[1-9]{2,} [0-9]{2,}(\-[0-9]+|)$"
Correct
"^(\\+[0-9]+ )[1-9]{2,} [0-9]{2,}(-[0-9]+|)$"
The double escaping is a requirement of JavaScript string literals. It has nothing to do with regex.
Upon parsing your program your string literal becomes "^(+[0-9]+ )[1-9]{2,} [0-9]{2,}(-[0-9]+|)$" in memory, because \+ (as opposed to, let's say, \n) has no meaning in JS strings.
At this time the regex engine complains about the lone + that follows nothing.
Note that the something-or-nothing (something|) is better written as (something)?.
Apart from that: Thou shalt not use regex to validate phone numbers.
EDIT: The proof is in the comments. ;)
I'm trying to find out if a string contains css code with this expression:
var pattern = new RegExp('\s(?[a-zA-Z-]+)\s[:]{1}\s*(?[a-zA-Z0-9\s.#]+)[;]{1}');
But I get "invalid regular expression" error on the line above...
What's wrong with it?
found the regex here: http://www.catswhocode.com/blog/10-regular-expressions-for-efficient-web-development
It's for PHP but it should work in javascript too, right?
What are the ? at the start of the two [a-zA-z-] blocks for? They look wrong to me.
The ? is unfortunately somewhat overload in regexp syntax, it can have three different meanings that I know of, and none of them match what I see in your example.
Also, your \s sequences need the backslash escaping because this is a string - they should look like \\s. To avoid escaping, just use the /.../ syntax instead of new Regexp("...").
That said, even that is insufficient - the regexp still produces an Invalid Group error in Chrome, probably related to the {1} sequences.
The ?'s are messing it up. I'm not sure what they are for.
/\s[a-zA-Z\-]+\s*:\s*[a-zA-Z0-9\s.#]+;/
worked for me (as far as compiling. I didn't test to see if it properly detected a CSS string).
Replace the quotes with / (slashes):
var pattern = /\s([a-zA-Z-]+)\s[:]{1}\s*([a-zA-Z0-9\s.#]+)[;]{1}/;
You also don't need the new RegExp() part either, which is why it's been removed; instead of using a quote or double quote to denote a string, JavaScript uses a slash / to denote a regular expression, which isn't a normal string.
That regular expression is very bad and I would avoid its source in the future. That said, I cleaned it up a bit and got the following result:
var pattern = /\s(?:[a-zA-Z-]+)\s*:\s*(?:[^;\n\r]+);/;
this matches something that looks like css, for example:
background-color: red;
Here's the fiddle to prove it, though I'd recommend to find a different solution to your problem. This is a very simple regex and it's not save to say that it is reliable.