I'm suresomeone of you know how to do this easily, I am trying to validate my inputs that only allows numbers, characters and/or an empty space which mean nothing in a string "", so I have /^[0-9a-zA-Z]+$/ only missing that empty space in there. Seems I can't find the express for it, so I tried to create a variabel var emptyStr = "" and put it in there but don't really know how to write the correct syntax.
Use * rather than +. * is 0 or more times. + is 1 or more times.
/^[0-9a-zA-Z]*$/
/^[A-z0-9\s]*$/
The \s will also capture spaces
Related
I was a bit surprised, that actually no one had the exact same issue in javascript...
I tried several different solutions none of them parse the content correctly.
The closest one I tried : (I stole its regex query from a PHP solution)
const test = `abc?aaa.abcd?.aabbccc!`;
const sentencesList = test.split("/(\?|\.|!)/");
But result just going to be
["abc?aaa.abcd?.aabbccc!"]
What I want to get is
['abc?', 'aaa.', 'abcd?','.', 'aabbccc!']
I am so confused.. what exactly is wrong?
/[a-z]*[?!.]/g) will do what you want:
const test = `abc?aaa.abcd?.aabbccc!`;
console.log(test.match(/[a-z]*[?!.]/g))
To help you out, what you write is not a regex. test.split("/(\?|\.|!)/"); is simply an 11 character string. A regex would be, for example, test.split(/(\?|\.|!)/);. This still would not be the regex you're looking for.
The problem with this regex is that it's looking for a ?, ., or ! character only, and capturing that lone character. What you want to do is find any number of characters, followed by one of those three characters.
Next, String.split does not accept regexes as arguments. You'll want to use a function that does accept them (such as String.match).
Putting this all together, you'll want to start out your regex with something like this: /.*?/. The dot means any character matches, the asterisk means 0 or more, and the questionmark means "non-greedy", or try to match as few characters as possible, while keeping a valid match.
To search for your three characters, you would follow this up with /[?!.]/ to indicate you want one of these three characters (so far we have /.*?[?!.]/). Lastly, you want to add the g flag so it searches for every instance, rather than only the first. /.*?[?!.]/g. Now we can use it in match:
const rawText = `abc?aaa.abcd?.aabbccc!`;
const matchedArray = rawText.match(/.*?[?!.]/g);
console.log(matchedArray);
The following code works, I do not think we need pattern match. I take that back, I have been answering in Java.
final String S = "An sentence may end with period. Does it end any other way? Ofcourse!";
final String[] simpleSentences = S.split("[?!.]");
//now simpleSentences array has three elements in it.
I am trying to edit a DateTime string in typescript file.
The string in question is 02T13:18:43.000Z.
I want to trim the first three characters including the letter T from the beginning of a string AND also all 5 characters from the end of the string, that is Z000., including the dot character. Essentialy I want the result to look like this: 13:18:43.
From what I found the following pattern (^(.*?)T) can accomplish only the first part of the trim I require, that leaves the initial result like this: 13:18:43.000Z.
What kind of Regex pattern must I use to include the second part of the trim I have mentioned? I have tried to include the following block in the same pattern (Z000.)$ but of course it failed.
Thanks.
Any help would be appreciated.
There is no need to use regular expression in order to achieve that. You can simply use:
let value = '02T13:18:43.000Z';
let newValue = value.slice(3, -5);
console.log(newValue);
it will return 13:18:43, assumming that your string will always have the same pattern. According to the documentation slice method will substring from beginIndex to endIndex. endIndex is optional.
as I see you only need regex solution so does this pattern work?
(\d{2}:)+\d{2} or simply \d{2}:\d{2}:\d{2}
it searches much times for digit-digit-doubleDot combos and digit-digit-doubleDot at the end
the only disadvange is that it doesn't check whether say there are no minutes>59 and etc.
The main reason why I didn't include checking just because I kept in mind that you get your dates from sources where data that are stored are already valid, ex. database.
Solution
This should suffice to remove both the prefix from beginning to T and postfix from . to end:
/^.*T|\..*$/g
console.log(new Date().toISOString().replace(/^.*T|\..*$/g, ''))
See the visualization on debuggex
Explanation
The section ^.*T removes all characters up to and including the last encountered T in the string.
The section \..*$ removes all characters from the first encountered . to the end of the string.
The | in between coupled with the global g flag allows the regular expression to match both sections in the string, allowing .replace(..., '') to trim both simultaneously.
I have searched StackOverflow and I can't find an answer as to how to check for regex of numeric inputs for a calculator app that will check for the following format with every keyup (jquery key up):
Any integer like: 34534
When a dot follows the integer when the user is about to enter a decimal number like this: 34534. Note that a dot can only be entered once.
Any float: 34534.093485
I don't plan to use commas to separate the thousands...but I would welcome if anyone can also provide a regex for that.
Is it possible to check the above conditions with just one regex? Thanks in advance.
Is a lone . a successful match or not? If it is then use:
\d+(\.\d*)?|\.\d*
If not then use:
\d+(\.\d*)?|\.\d+
Rather than incorporating commas into the regexes, I recommend stripping them out first: str = str.replace(/,/g, ''). Then check against the regex.
That wouldn't verify that digits are properly grouped into groups of three, but I don't see much value in such a check. If a user types 1,024 and then decides to add a digit (1,0246), you probably shouldn't force them to move the comma.
Let's write our your specifications, and develop from that.
Any integer: \d+
A comma, optionally followed by an integer: \.\d*
Combine the two and make the latter optional, and you get:
\d+\.?\d*
As for handling commas, I'd rather not go into it, as it gets very ugly very fast. You should simply strip all commas from input if you still care about them.
you can use in this way:
[/\d+./]
I think this can be used for any of your queries.
Whether it's 12445 or 1244. or 12445.43
I'm going to throw in a potentially downvoted answer here - this is a better solution:
function valid_float (num) {
var num = (num + '').replace(/,/g, ''), // don't care about commas, this turns `num` into a String
float_num = parseFloat(num);
return float_num == num || float_num + '.' == num; // allow for the decimal point, deliberately using == to ignore type as `num` is a String now
}
Any regex that does your job correctly will come with a big asterisk after it saying "probably", and if it's not spot on, it'll be an absolute pig to debug.
Sure, this answer isn't giving you the most awesomely cool one-liner that's going to make you go "Cool!", but in 6 months time when you realise it's going wrong somewhere, or you want to change it to do something slightly different, it's going to be a hell of a lot easier to see where, and to fix.
I'm using ^(\d)+(.(\d)+)+$ to capture each integer and to have an unlimited length, so long as the string begins and ends with integers and has dots between each integer group. I'm capturing the integer groups so that I can compare them.
var str='userkwd* type:"Office"';
How do I trim or substr or slice this string to only get 'userkwd'? Also the variable will have quotes as part of it..This one is tricky as if there is no userkwd .i.e. if
var str=' type:"Office"';
it should return null. The * gets appended with userkwd from inputbox..
str.slice(0,str.indexOf('*')); ???
str.split("*")[0]; ????
str.substring(0, str.indexOf('*')); ???
Which one?
str.replace(/\*? type.*/, '');
The answer is trivial. Think if userkwd can contain " ", for example "user kwd". If so, then you cannot use split(" "). If userkwd could contain additional * like user*kwd then you cannot use this character. If the keyword could contain both spaces and * then you should use different method, for example some more complicated regular expression, but try to evalute your suggested methods first.
By the way, in the scenario you have given var str=' type:"Office"'; there is no userkwd at all, there is also no *, so you definitely should not use * if this is valid example. Space seaprator however holds, so I would go for it if possible.
You can also try to find the last occurance of "type:" and get everything what is before it, if empty then change it to null, or something like this. There are really a lot of possibilities, we do not have enough information why you need this functionality and what could be the input...
I am trying to match a string containing a mix of digits and hyphenated digits, like a crossword answer specification, for example 1,2-2 or 1-1,3,4,2-2
/,?(([1-9]-[1-9])|([1-9]))/g is what I've come up to match the string
value = value.replace(/,?(([1-9]-[1-9])|([1-9]))/g, '');
replaces ok, and I've checked it out in an online tester.
What I really need is to negate this, so I can use it on a keyup event, examine the contents of a textarea and remove characters that don't fit, so it only allows through characters as in the example.
I've tried ^ where expected, but this it's not doing what I expect, how should I negate the regex so I remove everything that doesn't match?
If there is a better way of doing this I'm open to suggestions too.
var value = 'hello,1,2,3,4-6,1-1,3,test,4,2-2';
var pattern = /,?(([1-9]-[1-9])|([1-9]))/g;
value.replace(pattern, ''); // "hello,test"
You can use String#match. With /g flag, it returns an array of all the matches, then you can use Array#join to join them.
The problem is that String#match returns null when there is no match, so you have to handle that case and use an empty array so that it can join:
(value.match(pattern) || []).join(''); // ",1,2,3,4-6,1-1,3,4,2-2"
Note: It may better to check them on onblur rather than onkeyup. Messing with the text that the user is currently typing will make it annoying. Better to wait for the user to finish typing.
Didn't test it in JS, but this should return the valid string beginning from the left and as long as valid values are encountered (note that I used \d - if you'd like 1-9 only, then use your brackets).
(?:\d(?:-\d)?,)*\d(?:-\d)?
E.g. matching this regular expression with the string "0-1,1,2,3,4-4,2,,1,3--4" will return "0-1,1,2,3,4-4,2" as the first match.