Hi this is working now but I am confused at to why.
I am learning regex and need to pull the numbers out of strings like
'Discount 7.5%' should get 7.5
'Discount 15%' should get 15
'Discount 10%' should get 10
'Discount 5%' should get 5
etc.
/\d,?.\d?/ //works
/\d,?.\d,?/ //doesn't works
/\d?.\d?/ //doesn't works
I thought one of the second two would work could someone explain this.
Quick and dirty with easy to understand regex.
//Let the system to optimize the number parsing for efficiency
parseFloat(text.replace(/^\D+/, ""));
Demo: http://jsfiddle.net/DerekL/4bnp8381/
Try this. Make the second part with dot . optional with ?
(\d)*(\.\d*)?
IF you check this Regex, this is what you need so it will return the float number only if the word before it is Discount by making use of the lookbehind operator:
(?<=Discount\s)\d+(\.\d+)?
But the problem is that works with PHP(prce) I can't get it working with javascript.
EDIT:
As mentioned in here regex and commas Javascript does not support lookbehind regex so here is what I did to work around it JS Fiddle
var div = document.getElementById('test');
var text = div.innerHTML;
div.innerHTML = text.replace(/Discount (.*)%/ig,'$1');
<div id="test">
Discount 7.5%<br>
Discount 15%<br>
discount 10%<br>
Discount 5.433%<br>
Fee 11%<br>
Discount 22.7%
</div>
As you see it does match it only if it was followedd by the word Discount, the word Fee 11% does not match
You can use the following regex, which will match one or more digits, optionally followed by a decimal point and any number of digits:
^Discount\s(\d+(\.\d+)?)%$
Regex101
Related
I have the following text "2345dsds34.000" and i want the following value '234534.00'. I plan to do this via regex replace but somehow it doesnt limit the decimal places to 2.
I am using this "2345dsds34.000".replace(/[^\d+(\.\d{1,2})$]/g, '') but it keeps giving me 234534.000. How can i force it to limit it to 2 decimal points.
console.log("2345dsds34.000".replace(/[^\d+(\.\d{1,2})$]/g, ''))
Thanks
You can use toFixed JS Number function
Number( "2345dsds34.000".replace(/[^\d+(\.\d{1,2})$]/g, '') ).toFixed(2)
You can simply slice the last digit off with .slice(0, -1):
console.log("2345dsds34.000".replace(/[^\d+(\.\d{1,2})$]/g, '').slice(0, -1));
That regex is targetting the characters between some digits and some other digits having a decimal place.
Snapshot from https://regex101.com/
Then, the .replace() method is removing them. That all it does.
Since you don't know how many extra decimal you may have... I suggest you to use parseFloat() and .tofixed(2).
Please have a look at those two documentation links. ;)
var value = parseFloat("2345dsds34.000".replace(/[^\d+(\.\d{1,2})$]/g, '')).toFixed(2);
console.log(value);
You can simply use
[^\d.]
let str = "2345dsds34.000"
let op = parseFloat(str.replace(/[^\d.]+/g, '')).toFixed(2)
console.log(op)
By using a Regex with capturing parentheses It will look like this
var regex = /(\d{4})[a-z]+(\d{2})\.(\d{2}).*/;
var input_chain = "2345dsds34.000";
var output = input_chain.replace(regex, "$1$2.$3");
console.log(output);
Each capturing parentheses can be references in the second part of the replace method by their position number prefixed with a $
I'm extracting the phone numbers that begin with 9 followed by other 9 digits from tweets using JavaScript.
Here's the regex pattern I am using:
var numberPattern = /^9[0-9]{9}/;
Here's the pattern matching phase:
var numberstring = JSON.stringify(data[i].text);
if(numberPattern.test(data[i].text.toString()) == true){
var obj={
tweet : {
status : data[i].text
},
phone : numberstring.match(numberPattern)
}
//console.log(numberstring.match(numberPattern));
stringarray.push(obj);
The problem is it is working for few numbers and not all. Also, I want to modify the regex to accept +91 prefix to numbers as well and(or) reject a starting 0 in numbers. I'm a beginner in regex, so help is needed. Thanks.
Example:
#Chennai O-ve blood for #arun_scribbles 's friend's father surgery in few days. Pl call 9445866298. 15May. via #arun_scribbles
Your regex pattern seems to be designed to allow a 9 or 8 at the beginning, but it would be better to enclose that choice in parentheses: /^(9|8)[0-9]{9}/.
To allow an optional "+" at the beginning, follow it with a question mark to make it optional: /^\+?(9|8)[0-9]{9}/.
To allow any character except "0", replace the (9|8) with a construct to accept only 1-9: /^\+?[1-9][0-9]{9}/.
And in your example, the phone number doesn't come at the beginning of the line, so the caret will not find it. If you're looking for content in the middle of the line, you'll need to drop the caret: /\+?[1-9][0-9]{9}/.
var numberPattern = /([+]91)?9[0-9]{9}\b/;
Try this regex pattern: [\+]?[0-9]{1,4}[\s]?[0-9]{10}
It accepts any country code with +, a space, and then 10 digit number.
While i type the phone number in a field after 5 digits put a hypen and continue the digits
after hypen. It allows only 10 digits and hypen.
Output : 90000-00000
Can any one help me how to fix the issue.
Thanks
The ReGex for this is really simple, you should have been able to Google this and work it out for yourself. The RegEx you want is as follows:
^[\d]{5}-[\d]{5}$
Using that with javascript:
var input = "90000-00000";
var valid = /^[\d]{5}-[\d]{5}$/.test(input);
//valid is a boolean: true or false
Here is a working example
This works:
var.replace(/[^0-9]+/g, '');
That simple snippet will replace anything that is not a number with nothing.
But decimals are real too. So, I'm trying to figure out how to include a period.
I'm sure it's really simple, but my tests aren't working.
Simply: var.replace(/[^\d.-]+/g, '');
Replacing something that is not a number is a little trickier than replacing something that is a number.
Those suggesting to simply add the dot, are ignoring the fact that . is also used as a period, so:
This is a test. 0.9, 1, 2, 3 will become .0.9123.
The specific regex in your problem will depend a lot on the purpose. If you only have a single number in your string, you could do this:
var.replace(/.*?(([0-9]*\.)?[0-9]+).*/g, "$1")
This finds the first number, and replaces the entire string with the matched number.
Try this:
var.replace(/[^0-9\\.]+/g, '');
there's a lot of correct answers already, just pointing out that you might need to account for negative signs too.. "\-" add that to any existing answer to allow for negative numbers.
Try this:
var.replace(/[0-9]*\.?[0-9]+/g, '');
That only matches valid decimals (eg "1", "1.0", ".5", but not "1.0.22")
If you don't want to catch IP address along with decimals:
var.replace(/[^0-9]+\\.?[0-9]*/g, '');
Which will only catch numerals with one or zero periods
How about doing this:
var numbers = str.gsub(/[0-9]*\.?[0-9]+/, "#{0} ");
Sweet and short inline replacing of non-numerical characters in the ASP.Net Textbox:
<asp:TextBox ID="txtJobNo" runat="server" class="TextBoxStyle" onkeyup="this.value=this.value.replace(/[^0-9]/g,'')" />
Alter the regex part as you'ld like. Lots and lots of people complain about the cursor going straight to the end when using the arrow keys, but people tend to deal with this without noticing it for instance, arrow... arrow... arrow... okay then... backspace back space, enter the new chars.
Here are a couple of jQuery input class types I use:
$("input.intgr").keyup(function (e) { // Filter non-digits from input value.
if (/\D/g.test($(this).val())) $(this).val($(this).val().replace(/\D/g, ''));
});
$("input.nmbr").keyup(function (e) { // Filter non-numeric from input value.
var tVal=$(this).val();
if (tVal!="" && isNaN(tVal)){
tVal=(tVal.substr(0,1).replace(/[^0-9\.\-]/, '')+tVal.substr(1).replace(/[^0-9\.]/, ''));
var raVal=tVal.split(".")
if(raVal.length>2)
tVal=raVal[0]+"."+raVal.slice(1).join("");
$(this).val(tVal);
}
});
intgr allows only numeric - like other solutions here.
nmbr allows only positive/negative decimal. Negative must be the first character (you can add "+" to the filter if you need it), strips -3.6.23.333 to -3.623333
I'm putting nmbr up because I got tired of trying to find the way to keep only 1 decimal and negative in 1st position
This one just worked for -ve to +ve numbers
<input type="text" oninput="this.value = this.value.replace(/[^0-9\-]+/g, '').replace(/(\..*)\./g, '$1');">
I use this expression to exclude all non-numeric characters + keep negative numbers with minus sign.
variable.replace(/[^0-9.,\-]/g,'')
Friends,
I'm new to both Javascript and Regular Expressions and hope you can help!
Within a Javascript function I need to check to see if a comma(,) appears 1 or more times. If it does then there should be one or more numbers either side of it.
e.g.
1,000.00 is ok
1,000,00 is ok
,000.00 is not ok
1,,000.00 is not ok
If these conditions are met I want the comma to be removed so 1,000.00 becomes 1000.00
What I have tried so is:
var x = '1,000.00';
var regex = new RegExp("[0-9]+,[0-9]+", "g");
var y = x.replace(regex,"");
alert(y);
When run the alert shows ".00" Which is not what I was expecting or want!
Thanks in advance for any help provided.
strong text
Edit
strong text
Thanks all for the input so far and the 3 answers given. Unfortunately I don't think I explained my question well enough.
What I am trying to achieve is:
If there is a comma in the text and there are one or more numbers either side of it then remove the comma but leave the rest of the string as is.
If there is a comma in the text and there is not at least one number either side of it then do nothing.
So using my examples from above:
1,000.00 becomes 1000.00
1,000,00 becomes 100000
,000.00 is left as ,000.00
1,,000.00 is left as 1,,000.00
Apologies for the confusion!
Your regex isn't going to be very flexible with higher orders than 1000 and it has a problem with inputs which don't have the comma. More problematically you're also matching and replacing the part of the data you're interested in!
Better to have a regex which matches the forms which are a problem and remove them.
The following matches (in order) commas at the beginning of the input, at the end of the input, preceded by a number of non digits, or followed by a number of non digits.
var y = x.replace(/^,|,$|[^0-9]+,|,[^0-9]+/g,'');
As an aside, all of this is much easier if you happen to be able to do lookbehind but almost every JS implementation doesn't.
Edit based on question update:
Ok, I won't attempt to understand why your rules are as they are, but the regex gets simpler to solve it:
var y = x.replace(/(\d),(\d)/g, '$1$2');
I would use something like the following:
^[0-9]{1,3}(,[0-9]{3})*(\.[0-9]+)$
[0-9]{1,3}: 1 to 3 digits
(,[0-9]{3})*: [Optional] More digit triplets seperated by a comma
(\.[0-9]+): [Optional] Dot + more digits
If this regex matches, you know that your number is valid. Just replace all commas with the empty string afterwards.
It seems to me you have three error conditions
",1000"
"1000,"
"1,,000"
If any one of these is true then you should reject the field, If they are all false then you can strip the commas in the normal way and move on. This can be a simple alternation:
^,|,,|,$
I would just remove anything except digits and the decimal separator ([^0-9.]) and send the output through parseFloat():
var y = parseFloat(x.replace(/[^0-9.]+/g, ""));
// invalid cases:
// - standalone comma at the beginning of the string
// - comma next to another comma
// - standalone comma at the end of the string
var i,
inputs = ['1,000.00', '1,000,00', ',000.00', '1,,000.00'],
invalid_cases = /(^,)|(,,)|(,$)/;
for (i = 0; i < inputs.length; i++) {
if (inputs[i].match(invalid_cases) === null) {
// wipe out everything but decimal and dot
inputs[i] = inputs[i].replace(/[^\d.]+/g, '');
}
}
console.log(inputs); // ["1000.00", "100000", ",000.00", "1,,000.00"]