Regex trying to match characters before and after symbol - javascript

I'm trying to match characters before and after a symbol, in a string.
string: budgets-closed
To match the characters before the sign -, I do: ^[a-z]+
And to match the other characters, I try: \-(\w+) but, the problem is that my result is: -closed instead of closed.
Any ideas, how to fix it?
Update
This is the piece of code, where I was trying to apply the regex http://jsfiddle.net/trDFh/1/
I repeat: It's not that I don't want to use split; it's just I was really curious, and wanted to see, how can it be done the regex way. Hacking into things spirit
Update2
Well, using substring is a solution as well: http://jsfiddle.net/trDFh/2/ and is the one I chosed to use, since the if in question, is actually an else if in a more complex if syntax, and the chosen solutions seems to be the most fitted for now.

Use exec():
var result=/([^-]+)-([^-]+)/.exec(string);
result is an array, with result[1] being the first captured string and result[2] being the second captured string.
Live demo: http://jsfiddle.net/Pqntk/

I think you'll have to match that. You can use grouping to get what you need, though.
var str = 'budgets-closed';
var matches = str.match( /([a-z]+)-([a-z]+)/ );
var before = matches[1];
var after = matches[2];
For that specific string, you could also use
var str = 'budgets-closed';
var before = str.match( /^\b[a-z]+/ )[0];
var after = str.match( /\b[a-z]+$/ )[0];
I'm sure there are better ways, but the above methods do work.

If the symbol is specifically -, then this should work:
\b([^-]+)-([^-]+)\b
You match a boundry, any "not -" characters, a - and then more "not -" characters until the next word boundry.
Also, there is no need to escape a hyphen, it only holds special properties when between two other characters inside a character class.
edit: And here is a jsfiddle that demonstrates it does work.

Related

I need some help for a specific regex in javascript

I try to set a correct regex in my javascript code, but I'm a bit confused with this. My goal is to find any occurence of "rotate" in a string. This should be simple, but in fact I'm lost as my "rotate" can have multiple endings! Here are some examples of what I want to find with the regex:
rotate5
rotate180
rotate-1
rotate-270
The "rotate" word can be at the begining of my string or at the end, or even in the middle separated by spaces from other words. The regex will be used in a search-and-replace function.
Can someone help me please?
EDIT: What I tried so far (probably missing some of them):
/\wrotate.*/
/rotate.\w*/
/rotate.\d/
/\Srotate*/
I'm not fully understanding the regex mechanic yet.
Try this regex as a start. It will return all occurrences of a "rotate" string where a number (positive or negative) follows the "rotate".
/(rotate)([-]?[0-9]*)/g
Here is sample code
var aString = ["rotate5","rotate180","rotate-1","some text rotate-270 rotate-1 more text rotate180"];
for (var x = 0; x < 4; x++){
var match;
var regex = /(rotate)([-]?[0-9]*)/g;
while (match = regex.exec(aString[x])){
console.log(match);
}
}
In this example,
match[0] gives the whole match (e.g. rotate5)
match[1] gives the text "rotate"
match[2] gives the numerical text immediately after the word "rotate"
If there are multiple rotate stings in the string, this will return them all
If you just need to know if the 'word' is in the string so /rotate/ simply will be OK.
But if you want some matching about what coming before or after the #mseifert will be good
If you just want to replace the word rotate by another one
you can just use the string method String.replace use it like var str = "i am rotating with rotate-90"; str.repalace('rotate','turning')'
WHy your regex doesnt work ?
/\wrotate.*/
means that the string must start with a caracter [a-zA-Z0-9_] followed by rotate and another optional character
/rotate.\w*/
meanse rotate must be followed by a character and others n optional character
...............
Using your description:
The "rotate" word can be at the beginning of my string or at the end, or even in the middle separated by spaces from other words. The regex will be used in a search-and-replace function.
This regex should do the work:
const regex = /(^rotate|rotate$|\ {1}rotate\ {1})/gm;
You can learn more about regular expressions with these sites:
http://www.regular-expressions.info
regex101.com and btw here is an example using your requirements.

Regex to get the text between two characters?

I want to replace a text after a forward slash and before a end parantheses excluding the characters.
My text:
<h3>notThisText/IWantToReplaceThis)<h3>
$('h3').text($('h3').text().replace(regEx, 'textReplaced'));
Wanted result after replace:
notThisText/textReplaced)
I have tried
regex = /([^\/]+$)+/ //replaces the parantheses as well
regex = \/([^\)]+) //replaces the slash as well
but as you can see in my comments neither of these excludes both the slash and the end parantheses. Can someone help?
A pattern like /(?<=\/)[^)]+(?=\))/ won't work in JS as its regex engine does not support a lookbehind construct. So, you should use one of the following solutions:
s.replace(/(\/)[^)]+(\))/, '$1textReplaced$2')
s.replace(/(\/)[^)]+(?=\))/, '$1textReplaced')
s.replace(/(\/)[^)]+/, '$1textReplaced')
s.replace(/\/[^)]+\)/, '/textReplaced)')
The (...) forms a capturing group that can be referenced to with $ + number, a backreference, from the replacement pattern. The first solution is consuming / and ), and puts them into capturing groups. If you need to match consecutive, overlapping matches, use the second solution (s.replace(/(\/)[^)]+(?=\))/, '$1textReplaced')). If the ) is not required at the end, the third solution (replace(/(\/)[^)]+/, '$1textReplaced')) will do. The last solution (s.replace(/\/[^)]+\)/, '/textReplaced)')) will work if the / and ) are static values known beforehand.
You can use str.split('/')
var text = 'notThisText/IWantToReplaceThis';
var splited = text.split('/');
splited[1] = 'yourDesireText';
var output = splited.join('/');
console.log(output);
Try Following: In your case startChar='/', endChar = ')', origString=$('h3').text()
function customReplace(startChar, endChar, origString, replaceWith){
var strArray = origString.split(startChar);
return strArray[0] + startChar + replaceWith + endChar;
}
First of all, you didn't define clearly what is the format of the text which you want to replace and the non-replacement part. For example,
Does notThisText contain any slash /?
Does IWantToReplaceThis contain any parentheses )?
Since there are too many uncertainties, the answer here only shows up the pattern exactly matches your example:
yourText.replace(/(\/).*?(\))/g, '$1textReplaced$2')
var text = "notThisText/IWantToReplaceThis";
text = text.replace(/\/.*/, "/whatever");
output : "notThisText/whatever"`

I want to find numbers-include dot and comma- but Regex not working in javascript

I thought it was very simple to find out. But how many ways I tried still not work properly.
Below is the test snippet.
"100$ and 1.000,000EUR 1,00.0.000USD .90000000000000000000$ (09898)".replace(/[\.,\d]*/g, '{n}')
And I want the result like below.
{n}$ and {n}EUR {n}USD {n}$ ({n})
The * is your problem, change the regex to /[.,\d]+/g instead.
"100$ and 1.000,000EUR 1,00.0.000USD .90000000000000000000$ (09898)".replace(/[.,\d]+/g, '{n}');
Output
{n}$ and {n}EUR {n}USD {n}$ ({n})
JSFiddle Example Check console screen for the output.
The problem here is that [\.,\d]* can match an empty string. The first step would be to use [.,\d]+ so that at least one of these characters matches.
But a better regex would be \d[.,\d]* because it ensures the replaced characters begin with a digit, so it won't replace periods in sentences.
If you want to go further, you can also use (?=[.,\d]*\d)[.,\d]+ if to handle numbers starting with periods. This one would be the proper answer for your case. The lookahead ensures there's at least one digit anywhere in the replaced text.
Note that you don't need to escape the . inside a character class.
\.?\d[^\s]*\d
Try this.Replace with {n}.See demo.
http://regex101.com/r/kP8uF5/3
var re = /\.?\d[^\s]*\d/gm;
var str = '100$ and 1.000,000EUR 1,00.0.000USD .90000000000000000000$ (09898)';
var subst = '{n}';
var result = str.replace(re, subst);

Javascript regex expression to replace multiple strings?

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,"");

Split string in JavaScript using a regular expression

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*'([^']*)'

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