I have the below two strings. In both cases, I am trying to retrieve "foreclosure_defenses".
str = "client_profile[lead_profile_attributes][foreclosure_defenses_attributes][0][own_property]"
str2 = "client_profile[foreclosure_defenses_attributes][0][own_property]"
I'm close but I can't get a regex that will work with both of them.
This regex works for str2, but not for str:
regex = /\w+(?:\[(\w+)_attributes\]+)\[\d+\]\[own_property\]/g
regex.exec(str2)
["client_profile[foreclosure_defenses_attributes][0][own_property]", "foreclosure_defenses"]
This regex works for str, but not for str2:
regex = /\w+(?:\[(\w+)_attributes\]?)+\[\d+\]\[own_property\]/g
regex.exec(str);
["client_profile[lead_profile_attributes][foreclosure_defenses_attributes][0][own_property]", "foreclosure_defenses"]
The last one should work for both cases, but doesn't. It should look for one or many _attributes patterns and grab the last one.
What am I doing wrong?
I think you want
/\w+(?:\[(\w+)_attributes\])+\[\d+\]\[own_property\]/
Note that if you use global g flag and you attempt to match multiple strings, you will need to reset the index of the regex.
Why not just use this:
/.+?\[(foreclosure_defenses)_attributes\]/g
Demo
Related
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);
I'm trying to get a much deeper understanding of JS RegExp for a project I'm working on.
So if I were checking for all strings containing foo and then a character that is not a number, I would use /foo[^0-9]/. However, let's say I want to change all strings matching that pattern to foobar and then the original characters, how would I go about that?
str = foozip;
newStr = str.replace(/foo[^0-9]/, "foobar");
console.log(newStr);
//returns foobarip Note the lack of a z.
str = foozip;
newStr = str.replace(/foo/, "foobar");
console.log(newStr);
//this matches foo6zip, which is no good
Do I have to run a separate check to do this? Is there a way to carry unknown characters from one side of a replace to the other?
You have two options:
Use lookahead:
str.replace(/foo(?=[^0-9])/, "foobar")
Use capture groups:
str.replace(/foo([^0-9])/, "foobar$1")
I'm trying to match characters before and after a symbol, in a string.
string: budgets-closed
To match the characters before the sign -, I do: ^[a-z]+
And to match the other characters, I try: \-(\w+) but, the problem is that my result is: -closed instead of closed.
Any ideas, how to fix it?
Update
This is the piece of code, where I was trying to apply the regex http://jsfiddle.net/trDFh/1/
I repeat: It's not that I don't want to use split; it's just I was really curious, and wanted to see, how can it be done the regex way. Hacking into things spirit
Update2
Well, using substring is a solution as well: http://jsfiddle.net/trDFh/2/ and is the one I chosed to use, since the if in question, is actually an else if in a more complex if syntax, and the chosen solutions seems to be the most fitted for now.
Use exec():
var result=/([^-]+)-([^-]+)/.exec(string);
result is an array, with result[1] being the first captured string and result[2] being the second captured string.
Live demo: http://jsfiddle.net/Pqntk/
I think you'll have to match that. You can use grouping to get what you need, though.
var str = 'budgets-closed';
var matches = str.match( /([a-z]+)-([a-z]+)/ );
var before = matches[1];
var after = matches[2];
For that specific string, you could also use
var str = 'budgets-closed';
var before = str.match( /^\b[a-z]+/ )[0];
var after = str.match( /\b[a-z]+$/ )[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.
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*'([^']*)'