Given this string:
var str = 'A1=B2;C3,D0*E9+F6-';
I would like to retrieve the substring that goes from the beginning of the string up to 'D0*' (excluding), in this case:
'A1=B2;C3,'
I know how to achieve this using the combination of the substr and indexOf methods:
str.substr(0, str.indexOf('D0*'))
Live demo: http://jsfiddle.net/simevidas/XSu22/
However, this is obviously not the best solution since it contains a redundancy (the str name has to be written twice). This redundancy can be avoided by using the match method together with a regular expression that captures the substring:
str.match(/???/)[1]
Which regular expression literal do we have to pass into match to ensure that the correct substring is returned?
My guess is this: /(.*)D0\*/ (and that works), but my experience with regular expressions is rather limited, so I'm going to need a confirmation...
Try this:
/(.*?)D0\*/.exec(str)[1]
Or:
str.match(/(.*?)D0\*/)[1]
DEMO HERE
? directly following a quantifier makes the quantifier non-greedy (makes it match minimum instead of maximum of the interval defined).
Here's where that's from
/^(.+?)D0\*/
Try it here: http://rubular.com/r/TNTizJLSn9
/^.*(?=D0\*)/
more text to hit character limit...
You can do a number-group, like your example.
/^(.*?)foo/
It mean somethink like:
Store all in group, from start (the 0)
Stop, but don't store on found foo (the indexOf)
After that, you need match and get
'hello foo bar foo bar'.match(/^(.*?)foo/)[1]; // will return "hello "
It mean that will work on str variable and get the first (and unique) number-group existent. The [0] instead [1] mean that will get all matched code.
Bye :)
Related
I need to split an email address and take out the first character and the first character after the '#'
I can do this as follows:
'bar#foo'.split('#').map(function(a){ return a.charAt(0); }).join('')
--> bf
Now I was wondering if it can be done using a regex match, something like this
'bar#foo'.match(/^(\w).*?#(\w)/).join('')
--> bar#fbf
Not really what I want, but I'm sure I miss something here! Any suggestions ?
Why use a regex for this? just use indexOf to get the char at any given position:
var addr = 'foo#bar';
console.log(addr[0], addr[addr.indexOf('#')+1])
To ensure your code works on all browsers, you might want to use charAt instead of []:
console.log(addr.charAt(0), addr.charAt(addr.indexOf('#')+1));
Either way, It'll work just fine, and This is undeniably the fastest approach
If you are going to persist, and choose a regex, then you should realize that the match method returns an array containing 3 strings, in your case:
/^(\w).*?#(\w)/
["the whole match",//start of string + first char + .*?# + first string after #
"groupw 1 \w",//first char
"group 2 \w"//first char after #
]
So addr.match(/^(\w).*?#(\w)/).slice(1).join('') is probably what you want.
If I understand correctly, you are quite close. Just don't join everything returned by match because the first element is the entire matched string.
'bar#foo'.match(/^(\w).*?#(\w)/).splice(1).join('')
--> bf
Using regex:
matched="",
'abc#xyz'.replace(/(?:^|#)(\w)/g, function($0, $1) { matched += $1; return $0; });
console.log(matched);
// ax
The regex match function returns an array of all matches, where the first one is the 'full text' of the match, followed by every sub-group. In your case, it returns this:
bar#f
b
f
To get rid of the first item (the full match), use slice:
'bar#foo'.match(/^(\w).*?#(\w)/).slice(1).join('\r')
Use String.prototype.replace with regular expression:
'bar#foo'.replace(/^(\w).*#(\w).*$/, '$1$2'); // "bf"
Or using RegEx
^([a-zA-Z0-9])[a-zA-Z0-9.!#$%&'*+\/=?^_`{|}~-]+#([a-zA-Z0-9-])[a-zA-Z0-9-]+(?:\.[a-zA-Z0-9-]+)*$
Fiddle
Hopefully a simple one!
I've been trying to get this to work for several hours now but am having no luck, as I'm fairly new to regexp I may be missing something very obvious here and was hoping someone could point me in the right direction. The pattern I want to match is as follows: -
At least 1 or more numbers + "##" + at least 1 or more numbers + "##" + at least 1 or more numbers
so a few examples of valid combinations would be: -
1##2##3
123#123#123
0##0##0
A few invalid combinations would be
a##b##c
1## ##1
I've got the following regexp like so: -
[\d+]/#/#[\d+]/#/#[\d+]
And am using it like so (note the double slashes as its inside a string): -
var patt = new RegExp("[\\d+]/#/#[\\d+]/#/#[\\d+]");
if(newFieldValue!=patt){newFieldValue=="no match"}
I also tried these but still nothing: -
if(!patt.text(newFieldValue)){newFieldValue==""}
if(patt.text(newFieldValue)){}else{newFieldValue==""}
But nothing I try is matching, where am I going wrong here?
Any pointers gratefully received, cheers!
1) I can't see any reason to use the RegExp constructor over a RegExp literal for your case. (The former is used primarily where the pattern needs to by dynamic, i.e. is contributed to by variables.)
2) You don't need a character class if there's only one type of character in it (so \d+ not [\d+]
3) You are not actually checking the pattern against the input. You don't apply RegEx by creating an instance of it and using ==; you need to use test() or match() to see if a match is made (the former if you want to check only, not capture)
4) You have == where you mean to assign (=)
if (!/\d+##\d+##\d+/.test(newFieldValue)) newFieldValue = "no match";
You put + inside the brackets, so you're matching a single character that's either a digit or +, not a sequence of digits. I also don't understand why you have / before each #, your description doesn't mention anything about this character.
Use:
var patt = /\d+##\d+##\d+/;
You should use the test method of the pat regex
if (!patt.test(newFieldValue)){ newFieldValue=="no match"; }
once you have a valid regular expression.
Try this regex :
^(?:\d+##){2}\d+$
Demo: http://regex101.com/r/mE8aG7
With the following regex
[\d+]/#/#[\d+]/#/#[\d+]
You would only match things like:
+/#/#5/#/#+
+/#/#+/#/#+
0/#/#0/#/#0
because the regex engine sees it like on the schema below:
Something like:
((-\s)?\d+##)+\d+
I was given a task to do which requires a long time to do.
The image say it all :
This is what I have : (x100 times):
And I need to extract this value only
How can I capture the value ?
I have made it with this regex :
DbCommand.*?\("(.*?)"\);
As you can see it does work :
And after the replace function (replace to $1) I do get the pure value :
but the problem is that I need only the pure values and not the rest of the unmatched group :
Question : In other words :
How can I get the purified result like :
Eservices_Claims_Get_Pending_Claims_List
Eservices_Claims_Get_Pending_Claims_Step1
Here is my code at Online regexer
Is there any way of replacing "all besides the matched group" ?
p.s. I know there are other ways of doing it but I prefer a regex solution ( which will also help me to understand regex better)
Unfortunately, JavaScript doesn't understand lookbehind. If it did, you could change your regular expression to match .*? preceded (lookbehind) by DbCommand.*?\(" and followed (lookahead) by "\);.
With that solution denied, i believe the cleanest solution is to perform two matches:
// you probably want to build the regexps dynamically
var regexG = /DbCommand.*?\("(.*?)"\);/g;
var regex = /DbCommand.*?\("(.*?)"\);/;
var matches = str.match(regexG).map(function(item){ return item.match(regex)[1] });
console.log(matches);
// ["Eservices_Claims_Get_Pending_Claims_List", "Eservices_Claims_Get_Pending_Claims_Step1"]
DEMO: http://jsbin.com/aqaBobOP/2/edit
You should be able to do a global replace of:
public static DataTable.*?{.*?DbCommand.*?\("(.*?)"\);.*?}
All I've done is changed it to match the whole block including the function definition using a bunch of .*?s.
Note: Make sure your regex settings are such that the dot (.) matches all characters, including newlines.
In fact if you want to close up all whitespace, you can slap a \s* on the front and replace with $1\n:
\s*public static DataTable.*?{.*?DbCommand.*?\("(.*?)"\);.*?}
Using your test case: http://regexr.com?37ibi
You can use this (without the ignore case and multiline option, with a global search):
pattern: (?:[^D]+|\BD|D(?!bCommand ))+|DbCommand [^"]+"([^"]+)
replace: $1\n
Try simply replacing the whole document replacing using this expression:
^(?: |\t)*(?:(?!DbCommand).)*$
You will then only be left with the lines that begin with the string DbCommand
You can then remove the spaces in between by replacing:
\r?\n\s* with \n globally.
Here is an example of this working: http://regexr.com?37ic4
So a substring can take two parameters, the index to start at and the index to stop at like so
var str="Hello beautiful world!";
document.write(str.substring(3,7));
but is there a way to designate the start and stopping points as a set of characters to grab, so instead of the starting point being 3 I would want it to be "lo" and instead of the end point being 7 I would want it to be "wo" so I would be grabbing "lo beautiful wo". Is there a Javascript function that serves that purpose already?
Sounds like you want to use regular expressions and string.match() instead:
var str="Hello beautiful world!";
document.write(str.match(/lo.*wo/)[0]); // document.write("lo beautiful wo");
Note, match() returns an array of matches, which might be null if there is no match. So you should include a null check.
If you're not familiar with regexes, this is a pretty good source:
http://www.w3schools.com/jsref/jsref_obj_regexp.asp
use the method indexOf: document.write(str.substring(3,str.indexOf('wo')+2));
Yup, you can do this easily with regular expressions:
var substr = /lo.+wo/.exec( 'Hello beautiful world!' )[0];
console.log( substr ); //=> 'lo beautiful wo'
Use a regex brother:
if (/(lo.+wo)/.test("Hello beautiful world!")) {
document.write(RegExp.$1);
}
You need a backup plan in case the string does not match. Hence the use of test.
Regular expression may be able to achieve this to some extent, but there are many details that you must be aware of.
For example, if you want to find all the substrings that starts with "lo", and ends with the nearest "wo" after "lo". (If there are more than 1 match, the subsequent matches will pick up the first "lo" after the "wo" of last match).
"Hello beautiful world!".match(/lo.*?wo/g);
Using the RegExp constructor, you can make it more flexible (you can substitute "lo" and "wo" with the actual string you want to find):
"Hello beautiful world!".match(new RegExp("lo" + ".*?" + "wo", "g"));
Important: The downside of the RegExp approach above is that, you need to know what characters are special to escape them - otherwise, they will not match the actual substring you want to find.
It can also be achieve with indexOf, albeit a little bit dirty. For the first substring:
var startIndex = str.indexOf(startString);
var endIndex = str.indexOf(endString, startIndex);
if (startIndex >= 0 && endIndex >= 0)
str.substring(startIndex, endIndex + endString.length)
If you want to find the substring that starts with the first "lo" and ends with the last "wo" in the string, you can use indexOf and lastIndexOf to find it (with a small modification to the code above). RegExp can also do it, by changing .*? to .* in the two example above (there will be at most 1 match, so the "g" flag at the end is redundant).
I'm horrible at RegEx and I need a regex to test if a certain string ends a certain way. For example, if the RegEx tests for ending with foo, "somestringfoo" -> True and "someotherFoostring" -> False. It needs to be case sensitive and work with alphanumeric and underscore. Here is what I've got, but I can't get it to work:
var test = RegExp.test('/foo$/');
You would do it this way:
/foo$/.test("somestringfoo")
test is a method of the regexp object, so it would be /foo$/.test(someString) or new Regexp("foo$").test(someString).
However, testing a string for ending with a certain substring does not need regular expressions, see endsWith in JavaScript.
this should do the work:
function isFoo(string){
var pattern = /foo$/;
return pattern.test(string);
}