I have a value - toggle-save_2
I want to extract the 2 part. The part after the underscore will always be what i need but the length of the former part, which in this case is toggle-save_, may vary (eg. notoggle-val_45). Though this length may vary it will always be separated from the end number by an underscore.
Right now I am using this
var current = this.id.split('_');
current = current[1];
to select the number.
What would be cool is if I could pass a variable to the split to only give me the second index of the result from the split.
Just select the 2nd index when you do the split.
var current = this.id.split('_')[1];
The best solution here would be to use lastIndexOf and substring, like this
function getLastPart(strObject) {
return strObject.substring(strObject.lastIndexOf("_") + 1);
}
console.log(getLastPart("toggle-save_2"));
// 2
console.log(getLastPart("notoggle-save_45"));
// 45
It is better for this case because, you already know that the _ will be somewhere near the last position. Since lastIndexOf starts from the last position, it would find _ very soon and all we need to do is to get the rest of the string from the next position.
There are often times I am breaking up a string where I only need the very last value of the result of String.prototype.split and consider the rest to be garbage no matter how many values the split produced.
When those cases arise, I like to chain Array.prototype.pop off of the split
var s = 'toggle-save_2',
current = s.split('_').pop();
The split method can only be limited from the end, and it will always return an array.
You don't need to use split, you can use string operations to get part of the string:
var current = this.id.substr(this.id.indexOf('_') + 1);
Related
How do I find a string start and end position from a string. Is there a way to search the "m" and "g" coordinats or char position? given that there maybe multiple values. so far only upto start.
Find: myname.jpg
<img src="myname.jpg" id="men">
var start = x.indexOf("myname.jpg");
//10
You can use a combination of .indexOf() and .substr().
When there is a possibility of multiple matches, providing more context characters to .indexOf() can resolve the issue or you can use .lastIndexOf() to start the search from the end of the string. You can also pass regular expressions to many of the string related methods and that is probably the best solution for extracting the correct positions based on some search criteria.
var x = "thewholestring";
// Get position where "whole" starts
var start = x.indexOf("whole");
// Extract 5 characters starting at that position
var result = x.substr(start, 5);
console.log(result);
I have a little question which I don't find any documentation about it (or it doesn't exist, or I'm searching in a wrong way..)
I have an ID which I want to get the whole string before the last character I choose.
In this case:
str_stringId = "AB_generalForm_DateTime_calendar";
str_dateId = str_firstId.substring(str_firstId.lastIndexOf("_") +1);
In this case, str_dateId returns "calendar", but I want the whole string behind it.
Is there a way to reverse this operation or do I need to count the number of letters of the whole string and then subtract with the length of "str_dateId" and after that, substring with the result?
Thank you.
The definition is string.substring(start,end) so you can simply
str_dateId = str_firstId.substring(0, str_firstId.lastIndexOf("_"));
I am trying to extract the first character after the last underscore in a string with an unknown number of '_' in the string but in my case there will always be one, because I added it in another step of the process.
What I tried is this. I also tried the regex by itself to extract from the name, but my result was empty.
var s = "XXXX-XXXX_XX_DigitalF.pdf"
var string = match(/[^_]*$/)[1]
string.charAt(0)
So the final desired result is 'D'. If the RegEx can only get me what is behind the last '_' that is fine because I know I can use the charAt like currently shown. However, if the regex can do the whole thing, even better.
If you know there will always be at least one underscore you can do this:
var s = "XXXX-XXXX_XX_DigitalF.pdf"
var firstCharAfterUnderscore = s.charAt(s.lastIndexOf("_") + 1);
// OR, with regex
var firstCharAfterUnderscore = s.match(/_([^_])[^_]*$/)[1]
With the regex, you can extract just the one letter by using parentheses to capture that part of the match. But I think the .lastIndexOf() version is easier to read.
Either way if there's a possibility of no underscores in the input you'd need to add some additional logic.
I'm fairly sure after spending the night trying to find an answer that this isn't possible, and I've developed a work around - but, if someone knows of a better method, I would love to hear it...
I've gone through a lot of iterations on the code, and the following is just a line of thought really. At some point I was using the global flag, I believe, in order for match() to work, and I can't remember if it was necessary now or not.
var str = "#abc#def#ghi&jkl";
var regex = /^(?:#([a-z]+))?(?:&([a-z]+))?$/;
The idea here, in this simplified code, is the optional group 1, of which there is an unspecified amount, will match #abc, #def and #ghi. It will only capture the alpha characters of which there will be one or more. Group 2 is the same, except matches on & symbol. It should also be anchored to the start and end of the string.
I want to be able to back reference all matches of both groups, ie:
result = str.match(regex);
alert(result[1]); //abc,def,ghi
alert(result[1][0]); //abc
alert(result[1][1]); //def
alert(result[1][2]); //ghi
alert(result[2]); //jkl
My mate says this works fine for him in .net, unfortunately I simply can't get it to work - only the last matched of any group is returned in the back reference, as can be seen in the following:
(additionally, making either group optional makes a mess, as does setting global flag)
var str = "#abc#def#ghi&jkl";
var regex = /(?:#([a-z]+))(?:&([a-z]+))/;
var result = str.match(regex);
alert(result[1]); //ghi
alert(result[1][0]); //g
alert(result[2]); //jkl
The following is the solution I arrived at, capturing the whole portion in question, and creating the array myself:
var str = "#abc#def#ghi&jkl";
var regex = /^([#a-z]+)?(?:&([a-z]+))?$/;
var result = regex.exec(str);
alert(result[1]); //#abc#def#ghi
alert(result[2]); //jkl
var result1 = result[1].toString();
result[1] = result1.split('#')
alert(result[1][1]); //abc
alert(result[1][2]); //def
alert(result[1][3]); //ghi
alert(result[2]); //jkl
That's simply not how .match() works in JavaScript. The returned array is an array of simple strings. There's no "nesting" of capture groups; you just count the ( symbols from left to right.
The first string (at index [0]) is always the overall matched string. Then come the capture groups, one string (or null) per array element.
You can, as you've done, rearrange the result array to your heart's content. It's just an array.
edit — oh, and the reason your result[1][0] was "g" is that array indexing notation applied to a string gets you the individual characters of the string.
I need to do a lot of regex things in javascript but am having some issues with the syntax and I can't seem to find a definitive resource on this.. for some reason when I do:
var tesst = "afskfsd33j"
var test = tesst.match(/a(.*)j/);
alert (test)
it shows
"afskfsd33j, fskfsd33"
I'm not sure why its giving this output of original and the matched string, I am wondering how I can get it to just give the match (essentially extracting the part I want from the original string)
Thanks for any advice
match returns an array.
The default string representation of an array in JavaScript is the elements of the array separated by commas. In this case the desired result is in the second element of the array:
var tesst = "afskfsd33j"
var test = tesst.match(/a(.*)j/);
alert (test[1]);
Each group defined by parenthesis () is captured during processing and each captured group content is pushed into result array in same order as groups within pattern starts. See more on http://www.regular-expressions.info/brackets.html and http://www.regular-expressions.info/refcapture.html (choose right language to see supported features)
var source = "afskfsd33j"
var result = source.match(/a(.*)j/);
result: ["afskfsd33j", "fskfsd33"]
The reason why you received this exact result is following:
First value in array is the first found string which confirms the entire pattern. So it should definitely start with "a" followed by any number of any characters and ends with first "j" char after starting "a".
Second value in array is captured group defined by parenthesis. In your case group contain entire pattern match without content defined outside parenthesis, so exactly "fskfsd33".
If you want to get rid of second value in array you may define pattern like this:
/a(?:.*)j/
where "?:" means that group of chars which match the content in parenthesis will not be part of resulting array.
Other options might be in this simple case to write pattern without any group because it is not necessary to use group at all:
/a.*j/
If you want to just check whether source text matches the pattern and does not care about which text it found than you may try:
var result = /a.*j/.test(source);
The result should return then only true|false values. For more info see http://www.javascriptkit.com/javatutors/re3.shtml
I think your problem is that the match method is returning an array. The 0th item in the array is the original string, the 1st thru nth items correspond to the 1st through nth matched parenthesised items. Your "alert()" call is showing the entire array.
Just get rid of the parenthesis and that will give you an array with one element and:
Change this line
var test = tesst.match(/a(.*)j/);
To this
var test = tesst.match(/a.*j/);
If you add parenthesis the match() function will find two match for you one for whole expression and one for the expression inside the parenthesis
Also according to developer.mozilla.org docs :
If you only want the first match found, you might want to use
RegExp.exec() instead.
You can use the below code:
RegExp(/a.*j/).exec("afskfsd33j")
I've just had the same problem.
You only get the text twice in your result if you include a match group (in brackets) and the 'g' (global) modifier.
The first item always is the first result, normally OK when using match(reg) on a short string, however when using a construct like:
while ((result = reg.exec(string)) !== null){
console.log(result);
}
the results are a little different.
Try the following code:
var regEx = new RegExp('([0-9]+ (cat|fish))','g'), sampleString="1 cat and 2 fish";
var result = sample_string.match(regEx);
console.log(JSON.stringify(result));
// ["1 cat","2 fish"]
var reg = new RegExp('[0-9]+ (cat|fish)','g'), sampleString="1 cat and 2 fish";
while ((result = reg.exec(sampleString)) !== null) {
console.dir(JSON.stringify(result))
};
// '["1 cat","cat"]'
// '["2 fish","fish"]'
var reg = new RegExp('([0-9]+ (cat|fish))','g'), sampleString="1 cat and 2 fish";
while ((result = reg.exec(sampleString)) !== null){
console.dir(JSON.stringify(result))
};
// '["1 cat","1 cat","cat"]'
// '["2 fish","2 fish","fish"]'
(tested on recent V8 - Chrome, Node.js)
The best answer is currently a comment which I can't upvote, so credit to #Mic.