In Javascript, i want the below original string:
I want to replace \"this\" and \"that\" words, but NOT the one "here"
.. to become like:
I want to replace ^this^ and ^that^ words, but NOT the one "here"
I tried something like:
var str = 'I want to replace \"this\" and \"that\" words, but NOT the one "here"';
str = str.replace(/\"/g,"^");
console.log( str );
Demo: JSFiddle here.
But still .. the output is:
I want to replace ^this^ and ^that^ words, but NOT the one ^here^
Which means i wanted to replace only the \" occurrences but NOT the " alone. But i cannot.
Please help.
As #adeneo's comment, your string was created wrong and not exactly like your expectation. Please try this:
var str = 'I want to replace \\"this\\" and \\"that\\" words, but not the one "here"';
str = str.replace(/\\\"/g,"^");
console.log(str);
You can use RegExp /(")/, String.prototype.lastIndexOf(), String.prototype.slice() to check if matched character is last or second to last match in input string. If true, return original match, else replace match with "^" character.
var str = `I want to replace \"this\" and \"that\" words, but NOT the one "here"`;
var res = str.replace(/(")/g, function(match, _, index) {
return index === str.lastIndexOf(match)
|| index === str.slice(0, str.lastIndexOf(match) -1)
.lastIndexOf(match)
? match
: "^"
});
console.log(res);
The problem with String.prototype.replace is that it only replaces the first occurrence without Regular Expression. To fix this, you need to add a g and the end of the RegEx, like so:
var mod = str => str.replace(/\\\"/g,'^');
mod('I want to replace \\"this\\" and \\"that\\" words, but NOT the one "here"');
A less effective but easier to understand to do what you wanted is to split the string with the delimiter and then join it with the replacement, like so:
var mod = str => str.split('\\"').join('^');
mod('I want to replace \\"this\\" and \\"that\\" words, but NOT the one "here"');
Note: You can wrap a string with either ' or ". Suppose your string contains ", i.e. a"a, you will need to put an \ in front of the " as "a"a" causes syntax error. 'a"a' won't cause syntax error as the parser knows the " is part of the string, but when you put a \ in front of " or any other special characters, it means the following character is a special character. So 'a\"a' === 'a"a' === "a\"a". If you want to store \, you will need to use \ regardless of the type of quote you use, so to store \", you will need to use '\\"', '\\\"' or "\\\"".
Related
I have a huge string with this inside:
linha[0] = '12/2010 281R$ 272.139,05 ';
linha[0] = '13SL 1R$ 226.185,81 ';
Both lines are separate, and I need get the last occurrence from both. I'm using the following regex to match the first one:
/linha\[0]\s=\s'(.*)';/
I would like to get the second "linha..." too, but I don't know exactly how.
That's how i'm using this regex to get the first "linha...":
string.match(/linha\[0]\s=\s'(.*)';/);
output:
linha[0] = '12/2010 281R$ 272.139,05 ';
Also, i can't do extra work, i need get the second occurrence using only regex.
If you want to get the last occurrence of your regex in a string (and assuming it exists), you can do
var str = hugeString.match(/linha\[0]\s=\s'([^']*)';/g).pop();
(yes, I changed .* to [^']* for a better efficiency, ignore that if you have quotes in your inner string)
Now, if you want to extract just the submatch, you can do
var regex = /linha\[0]\s=\s'([^']*)';/g,
arr,
str;
while (arr = regex.exec(hugeString)) str = arr[1];
I have the following string:
var str = '\x27';
I have no control on it, so I cannot write it as '\\x27' for example. Whenever I print it, i get:
'
since 27 is the apostrophe. When I call .length on it, it gives me 1. This is of course correct, but how can I treat it like a not escaped string and have it print literally
\x27
and give me a length of 4?
I'm not sure if you should do what you are trying to do, but this is how it works:
var s = '\x27';
var sEncoded = '\\x' + s.charCodeAt(0).toString(16);
s is a string that contains one character, the apostrophe. The character code as a hexadecimal number is 27.
After the assignment var str = '\x27';, you can't tell where the contents of str came from. There's no way to find out whether a string literal was assigned, or whether the string literal contained an escape sequence. All you have is a string containing a single apostrophe character (Unicode code point U+0027). The original assignment could have been
var str = '\x27'; // or
var str = "'"; // or
var str = String.fromCodePoint(3 * 13);
There's simply no way to tell.
That said, your question looks like an XY problem. Why are you trying to print \x27 in the first place?
I'm trying to build a regular expression that parses a string and skips things in brackets.
Something like
string = "A bc defg hi [hi] jkl mnop.";
The .match() should return "hi" but not [hi]. I've spent 5 hours running through RE's but I'm throwing in the towel.
Also this is for javascript or jquery if that matters.
Any help is appreciated. Also I'm working on getting my questions formatted correctly : )
EDIT:
Ok I just had a eureka moment and figured out that the original RegExp I was using actually did work. But when I was replaces the matches with the [matches] it simply replaced the first match in the string... over and over. I thought this was my regex refusing to skip the brackets but after much time of trying almost all of the solutions below, I realized that I was derping Hardcore.
When .replace was working its magic it was on the first match, so I quite simply added a space to the end of the result word as follows:
var result = string.match(regex);
var modifiedResult = '[' + result[0].toString() + ']';
string.replace(result[0].toString() + ' ', modifiedResult + ' ');
This got it to stop targeting the original word in the string and stop adding a new set of brackets to it with every match. Thank you all for your help. I am going to give answer credit to the post that prodded me in the right direction.
preprocess the target string by removing everything between brackets before trying to match your RE
string = "A bc defg hi [hi] jkl mnop."
tmpstring = string.replace(/\[.*\]/, "")
then apply your RE to tmpstring
correction: made the match for brackets eager per nhahtd comment below, and also, made the RE global
string = "A bc defg hi [hi] jkl mnop."
tmpstring = string.replace(/\[.*?\]/g, "")
You don't necessarily need regex for this. Simply use string manipulation:
var arr = string.split("[");
var final = arr[0] + arr[1].split("]")[1];
If there are multiple bracketed expressions, use a loop:
while (string.indexOf("[") != -1){
var arr = string.split("[");
string = arr[0] + arr.slice(1).join("[").split("]").slice(1).join("]");
}
Using only Regular Expressions, you can use:
hi(?!])
as an example.
Look here about negative lookahead: http://www.regular-expressions.info/lookaround.html
Unfortunately, javascript does not support negative lookbehind.
I used http://regexpal.com/ to test, abcd[hi]jkhilmnop as test data, hi(?!]) as the regex to find. It matched 'hi' without matching '[hi]'. Basically it matched the 'hi' so long as there was not a following ']' character.
This of course, can be expanded if needed. This has a benefit of not requiring any pre-processing for the string.
r"\[(.*)\]"
Just play arounds with this if you wanto to use regular expressions.
What do yo uwant to do with it? If you want to selectively replace parts like "hi" except when it's "[hi]", then I often use a system where I match what I want to avoid first and then what I want to watch; if it matches what I want to avoid then I return the match, otherwise I return the processed match.
Like this:
return string.replace(/(\[\w+\])|(\w+)/g, function(all, m1, m2) {return m1 || m2.toUpperCase()});
which, with the given string, returns:
"A BC DEFG HI [hi] JKL MNOP."
Thus: it replaces every word with uppercase (m1 is empty), except if the word is between square brackets (m1 is not empty).
This builds an array of all the strings contained in [ ]:
var regex = /\[([^\]]*)\]/;
var string = "A bc defg hi [hi] [jkl] mnop.";
var results=[], result;
while(result = regex.exec(string))
results.push(result[1]);
edit
To answer to the question, this regex returns the string less all is in [ ], and trim whitespaces:
"A bc defg [hi] mnop [jkl].".replace(/(\s{0,1})\[[^\]]*\](\s{0,1})/g,'$1')
Instead of skipping the match you can probably try something different - match everything but do not capture the string within square brackets (inclusive) with something like this:
var r = /(?:\[.*?[^\[\]]\])|(.)/g;
var result;
var str = [];
while((result = r.exec(s)) !== null){
if(result[1] !== undefined){ //true if [string] matched but not captured
str.push(result[1]);
}
}
console.log(str.join(''));
The last line will print parts of the string which do not match the [string] pattern. For example, when called with the input "A [bc] [defg] hi [hi] j[kl]u m[no]p." the code prints "A hi ju mp." with whitespaces intact.
You can try different things with this code e.g. replacing etc.
I'm sorry if it is a confusing question. I was trying to find a way to do this but couldn't find it so, if it is a repeated question, my apologies!
I have a text something like this: something:"askjnqwe234"
I want to be able to get askjnqwe234 using a RegExp. You can notice I want to omit the quotes. I was trying this using /[^"]+(?=(" ")|"$)/g but it returns an array. I want a RegExt to return a single string, not an array.
I don't know if it's possible but I do not want to specify the position of the array; something like this:
var x = string.match(/[^"]+(?=(" ")|"$)/g)[0];
Thanks!
Try:
/"([^"]*)"/g
in English: look for " the match and record anything that isn't " till you see another "".
match and exec always return an array or null, so, assuming you have a single double-quoted value and no newlines in the string, you could use
var x;
var str = 'something:"askjnqwe234"';
x = str.replace( /^[^"]*"|".*/g, '' );
// "askjnqwe234"
Or, if you may have other quoted values in the string
x = str.replace( /.*?something:"([^"]*)".*/, '$1' );
where $1 refers to the substring captured by the sub-pattern [^"]* between the ().
Further explanation on request.
Notwithstanding the above, I recommend that you tolerate the array indexing and just use match.
You can capture the information inside quotes like this, assuming it matches:
var x = string.match(/something:"([^"]*)"/)[1];
The memory capture at index 1 is the part inside the double quotes.
If you're not sure it will match:
var match = string.match(/something:"([^"]*)"/);
if (match) {
// use match[1] here
}
I have been looking for this for a while, and while I have found many responses for changing a space into a dash (hyphen), I haven't found any that go the other direction.
Initially I have:
var str = "This-is-a-news-item-";
I try to replace it with:
str.replace("-", ' ');
And simply display the result:
alert(str);
Right now, it doesn't do anything, so I'm not sure where to turn. I tried reversing some of the existing ones that replace the space with the dash, and that doesn't work either.
Thanks for the help.
This fixes it:
let str = "This-is-a-news-item-";
str = str.replace(/-/g, ' ');
alert(str);
There were two problems with your code:
First, String.replace() doesn’t change the string itself, it returns a changed string.
Second, if you pass a string to the replace function, it will only replace the first instance it encounters. That’s why I passed a regular expression with the g flag, for 'global', so that all instances will be replaced.
replace() returns an new string, and the original string is not modified. You need to do
str = str.replace(/-/g, ' ');
I think the problem you are facing is almost this: -
str = str.replace("-", ' ');
You need to re-assign the result of the replacement to str, to see the reflected change.
From MSDN Javascript reference: -
The result of the replace method is a copy of stringObj after the
specified replacements have been made.
To replace all the -, you would need to use /g modifier with a regex parameter: -
str = str.replace(/-/g, ' ');
var str = "This-is-a-news-item-";
while (str.contains("-")) {
str = str.replace("-", ' ');
}
alert(str);
I found that one use of str.replace() would only replace the first hyphen, so I looped thru while the input string still contained any hyphens, and replaced them all.
http://jsfiddle.net/LGCYF/
In addition to the answers already given you probably want to replace all the occurrences. To do this you will need a regular expression as follows :
str = str.replace(/-/g, ' '); // Replace all '-' with ' '
Use replaceAll() in combo with trim() may meet your needs.
const str = '-This-is-a-news-item-';
console.log(str.replaceAll('-', ' ').trim());
Imagine you end up with double dashes, and want to replace them with a single character and not doubles of the replace character. You can just use array split and array filter and array join.
var str = "This-is---a--news-----item----";
Then to replace all dashes with single spaces, you could do this:
var newStr = str.split('-').filter(function(item) {
item = item ? item.replace(/-/g, ''): item
return item;
}).join(' ');
Now if the string contains double dashes, like '----' then array split will produce an element with 3 dashes in it (because it split on the first dash). So by using this line:
item = item ? item.replace(/-/g, ''): item
The filter method removes those extra dashes so the element will be ignored on the filter iteration. The above line also accounts for if item is already an empty element so it doesn't crash on item.replace.
Then when your string join runs on the filtered elements, you end up with this output:
"This is a news item"
Now if you were using something like knockout.js where you can have computer observables. You could create a computed observable to always calculate "newStr" when "str" changes so you'd always have a version of the string with no dashes even if you change the value of the original input string. Basically they are bound together. I'm sure other JS frameworks can do similar things.
if its array like
arr = ["This-is-one","This-is-two","This-is-three"];
arr.forEach((sing,index) => {
arr[index] = sing.split("-").join(" ")
});
Output will be
['This is one', 'This is two', 'This is three']