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So I have a string that will have basically any kind of text in it and I want to capture all the text with regular expression and replace it.
var img_url = "http://grfx.domain.com/photos/dir/school/dir/m-dir/auto_original/123456.jpeg?123456";
img_url.replace(/regexp/,newvalue);
Just trying to capture everything in the img_url var and replace it.
Thanks!
Well, if you want to replace everything, you don't need regex. Simply write:
img_url = newvalue;
If you are really determined to do this with a regular expression, this could help:
img_url = img_url.replace(/.*/, newvalue);
If you're reassigning, just use =
If you want to replace all matches in one go, use the g flag, in which case this question is a duplicate of this post
If that expression needs to be dynamic:
var expr = new RegExp(replaceVar,'gi');//globally, and case-insensitively
var newUrl = oldUrl.replace(expr,replacement);
Note that, here, all back-slashes like you'd use for \b have to be escaped, too: \\b and \\/ for forward slashes. and \\\\ for an escaped backslash
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Sorry i can't speak english :(
I write Supplant I have a regex
({{([']).*\2}})
needs regex to detect what is in a rectangle
I need it rigidly between {{' - '}} because there can be any character between the quotation marks (e.g. one { )
best to use it \1 because regex is more longer
Thanks for the help. Regards
This can work for you:
/\{\{'((?:(?!\{\{').)*?)'\}\}/g
\{\{' - start by matching {{'.
( ... ) - if you're writing a template engine you probably care about what's inside the curly braces and quotes. This captures the string inside the quotes so you'll be able to use it.
(?: ... ) - a non-captureing group (the value here will not be used).
(?!\{\{'). - match anything, except if we're seeing another {{'.
(?!\{\{').)*? - *? is a lazy match, so we'll stop at the first '}}
and finally, the closing '}}
as code it will looks something like this - I included a function to set the replaced value, because typically that's what you'd do in a template engine:
let s = "n {{'a{123456789}'}} n {{\"a{123456789}\"}} n {{'a{1234{{'a{12345{{'a{123456789}'}}6789}'}}56789}'}} ";
s = s.replace(/\{\{'((?:(?!\{\{').)*?)'\}\}/g, (wholeMatch, capturedKey) => {
console.log('captured key:', capturedKey);
return "REPLACED " + capturedKey;
});
console.log(s);
if you want to support double quotes it becomes a bit more complicated:
s = s.replace(/\{\{(["'])((?:(?!\{\{["']).)*?)\1\}\}/g,
(wholeMatch, quote, capturedKey) => { ... }
);
Try this:/(?<={{')(?!.+?{{)[^']+/gm
This basically looks for the {{' which doesn't have any other {{ in front of it then, matches everything till the first '.
Test here: https://regex101.com/r/rw6kkP/2
var myString = `{{'a{content 1}a'}}
{{'a{every{{'a{everysign}a'}}sign}a'}}
{{'a{every{{'a{every{{'a{inside content}a'}}sign}a'}}sign}a'}} `;
var myRegexp = /(?<={{')(?!.+?{{)[^']+/gm;
console.log(myString.match(myRegexp))
// [ "a{content 1}a", "a{everysign}a", "a{inside content}a" ]
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I try to regex for not contain some character.
I need to show /%7(.*?);/g which dose not contain "=".
I try to input
?!xx=1
and change to
?!( (.?)=(.?) )
But it dose not work.
Please help. Thanks.
//Here is my simple regex
reg = /%7((?!xx=1).*?);/g ;
//Here is my string
str = "%7aa; %7bb=11; %7cc=123; %7xx=1; %7yy; %7zz=2;"
//I need
%7aa; and %7yy;
Instead of using a negative lookahead, try using a ^ block:
const reg = /%7([^=;]+);/g;
The ([^=;]+) bit matches any non-=, the condition you're looking for, and non-;, the character at the end of your regex.
I left the capture group in since your question's regex also contains it.
const reg = /%7([^=;]+);/g;
const str = "%7aa; %7bb=11; %7cc=123; %7xx=1; %7yy; %7zz=2;"
const matches = str.match(reg);
console.log(matches);
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I have a string 'OR-xxxxxxxx-001-01'.
I need to split it as 'OR-xxxxxxx-001' and '01'. Is it possible without split by '-' and concatenating again ??
In Java, use lastIndexOf:
String string = "OR-xxxxxxxx-001-01";
int lastDash = string.lastIndexOf('-');
String prefix = string.substring(0, lastDash); // OR-xxxxxxx-001
String suffix = string.substring(lastDash + 1); // 01
This solution is for JavaScript (Java and JavaScript is not the same thing).
Since you don't want to split by - it's possible to do it with indexes. You haven't specified if it's of fixed length or not (which this solution depends on), but this should do the trick:
t = 'OR-xxxxxxxx-001-01'
str = t.slice(0, 12) + t.slice(16);
Will give str = 'OR-xxxxxxxx-01'
You could use a split with a positive look-ahead:
String str = "OR-xxxxxxxx-001-01";
String[] split = str.split("-(?=[^-]+$)");
[^-]: any character other than -.
+: repeated one or more times
$: at the end of the String
Then the (?= ... ) is used for the positive look-ahead, so it won't be removed by the split. This means only the - outside the parenthesis act as delimiter to split on.
Try it online.
PS: In Java.
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I don't know how to replace with capital letters in this input text, for example I want to put on bold the letters which they are de same on the input, but if is capital letters doesn't put on bold
this is my line of code $texto_option = $texto_option.replace($(input).val(),'<b>'+$(input).val()+'</b>');
Generate a regular expression that ignores the letters' case:
var search = $(input).val();
var re = new RegExp(search, "i");
$texto_option = $texto_option.replace(re, "<b>$&</b>");
This is an easy answer. But keep in mind that your input has to be sanitized, as some characters are control characters for regular expressions too:
search = search.replace(/[\-\[\]\/\{\}\(\)\*\+\?\.\\\^\$\|]/g, "\\$&");
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I am looking for a method to get the regex rules, for a capture group.
So if I had: /(\w) (\d)/ I would want $1 = /\w/ and $2 = /\d/.
Is there a method for this?
Provided that the regex is valid and no nested groups are used, then you can try this out.
var arr = (regex + "").match(/\(.*?\)/g).map(function(rule){
return rule.replace(/[()]/g, "");
});
Now, arr[0] will have rule of group1, arr[1] will be group2 and so on.
Although I don't know of any direct way to do this on a regex rule given by /(rule1)(rule2)/ you could build the regex rule using strings and specify your grouping in different strings.
var group1 = '([A-Za-z]+)',
group2 = '(\\d+)',
r = new RegExp(group1+group2);
r.test('hello1234');
To get the groups from a regex string you could run a regex on that regex string to extract the groups. If your regex string is "(\w+)(\d+)" then you'd have a regex to extract the groups as /(\([^\)\))+/