All I need to do here is to add a variable before each specific string.
Example:
var exampleString = "blabla:test abcde 123test:123";
var formattedString = "el.blabla:test abcde el.123test:123";
As you can see, when I have something like "XXX:XXX", I need to add a variable before it.
I have the Regex to find "XXX:"
var regex = new RegExp(/\w+([aA-zZ]:)/g)
But when I try to replace it, it replaces all instead of adding the variable "el."
var exampleString = "blabla:test abcde 123test:123";
var formattedString = exampleString.replace(new RegExp(/\w+([aA-zZ]:)/g), 'el.');
// formattedString is now "el.test abcde el.123"
// Instead of "el.blabla:test abcde el.123test:123"
Could anyone makes this work ? Thanks :)
Source: Javascript Regex: How to put a variable inside a regular expression?
var exampleString = "blabla:test abcde 123test:123";
var formattedString = exampleString.replace(/\w*:\w*/gi, 'el.$&');
console.log(formattedString);
Regex use and Explanation Here https://regex101.com/r/U2KeXi/3
Sample Fiddle here https://jsfiddle.net/a8wyLb0g/2/
You need to use ^ to match only at the beginning. And remove the g modifier, since you only want to replace once, not every time.
There's also no reason to use new RegExp(), just use a RegExp literal.
In the replacement string, you need to use $& to copy the original string into the replacement.
var exampleString = "blabla:test abcde 123test:123";
var formattedString = exampleString.replace(/^\w+[a-z]:/i, 'el.$&');
console.log(formattedString);
Also, the proper way to match all letters in either case is with [A-Za-z], not [aA-zZ], or use the i modifier to make the regexp case-insensitive. Your regexp matches all characters in the range A-z, which includes lots of punctuation characters that are between the uppercase letters and lowercase letters in the ASCII code.
Just use this
exampleString.replace(/(\w*):(\w*)/gi, 'el.$1:$2');
REGEXP explanation :
capturing group (\w*) is for capturing any alphabets in any number of occurance,
$1 and $2 specifies the first and second capturing group.
You should use a function like insertAt instead replace, see following example:
String.prototype.insertAt=function(index, string) {
return this.substr(0, index) + string + this.substr(index);
}
var exampleString = "blabla:test abcde 123test:123";
var regex = new RegExp(/\w+([aA-zZ]:)/g)
var formattedString = exampleString;
while ( (result = regex.exec(exampleString)) ) {
formattedString = formattedString.insertAt(result.index, "el.");
}
console.log(formattedString);
I hope it helps you, bye.
Related
I want to find ++ or -- or // or ** sign in in string can anyone help me?
var str = document.getElementById('screen').innerHTML;
var res = str.substring(0, str.length);
var patt1 = ++,--,//,**;
var result = str.match(patt1);
if (result)
{
alert("you cant do this :l");
document.getElementById('screen').innerHTML='';
}
This finds doubles of the characters by a backreference:
/([+\/*-])\1/g
[from q. comments]: i know this but when i type var patt1 = /[++]/i; code find + and ++
[++] means one arbitrary of the characters. Normally + is the qantifier "1 or more" and needs to be escaped by a leading backslash when it should be a literal, except in brackets where it does not have any special meaning.
Characters that do need to be escaped in character classes are e.g. the escape character itself (backslash), the expression delimimiter (slash), the closing bracket and the range operator (dash/minus), the latter except at the end of the character class as in my code example.
A character class [] matches one character. A quantifier, e.g. [abc]{2} would match "aa", "bb", but "ab" as well.
You can use a backreference to a match in parentheses:
/(abc)\1
Here the \1 refers to the first parentheses (abc). The entire expression would match "abcabc".
To clarify again: We could use a quantifier on the backreference:
/([+\/*-])\1{9}/g
This matches exactly 10 equal characters out of the class, the subpattern itself and 9 backreferences more.
/.../g finds all occurrences due to the modifier global (g).
test-case on regextester.com
Define your pattern like this:
var patt1 = /\+\+|--|\/\/|\*\*/;
Now it should do what you want.
More info about regular expressions: https://developer.mozilla.org/en-US/docs/Web/JavaScript/Guide/Regular_Expressions
You can use:
/\+\+|--|\/\/|\*\*/
as your expression.
Here I have escaped the special characters by using a backslash before each (\).
I've also used .test(str) on the regular expression as all you need is a boolean (true/false) result.
See working example below:
var str = document.getElementById('screen').innerHTML;
var res = str.substring(0, str.length);
var patt1 = /\+\+|--|\/\/|\*\*/;
var result = patt1.test(res);
if (result) {
alert("you cant do this :l");
document.getElementById('screen').innerHTML = '';
}
<div id="screen">
This is some++ text
</div>
Try this:-
As
n+:- Matches any string that contains at least one n
n* Matches any string that contains zero or more occurrences of n
We need to use backslash before this special characters.
var str = document.getElementById('screen').innerHTML;
var res = str.substring(0, str.length);
var patt1 = /\+\+|--|\/\/|\*\*/;
var result = str.match(patt1);
if (result)
{
alert("you cant do this :l");
document.getElementById('screen').innerHTML='';
}
<div id="screen">2121++</div>
I have a string (url) like this:
https://8.random.url.com/g/DFGTER5675/test1/undefined/codec/
Here is my regex:
/https\:\/\/(?:.*)\/g\/(?:.*)\/(?:.*)\/(.*)\/codec\/(?:.*)/gi
And my code:
var string = "https://8.random.url.com/g/DFGTER5675/test1/undefined/codec/";
var myRegexp = /https\:\/\/(?:.*)\/g\/(?:.*)\/(?:.*)\/(.*)\/codec\/(?:.*)/gi
var match = string.replace(myRegexp, "OMEGA3");
When I do console.log(match) it returns only "OMEGA3". What I want is just my string with "undefined" replaced by "OMEGA3". What am I doing wrong? Thanks.
You can use this regex with capturing groups and back-reference:
url = url.replace(/(https?:\/\/[^\/]*\/g\/[^\/]*\/[^\/]*\/).*(\/codec\/)/gi, '$1OMEGA3$2');
RegEx Demo
You have the use of the capture group backwards. You should be capturing the parts of the pattern that you want to keep, not the part you want to replace. Then use $1, $2, etc. to copy those to the replacement.
You also have several non-capturing groups that aren't needed at all.
var myRegexp = /(https:\/\/.*\/g\/.*\/).*(\/codec\/)/gi
var match = string.replace(myRegexp, "$1OMEGA3$2");
why not use /undefined/gi
var string = "https://8.random.url.com/g/DFGTER5675/test1/undefined/codec/";
var myRegexp = /(https\:\/\/.*?\/.*?\/.*?\/.*?\/).*?(\/.*?\/)/gi
var match = string.replace(myRegexp, "$1OMEGA3$2");
console.log(match)
Let's say I have a string that starts by 7878 and ends by 0d0a or 0D0A such as:
var string = "78780d0101234567890123450016efe20d0a";
var string2 = "78780d0101234567890123450016efe20d0a78780d0103588990504943870016efe20d0a";
var string 3 = "78780d0101234567890123450016efe20d0a78780d0103588990504943870016efe20d0a78780d0101234567890123450016efe20d0a"
How can I split it by regex so it becomes an array like:
['78780d0101234567890123450016efe20d0a']
['78780d0101234567890123450016efe20d0a','78780d0101234567890123450016efe20d0a']
['78780d0101234567890123450016efe20d0a','78780d0101234567890123450016efe20d0a','78780d0101234567890123450016efe20d0a']
You can split the string with a positive lookahead (?=7878). The regex isn't consuming any characters, so 7878 will be part of the string.
var rgx = /(?=7878)/;
console.log(string1.split(rgx));
console.log(string2.split(rgx));
console.log(string3.split(rgx));
Another option is to split on '7878' and then take all the elements except first and add '7878' to each of them. For example:
var arr = string3.split('7878').slice(1).map(function(str){
return '7878' + str;
});
That works BUT it also matches strings that do NOT end on 0d0a. How
can I only matches those ending on 0d0a OR 0D0A?
Well, then you can use String.match with a plain regex.
console.log(string3.match(/7878.*?0d0a/ig));
I'm sure this is an easy one, but I can't find it on the net.
This code:
var new_html = "foo and bar(arg)";
var bad_string = "bar(arg)";
var regex = new RegExp(bad_string, "igm");
var bad_start = new_html.search(regex);
sets bad_start to -1 (not found). If I remove the (arg), it runs as expected (bad_start == 8). Is there something I can do to make the (very handy) "new Regexp" syntax work, or do I have to find another way? This example is trivial, but in the real app it would be doing global search and replace, so I need the regex and the "g". Or do I?
TIA
Escape the brackets by double back slashes \\. Try this.
var new_html = "foo and bar(arg)";
var bad_string = "bar\\(arg\\)";
var regex = new RegExp(bad_string, "igm");
var bad_start = new_html.search(regex);
Demo
Your RegEx definition string should be:
var bad_string = "bar\\(arg\\)";
Special characters need to be escaped when using RegEx, and because you are building the RegEx in a string you need to escape your escape character :P
http://www.regular-expressions.info/characters.html
You need to escape the special characters contained in string you are creating your Regex from. For example, define this function:
function escapeRegex(string) {
return string.replace(/[/\-\\^$*+?.()|[\]{}]/g, '\\$&');
}
And use it to assign the result to your bad_string variable:
let bad_string = "bar(arg)"
bad_string = escapeRegex(bad_string)
// You can now use the string to create the Regex :v:
I have a string like foobar1, foobaz2, barbar23, nobar100 I want only foobar, foobaz, barbar, nobar and ignoring the number part.
If you want to strip out things that are digits, a regex can do that for you:
var s = "foobar1";
s = s.replace(/\d/g, "");
alert(s);
// "foobar"
(\d is the regex class for "digit". We're replacing them with nothing.)
Note that as given, it will remove any digit anywhere in the string.
This can be done in JavaScript:
/^[^\d]+/.exec("foobar1")[0]
This will return all characters from the beginning of string until a number is found.
var str = 'foobar1, foobaz2, barbar23, nobar100';
console.log(str.replace(/\d/g, ''));
Find some more information about regular expressions in javascript...
This should do what you want:
var re = /[0-9]*/g;
var newvalue= oldvalue.replace(re,"");
This replaces al numbers in the entire string. If you only want to remove at the end then use this:
var re = /[0-9]*$/g;
I don't know how to do that in JQuery, but in JavaScript you can just use a regular expression string replace.
var yourString = "foobar1, foobaz2, barbar23, nobar100";
var yourStringMinusDigits = yourString.replace(/\d/g,"");