I am trying to replace all occurrences of http://xyz.yzx.com/abc/def/ by /qwe/ in String in Node js. I am very new to Node js. So, I think I am doing some mistake in syntax. Following is the code with which I am trying.
var newString = originalString.replace("http:\/\/xyz\.yzx\.com\/abc\/def\//g", "/qwe/");
But this is not doing any replace. Can someone suggest what I am doing wrong? I have tried lot of tweaks but somehow I am not able to achieve the replace all occurrences.
Any suggestions you can give would be really appreciated.
If a string is passed as replaces first argument it will only replace the first occurence ( which is badly implemented in my opinion). So we either use a small workaround:
var newString = originalString
.split("http:\/\/xyz\.yzx\.com\/abc\/def")
.join("/qwe/");
Or we need to remove the string literal and escape every / with a \/ ... :/
This works fine, have a look:
originalString.replace(/\http:\/\/xyz\.yzx\.com\/abc\/def\//g, '/qwe/')
Related
I have this in a javascript/jQuery string (This string is grabbed from an html ($('#shortcode')) elements value which could be changed if user clicks some buttons)
[csvtohtml_create include_rows="1-10"
debug_mode="no" source_type="visualizer_plugin" path="map"
source_files="bundeslander_staple.csv" include cols="1,2,4" exclude cols="3"]
In a textbox (named incl_sc) I have the value:
include cols="2,4"
I want to replace include_cols="1,2,4" from the above string with the value from the textbox.
so basically:
How do I replace include_cols values here? (include_cols="2,4" instead of include_cols="1,2,4") I'm great at many things but regex is not one of them. I guess regex is the thing to use here?
I'm trying this:
var s = $('#shortcode').html();
//I want to replace include cols="1,2,4" exclude cols="3"
//with include_cols="1,2" exclude_cols="3" for example
s.replace('/([include="])[^]*?\1/g', incl_sc.val() );
but I don't get any replacement at all (the string s is same string as $("#shortcode").html(). Obviously I'm doing something really dumb. Please help :-)
In short what you will need is
s.replace(/include cols="[^"]+"/g, incl_sc.val());
There were a couple problems with your code,
To use a regex with String.prototype.replace, you must pass a regex as the first argument, but you were actually passing a string.
This is a regex literal /regex/ while this isn't '/actually a string/'
In the text you supplied in your question include_cols is written as include cols (with a space)
And your regex was formed wrong. I recomend testing them in this website, where you can also learn more if you want.
The code above will replace the part include cols="1,2,3" by whatever is in the textarea, regardless of whats between the quotes (as long it doesn't contain another quote).
First of all I think you need to remove the quotes and fix a little bit the regex.
const r = /(include_cols=\")(.*)(\")/g;
s.replace(r, `$1${incl_sc.val()}$3`)
Basically, I group the first and last part in order to include them at the end of the replacement. You can also avoid create the first and last group and put it literally in the last argument of the replace function, like this:
const r = /include_cols=\"(.*)\"/g;
s.replace(r, `include_cols="${incl_sc.val()}"`)
There may be a very simple answer to this, probably because of my familiarity (or possibly lack thereof) of the replace method and how it works with regex.
Let's say I have the following string: abcdefHellowxyz
I just want to strip the first six characters and the last four, to return Hello, using regex... Yes, I know there may be other ways, but I'm trying to explore the boundaries of what these methods are capable of doing...
Anyway, I've tinkered on http://regex101.com and got the following Regex worked out:
/^(.{6}).+(.{4})$/
Which seems to pass the string well and shows that abcdef is captured as group 1, and wxyz captured as group 2. But when I try to run the following:
"abcdefHellowxyz".replace(/^(.{6}).+(.{4})$/,"")
to replace those captured groups with "" I receive an empty string as my final output... Am I doing something wrong with this syntax? And if so, how does one correct it, keeping my original stance on wanting to use Regex in this manner...
Thanks so much everyone in advance...
The code below works well as you wish
"abcdefHellowxyz".replace(/^.{6}(.+).{4}$/,"$1")
I think that only use ()to capture the text you want, and in the second parameter of replace(), you can use $1 $2 ... to represent the group1 group2.
Also you can pass a function to the second parameter of replace,and transform the captured text to whatever you want in this function.
For more detail, as #Akxe recommend , you can find document on https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/String/replace.
You are replacing any substring that matches /^(.{6}).+(.{4})$/, with this line of code:
"abcdefHellowxyz".replace(/^(.{6}).+(.{4})$/,"")
The regex matches the whole string "abcdefHellowxyz"; thus, the whole string is replaced. Instead, if you are strictly stripping by the lengths of the extraneous substrings, you could simply use substring or substr.
Edit
The answer you're probably looking for is capturing the middle token, instead of the outer ones:
var str = "abcdefHellowxyz";
var matches = str.match(/^.{6}(.+).{4}$/);
str = matches[1]; // index 0 is entire match
console.log(str);
I need some help with Regex.
I have this string: \\lorem\ipsum\dolor,\\sit\amet\conseteteur,\\sadipscing\elitr\sed\diam
and want to get the result: ["dolor", "conseteteur", "diam"]So in words the word between the last backslash and a comma or the end.
I've already figured out a working test, but because of reasons it won't work in neitherChrome (v44.0.2403.130) nor IE (v11.0.9600.17905) console.There i'm getting the result: ["\loremipsumdolor,", "\sitametconseteteur,", "\sadipscingelitrseddiam"]
Can you please tell me, why the online testers aren't working and how i can achieve the right result?
Thanks in advance.
PS: I've tested a few online regex testers with all the same result. (regex101.com, regexpal.com, debuggex.com, scriptular.com)
The string
'\\lorem\ipsum\dolor,\\sit\amet\conseteteur,\\sadipscing\elitr\sed\diam'
is getting escaped, if you try the following in the browser's console you'll see what happens:
var s = '\\lorem\ipsum\dolor,\\sit\amet\conseteteur,\\sadipscing\elitr\sed\diam'
console.log(s);
// prints '\loremipsumdolor,\sitametconseteteur,\sadipscingelitrseddiam'
To use your original string you have to add additional backslashes, otherwise it becomes a different one because it tries to escape anything followed by a single backslash.
The reason why it works in regexp testers is because they probably sanitize the input string to make sure it gets evaluated as-is.
Try this (added an extra \ for each of them):
str = '\\\\lorem\\ipsum\\dolor,\\\\sit\\amet\\conseteteur,\\\\sadipscing\\elitr\\sed\\diam'
re = /\\([^\\]*)(?:,|$)/g
str.match(re)
// should output ["\dolor,", "\conseteteur,", "\diam"]
UPDATE
You can't prevent the interpreter from escaping backslashes in string literals, but this functionality is coming with EcmaScript6 as String.raw
s = String.raw`\\lorem\ipsum\dolor,\\sit\amet\conseteteur,\\sadipscing\elitr\sed\diam`
Remember to use backticks instead of single quotes with String.raw.
It's working in latest Chrome, but I can't say for all other browsers, if they're moderately old, it probably isn't implemented.
Also, if you want to avoid matching the last backslash you need to:
remove the \\ at the start of your regexp
use + instead of * to avoid matching the line end (it will create an extra capture)
use a positive lookahead ?=
like this
s = String.raw`\\lorem\ipsum\dolor,\\sit\amet\conseteteur,\\sadipscing\elitr\sed\diam`;
re = /([^\\]+)(?=,|$)/g;
s.match(re);
// ["dolor", "conseteteur", "diam"]
You may try this,
string.match(/[^\\,]+(?=,|$)/gm);
DEMO
My current code is:
var user_pattern = this.settings.tag;
user_pattern = user_pattern.replace(/[\-\[\]\/\{\}\(\)\*\+\?\.\\\^\$\|]/g, "\\$&"); // escape regex
var pattern = new RegExp(user_pattern.replace(/%USERNAME%/i, "(\\S+)"), "ig");
Where this.settings.tag is a string such as "[user=%USERNAME%]" or "#%USERNAME%". The code uses pattern.exec(str) to find any username in the corresponding tag and works perfectly fine. For example, if str = "Hello, [user=test]" then pattern.exec(str) will find test.
This works fine, but I want to be able to stop it from matching if the string is wrapped in [nocode][/nocode] tags. For example, if str = "[nocode]Hello, [user=test], how are you?[/nocode]" thenpattern.exec(str)` should not match anything.
I'm not quite sure where to start. I tried using a (?![nocode]) before and after the pattern, but to no avail. Any help would be great.
I would just test if the string starts with [nocode] first:
/^\[nocode\]/.test('[nocode]');
Then simply do not process it.
Maybe filter out [nocode] before trying to find the username(s)?
pattern.exec(str.replace(/\[nocode\](.*)\[\/nocode\]/g,''));
I know this isn't exactly what you asked for because now you have to use two separate regular expressions, however code readability is important too and doing it this way is definitely better in that aspect. Hope this helps 😉
JSFiddle: http://jsfiddle.net/1f485Lda/1/
It's based on this: Regular Expression to get a string between two strings in Javascript
in my regular expression, I tried to remove all "{" and "}"s from a string.
Pushing the script with packer/minimizer scripts, breaks them.
That's why I'd like to know about a better and more compatible way of writing:
mystring.replace(/\{/g,"");?
You can just use a string instead of a regex. I'm not sure if this is "better" but it should not break when minified. If you provide the minified example, we may be able to help with that.
mystring.replace("}", "").replace("{", "");
Edit:
If the curly bracket is causing the problem, perhaps this would work...
var reg = new RegExp("\\{|\\}", "g");
mystring.replace(reg, "");
Example from the console...
> var mystring = "test{foo}bar{baz}";
> var reg = new RegExp("\\{|\\}", "g");
> mystring.replace(reg, "");
"testfoobarbaz"
Lastly, you could do this:
If a regex really wont work for you, this will replace all {'s and }'s
It is probably a horrible solution, considering performance, but...
mystring.split("}").join("").split("{").join("");
You could try
mystring.replace(/\u007B/g,"");
This uses unicode rather than the actual symbol, so your packer won't get confused. If you want to replace more than one character in a single statement, you can use the "or" pipe:
mystring.replace(/\u007B|\u007D/g,"");
{ = \u007B
} = \u007D
For more unicode codes see:
http://www.unicode.org/charts/PDF/U0000.pdf
After re-reading the question, it sounds like you've found a bug with the minifier/packer. My first suggestion would be to use a better minimizer that doesn't have these issues, but if you're stuck with what you're using, you could try using the unicode escape sequence in the regular expression:
mystring.replace(/\u007b/g, '');
Alternatively, you could try String.prototype.split and Array.prototype.join:
mystring.split('{').join('');