Develop Reference

prevent javascript to encode url

I am trying to change url with pushstate in java script and my url doesn't have any space or bad character for url but java script encode it and add some character in it.
my code is:
name= name.trim();
const nextState = { additionalInformation: name };
window.history.pushState(nextState, "", my_domain()+"/brands/" + name);
my url is:
http://localhost/brands/Brilliance
but it show as:
http://localhost/brands/Brilliance%E2%80%8C

The '%E2%80%8C' at the end of your URL is an invisible Unicode/ASCII character that you are likely copying when you have pasted in the URL, or maybe a package is causing it. In either case, here are two ways you can solve this:
You can paste your link into a hex editor and remove the invisible character manually before copy-pasting back into your code editor.
You can use this javascript solution to remove the characters:
function remove_non_ascii(str) {
if ((str===null) || (str===''))
return false;
else
str = str.toString();
return str.replace(/[^\x20-\x7E]/g, '');
}
console.log(remove_non_ascii('äÄçÇéÉêHello-WorldöÖÐþúÚ'));

The provided code has no reason to not work, but I have a suspicion that in fact name is malformed. If I get the %E2%80%8C suffix in the URL and run it trough decodeURI, I get an empty string. However, if I manually convert it to the string \xe2\x80\x8c, I get the following: â\x80\x8C. This most probably means corrupted data.

Change your URL using encodeURIComponent
const nextState = encodeURIComponent('http://localhost/brands/Brilliance');
history.pushState({}, '', nextState);
This will change your url and the URL will not have any spaces and bad characters

How can I use a url string parameter when the url doesn't use /

How can I use a url string parameter when the url doesn't use /? I'm used to using /:parameter_here, but how can I use that variable syntax when using ? and &?
URL (Doesn't Work)
www.test.com/users?status=ACCEPTED&party_id=:partyId

www.test.com/users/parameter_here?status=ACCEPTED&party_id=partyId
parameter_here will be parameter_here
status will be ACCEPTED
and party_id will be partyId
You don't need the &, ? or = for parameters in the URL.

How to escape slash in URL parameter

$.ajax({
url: "/api/v1/cases/annotations/" + case_id + "/" + encodeURIComponent(current_plink),
I am using encodeURIComponent to escape slash. But it does not work for me. This code convert "slash" to "%2F" but apache does not recognize it.
My php part like this :
$app->get('/cases/annotations/:case_id/:prep_name', 'authenticatePathologist', function($case_id, $prep_name) use ($app) {
If i try to send parameter which include slash, it returns page not found.
But if i try to send parameter which does not include slash, it returns OK.

You should encode it twice with encodeURIComponent, i.e. encodeURIComponent(encodeURIComponent(current_plink)). If you encode it only once, the server decodes it, and it's the same as not encoding it at all.

you should follow
AllowEncodedSlashes Directive

I too had a similar issue with url like:
?filters=new:item:text
where it required to escape the colons. for that I tried replacing it to format
?filters/:items/:text but that couldn't work as the Chrome was NOT encoding the '/' value.
Tried the js [escape() method][1] and it worked like charm.
let url = 'https://dev.test.something?filters=';
let filterString = 'new:item:text';
url = url + escape(filterString);
Output:
https://dev.test.something?filters=new%3Aitem%3Atext
Hope this finds helpful!!

Encoding URL in JavaScript

I'm unable to encoding data URI:
var uri ="data:image/jpeg;base64,/9j/4AAQSkZJRgABAQAAAQABAAD/2wBDAAMCAgMCAgMDAwMEAwMEBQgFBQQEBQo...";
var res = encodeURI(uri);
document.location.href = 'display.jsp?img='+res;
After encoding, I'm getting the same uri. display.jsp is landing as am empty page.

There is no encoding happening because what you have there is already a valid, completely encoded URI.
If you want to use that as a parameter in an other URI, you should use encodeURIComponent:
document.location.href = 'display.jsp?img='+encodeURIComponent(uri);

It is not correct to use encodeURI() as this function encodes special character except: , / ? : # & = + $ #
Use encodeURIComponent() to encode these characters.
For more info check below link:
http://www.w3schools.com/jsref/jsref_encodeuri.asp

Your problem is that the encodeURI function is for making a URL valid for a browser, not for formatting content into a URL (which is what you're doing). A base64 string is already formatted in such a way that it registers as valid. To encode it as part of the URL, you need to use encodeURLComponent.
Basically, just use:
var uri ="data:image/jpeg;base64,/9j/4AAQSkZJRgABAQAAAQABAAD/2wBDAAMCAgMCAgMDAwMEAwMEBQgFBQQEBQo...";
var res = encodeURIComponent(uri);
document.location.href = 'display.jsp?img='+res;
For more info, check out: When are you supposed to use escape instead of encodeURI / encodeURIComponent?

How to prepare encoded POST data on javascript?

I need to send data by POST method.
For example, I have the string "bla&bla&bla". I tried using encodeURI and got "bla&bla&bla" as the result. I need to replace "&" with something correct for this example.
What kind of method should I call to prepare correct POST data?
UPDATED:
I need to convert only charachters which may broke POST request. Only them.

>>> encodeURI("bla&bla&bla")
"bla&bla&bla"
>>> encodeURIComponent("bla&bla&bla")
"bla%26bla%26bla"

You can also use escape() function.The escape() function encodes a string.
This function makes a string portable, so it can be transmitted across any network to any computer that supports ASCII characters.This function encodes special characters, with the exception of: * # - _ + . /
var queryStr = "bla&bla&bla";
alert(queryStr); //bla&bla&bla
alert(escape(queryStr)); //bla%26bla%26bla
Use unescape() to decode a string.
var newQueryStr=escape(queryStr);
alert(unescape(newQueryStr)); //bla&bla&bla
Note:
escape() will not encode: #*/+
encodeURI() will not encode: ~!##$&*()=:/,;?+'
encodeURIComponent() will not encode: ~!*()'
After some search on internet, I got the following:
escape()
Don't use it.
encodeURI()
Use encodeURI when you want a working URL. Make this call:
encodeURI("http://www.google.com/a file with spaces.html")
to get:
http://www.google.com/a%20file%20with%20spaces.html
Don't call encodeURIComponent since it would destroy the URL and return
http%3A%2F%2Fwww.google.com%2Fa%20file%20with%20spaces.html
encodeURIComponent()
Use encodeURIComponent when you want to encode a URL parameter.
param1 = encodeURIComponent("http://xyz.com/?a=12&b=55")
Then you may create the URL you need:
url = "http://domain.com/?param1=" + param1 + "&param2=99";
And you will get this complete URL:
http://www.domain.com/?param1=http%3A%2F%2Fxyz.com%2F%Ffa%3D12%26b%3D55¶m2=99
Note that encodeURIComponent does not escape the ' character. A common bug is to use it to create html attributes such as href='MyUrl', which could suffer an injection bug. If you are constructing html from strings, either use " instead of ' for attribute quotes, or add an extra layer of encoding (' can be encoded as %27).
REF:When are you supposed to use escape instead of encodeURI / encodeURIComponent?
Also, as you are using JQuery, take a look at this built-in function.

Use encodeURIComponent() as encodeURI() will not encode: ~!##$&*()=:/,;?+'
This has been explained quite well at the following link:
http://xkr.us/articles/javascript/encode-compare/

More recent DOM APIs for URLSearchParams (and via URL, possibly others too) handle encoding in some cases. For example, create or use an existing URL object (like from an anchor tag) I map entries of an object as key value pairs for URL encoded params (to use for GET/POST/etc in application/x-www-form-urlencoded mimetype). Note how the emoji, ampersand and double quotes are encoded without any special handling (copied from the Chrome devtools console):
var url = new URL(location.pathname, location.origin);
Object.entries({a:1,b:"🍻",c:'"stuff&things"'}).forEach(url.searchParams.set, url.searchParams);
url.search;
"?a=1&b=%F0%9F%8D%BB&c=%22stuff%26things%22"
fetch(url.pathname, {
method: 'POST',
headers: new Headers({
"Content-type": "application/x-www-form-urlencoded"
}),
// same format as GET url search params a&b&c
body: url.searchParams
}).then((res)=>{ console.log(res); return res }).catch((res)=>{ console.warn(res); return res; });

I want POST the javascript-created hidden form.
So the question is if encodeURIComponent() should be used on each POST variable.
I haven't found the answer for Dmitry's (and my) question in this thread.
But I have found the answer in this thread.
In case of form/POST where you have upload field(s) you must use <form enctype="multipart/form-data">, if no upload field is used, you should choose yourself as described here.
Submitting the form should do the job completly, so there is no need to use encodeURIComponent() explicitly.
If you create a Http POST without using a form or without some library which creates a Http POST from your data, then you need choose an enctype= and join data yourselves.
This will be easy for application/x-www-form-urlencoded, where you will use encodeURIComponent() on each value and join them exactly as for GET request.
If you decide use multipart/form-data then ....? You should google more how to encode and join them in such case.