I need to write a function to perform an action only if the URL has a specific string. The issue that I am finding is that the string can come up in multiple instances as part of another string. I need the function to run when the string is ONLY "?page=1". What I am finding is that the function is also being run when the string contains a string like "?page=10" , "?page=11" , "?page=12" , etc... I only need it to be done if the string is "?page=1" - that's it. How do I do that? I've tried a couple of different ways, but it does not work. Any help is appreciated. Here is the latest code that I have used that is close...but no cigar.
var location = window.location.href;
if (location.indexOf("?page=1") > -1){
//Do something
};
?page is a GET parameter. It doesn't necessarily have to be first in the URL string. I suggest you properly decode the GET params and then base your logic on that. Here's how you can do that:
function unparam(qs) {
var params = {},
e,
a = /\+/g,
r = /([^&=]+)=?([^&]*)/g,
d = function (s) { return decodeURIComponent(s.replace(a, " ")); };
while (e = r.exec(qs)) {
params[d(e[1])] = d(e[2]);
}
return params;
}
var urlParams = unparam(window.location.search.substring(1));
if(urlParams['page'] == '1') {
// code here
}
Alternatively, a regex with word boundaries would have worked:
if(/\bpage=1\b/.test(window.location.search)) {
// code here
}
if(location .indexOf("?page=1&") != -1 || (location .indexOf("?page=1") + 7 == i.length) ) {
}
You could look at the character immediately following the string "?page=1" in the url. If it's a digit,you don't have a match otherwise you do. You could trivially do something like this:
var index = location.indexOf("?page=1"); //Returns the index of the string
var number = location.charCodeAt(index+x); //x depends on the search string,here x = 7
//Unicode values for 0-9 is 48-57, check if number lies within this range
Now that you have the Unicode value of the next character, you can easily deduce if the url contains the string you require or not. I hope this points you in the right direction.
I am so close to getting this, but it just isn't right.
All I would like to do is remove the character r from a string.
The problem is, there is more than one instance of r in the string.
However, it is always the character at index 4 (so the 5th character).
Example string: crt/r2002_2
What I want: crt/2002_2
This replace function removes both r
mystring.replace(/r/g, '')
Produces: ct/2002_2
I tried this function:
String.prototype.replaceAt = function (index, char) {
return this.substr(0, index) + char + this.substr(index + char.length);
}
mystring.replaceAt(4, '')
It only works if I replace it with another character. It will not simply remove it.
Any thoughts?
var mystring = "crt/r2002_2";
mystring = mystring.replace('/r','/');
will replace /r with / using String.prototype.replace.
Alternatively you could use regex with a global flag (as suggested by Erik Reppen & Sagar Gala, below) to replace all occurrences with
mystring = mystring.replace(/\/r/g, '/');
EDIT:
Since everyone's having so much fun here and user1293504 doesn't seem to be coming back any time soon to answer clarifying questions, here's a method to remove the Nth character from a string:
String.prototype.removeCharAt = function (i) {
var tmp = this.split(''); // convert to an array
tmp.splice(i - 1 , 1); // remove 1 element from the array (adjusting for non-zero-indexed counts)
return tmp.join(''); // reconstruct the string
}
console.log("crt/r2002_2".removeCharAt(4));
Since user1293504 used the normal count instead of a zero-indexed count, we've got to remove 1 from the index, if you wish to use this to replicate how charAt works do not subtract 1 from the index on the 3rd line and use tmp.splice(i, 1) instead.
A simple functional javascript way would be
mystring = mystring.split('/r').join('/')
simple, fast, it replace globally and no need for functions or prototypes
There's always the string functions, if you know you're always going to remove the fourth character:
str.slice(0, 4) + str.slice(5, str.length)
Your first func is almost right. Just remove the 'g' flag which stands for 'global' (edit) and give it some context to spot the second 'r'.
Edit: didn't see it was the second 'r' before so added the '/'. Needs \/ to escape the '/' when using a regEx arg. Thanks for the upvotes but I was wrong so I'll fix and add more detail for people interested in understanding the basics of regEx better but this would work:
mystring.replace(/\/r/, '/')
Now for the excessive explanation:
When reading/writing a regEx pattern think in terms of: <a character or set of charcters> followed by <a character or set of charcters> followed by <...
In regEx <a character or set of charcters> could be one at a time:
/each char in this pattern/
So read as e, followed by a, followed by c, etc...
Or a single <a character or set of charcters> could be characters described by a character class:
/[123!y]/
//any one of these
/[^123!y]/
//anything but one of the chars following '^' (very useful/performance enhancing btw)
Or expanded on to match a quantity of characters (but still best to think of as a single element in terms of the sequential pattern):
/a{2}/
//precisely two 'a' chars - matches identically as /aa/ would
/[aA]{1,3}/
//1-3 matches of 'a' or 'A'
/[a-zA-Z]+/
//one or more matches of any letter in the alphabet upper and lower
//'-' denotes a sequence in a character class
/[0-9]*/
//0 to any number of matches of any decimal character (/\d*/ would also work)
So smoosh a bunch together:
var rePattern = /[aA]{4,8}(Eat at Joes|Joes all you can eat)[0-5]+/g
var joesStr = 'aaaAAAaaEat at Joes123454321 or maybe aAaAJoes all you can eat098765';
joesStr.match(rePattern);
//returns ["aaaAAAaaEat at Joes123454321", "aAaAJoes all you can eat0"]
//without the 'g' after the closing '/' it would just stop at the first match and return:
//["aaaAAAaaEat at Joes123454321"]
And of course I've over-elaborated but my point was simply that this:
/cat/
is a series of 3 pattern elements (a thing followed by a thing followed by a thing).
And so is this:
/[aA]{4,8}(Eat at Joes|Joes all you can eat)[0-5]+/
As wacky as regEx starts to look, it all breaks down to series of things (potentially multi-character things) following each other sequentially. Kind of a basic point but one that took me a while to get past so I've gone overboard explaining it here as I think it's one that would help the OP and others new to regEx understand what's going on. The key to reading/writing regEx is breaking it down into those pieces.
Just fix your replaceAt:
String.prototype.replaceAt = function(index, charcount) {
return this.substr(0, index) + this.substr(index + charcount);
}
mystring.replaceAt(4, 1);
I'd call it removeAt instead. :)
For global replacement of '/r', this code worked for me.
mystring = mystring.replace(/\/r/g,'');
This is improvement of simpleigh answer (omit length)
s.slice(0, 4) + s.slice(5)
let s = "crt/r2002_2";
let o = s.slice(0, 4) + s.slice(5);
let delAtIdx = (s, i) => s.slice(0, i) + s.slice(i + 1); // this function remove letter at index i
console.log(o);
console.log(delAtIdx(s, 4));
let str = '1234567';
let index = 3;
str = str.substring(0, index) + str.substring(index + 1);
console.log(str) // 123567 - number "4" under index "3" is removed
return this.substr(0, index) + char + this.substr(index + char.length);
char.length is zero. You need to add 1 in this case in order to skip character.
Maybe I'm a noob, but I came across these today and they all seem unnecessarily complicated.
Here's a simpler (to me) approach to removing whatever you want from a string.
function removeForbiddenCharacters(input) {
let forbiddenChars = ['/', '?', '&','=','.','"']
for (let char of forbiddenChars){
input = input.split(char).join('');
}
return input
}
Create function like below
String.prototype.replaceAt = function (index, char) {
if(char=='') {
return this.slice(0,index)+this.substr(index+1 + char.length);
} else {
return this.substr(0, index) + char + this.substr(index + char.length);
}
}
To replace give character like below
var a="12346";
a.replaceAt(4,'5');
and to remove character at definite index, give second parameter as empty string
a.replaceAt(4,'');
If it is always the 4th char in yourString you can try:
yourString.replace(/^(.{4})(r)/, function($1, $2) { return $2; });
It only works if I replace it with another character. It will not simply remove it.
This is because when char is equal to "", char.length is 0, so your substrings combine to form the original string. Going with your code attempt, the following will work:
String.prototype.replaceAt = function (index, char) {
return this.substr(0, index) + char + this.substr(index + 1);
// this will 'replace' the character at index with char ^
}
DEMO
You can use this: if ( str[4] === 'r' ) str = str.slice(0, 4) + str.slice(5)
Explanation:
if ( str[4] === 'r' )
Check if the 5th character is a 'r'
str.slice(0, 4)
Slice the string to get everything before the 'r'
+ str.slice(5)
Add the rest of the string.
Minified: s=s[4]=='r'?s.slice(0,4)+s.slice(5):s [37 bytes!]
DEMO:
function remove5thR (s) {
s=s[4]=='r'?s.slice(0,4)+s.slice(5):s;
console.log(s); // log output
}
remove5thR('crt/r2002_2') // > 'crt/2002_2'
remove5thR('crt|r2002_2') // > 'crt|2002_2'
remove5thR('rrrrr') // > 'rrrr'
remove5thR('RRRRR') // > 'RRRRR' (no change)
If you just want to remove single character and
If you know index of a character you want to remove, you can use following function:
/**
* Remove single character at particular index from string
* #param {*} index index of character you want to remove
* #param {*} str string from which character should be removed
*/
function removeCharAtIndex(index, str) {
var maxIndex=index==0?0:index;
return str.substring(0, maxIndex) + str.substring(index, str.length)
}
I dislike using replace function to remove characters from string. This is not logical to do it like that. Usually I program in C# (Sharp), and whenever I want to remove characters from string, I use the Remove method of the String class, but no Replace method, even though it exists, because when I am about to remove, I remove, no replace. This is logical!
In Javascript, there is no remove function for string, but there is substr function. You can use the substr function once or twice to remove characters from string. You can make the following function to remove characters at start index to the end of string, just like the c# method first overload String.Remove(int startIndex):
function Remove(str, startIndex) {
return str.substr(0, startIndex);
}
and/or you also can make the following function to remove characters at start index and count, just like the c# method second overload String.Remove(int startIndex, int count):
function Remove(str, startIndex, count) {
return str.substr(0, startIndex) + str.substr(startIndex + count);
}
and then you can use these two functions or one of them for your needs!
Example:
alert(Remove("crt/r2002_2", 4, 1));
Output: crt/2002_2
Achieving goals by doing techniques with no logic might cause confusions in understanding of the code, and future mistakes, if you do this a lot in a large project!
The following function worked best for my case:
public static cut(value: string, cutStart: number, cutEnd: number): string {
return value.substring(0, cutStart) + value.substring(cutEnd + 1, value.length);
}
The shortest way would be to use splice
var inputString = "abc";
// convert to array and remove 1 element at position 4 and save directly to the array itself
let result = inputString.split("").splice(3, 1).join();
console.log(result);
This problem has many applications. Tweaking #simpleigh solution to make it more copy/paste friendly:
function removeAt( str1, idx) {
return str1.substr(0, idx) + str1.substr(idx+1)
}
console.log(removeAt('abbcdef', 1)) // prints: abcdef
Using [index] position for removing a specific char (s)
String.prototype.remplaceAt = function (index, distance) {
return this.slice(0, index) + this.slice(index + distance, this.length);
};
credit to https://stackoverflow.com/users/62576/ken-white
So basically, another way would be to:
Convert the string to an array using Array.from() method.
Loop through the array and delete all r letters except for the one with index 1.
Convert array back to a string.
let arr = Array.from("crt/r2002_2");
arr.forEach((letter, i) => { if(letter === 'r' && i !== 1) arr[i] = "" });
document.write(arr.join(""));
In C# (Sharp), you can make an empty character as '\0'.
Maybe you can do this:
String.prototype.replaceAt = function (index, char) {
return this.substr(0, index) + char + this.substr(index + char.length);
}
mystring.replaceAt(4, '\0')
Search on google or surf on the interent and check if javascript allows you to make empty characters, like C# does. If yes, then learn how to do it, and maybe the replaceAt function will work at last, and you'll achieve what you want!
Finally that 'r' character will be removed!
I'm trying to do a URL GET variable replace, however the regular expression for checking whether the variable exists in amongst other GET variables is returning true when I am expecting it to return false.
The pattern I am using is: &sort=.*&
Test URL: http://localhost/search?location=any&sort=asc
Am I right to believe that this pattern should be returning false on the basis that their is no ampersand character following the sort parameter's value?
Full code:
var sort = getOptionValue($(this).attr('id'));
var url = document.URL;
if(url.indexOf('?') == -1) {
url = url+'?sort='+sort;
} else {
if(url.search('/&sort=.*&/i')) {
url.replace('/&sort=.*&/i','&sort='+sort+'&');
}
else if(url.search('/&sort=.*/i')) {
url.replace('/&sort=.*/i','&sort='+sort);
}
}
Am I right to believe that this pattern should be returning false on the basis that their is no ampersand character following the sort parameter's value?
Well, you are using String.search, which, according to the linked documentation:
If successful, search returns the index of the regular expression inside the string. Otherwise, it returns -1.
So it will return -1, or 0 or greater when there is a match. So you should test for -1, not truthiness.
Also, there is no need to pass the regexes as strings, you might as well use:
url.replace(/&sort=.*&/i,'&sort='+sort+'&');
Further, keep in mind that replace will create a new string, not replace in the string (strings in Javascript are immutable).
Finally, I don't see the need for searching for the string, and then replacing it -- it seems that you always want to replace &sort=SOMETHING with &sort=SOMETHING_ELSE, so just do that:
if(url.indexOf('?') == -1) {
url = url+'?sort='+sort;
} else {
url = url.replace(/&sort=[^&]*/i, '&sort=' + sort);
}
The javascript string function search() returns -1 if not found, not false. Your code should read:
if(url.search('/&sort=.*&/i') != -1) {
url.replace('/&sort=.*&/i','&sort='+sort+'&');
}
else if(url.search('/&sort=.*/i') != -1) {
url.replace('/&sort=.*/i','&sort='+sort);
}
You should check
if(url.search('/&sort=.*&/i') >= 0)
then it should work
You could use this code
var url = 'http://localhost/search?location=any&sort=asc';
var vars = {};
var parts = url.replace(/[?&]+([^=&]+)=([^&]*)/gi, function(m,key,value) {
vars[key] = value;
});
console.log(vars);
//vars is an object with two properties: location and sort
This can be done by using
url.replace(/([?&])(sort=)([^&?]*)/, "$1$2" + sort);
The match broken down
Group 1 matches for ? or &
Group 2 matches sort=
Group 3 matches anything that is not a & or ?
Then "$1$2" + sort will replace all 3 group matches with the first 2 + your variable
examples using string "REPLACE" instead of your sort variable
url = "http://localhost/search?location=any&sort=asc&a=z"
url.replace(/([?&])(sort=)([^&?]*)/, "$1$2" + "REPLACE");
// => "http://localhost/search?location=any&sort=REPLACE&a=z"
url = "http://localhost/search?location=any&sort=asc"
url.replace(/([?&])(sort=)([^&?]*)/, "$1$2" + "REPLACE");
// => "http://localhost/search?location=any&sort=REPLACE"
url = "http://localhost/search?sort=asc"
url.replace(/([?&])(sort=)([^&?]*)/, "$1$2" + "REPLACE");
// => "http://localhost/search?sort=REPLACE"
url = "http://localhost/search?sort=asc&z=y"
url.replace(/([?&])(sort=)([^&?]*)/, "$1$2" + "REPLACE");
// => "http://localhost/search?sort=REPLACE&z=y"
The pattern I am using is: &sort=.*& Test URL:
http://localhost/search?location=any&sort=asc
Am I right to believe that this pattern should be returning false on
the basis that their is no ampersand character following the sort
parameter's value?
you are assuming right. But in your code you have else if(url.search('/&sort=.*/i')) which will match and thus still replace the value.
You should also note that your code would turn http://localhost/search?sort=asc&location=any&some=more into http://localhost/search?sort=asc&some=more. that's because because .* is greedy (trying to match as much as possible). You can avoid that by telling it to match as little as possible by appending a ? like so .*?.
That said, I believe you may be better off with a library that knows how URLs actually work. You're not compensating for parameter position, possible escaped values etc. I suggest you have a look at URI.js and replace your wicked regex with
var uri = URI(document.URL),
data = uri.query(true);
data.foo = 'bazbaz';
uri.query(data);
The replace function returns the new string with the replaces, but if there weren't any words to replace, then the original string is returned. Is there a way to know whether it actually replaced anything apart from comparing the result with the original string?
A simple option is to check for matches before you replace:
var regex = /i/g;
var newStr = str;
var replaced = str.search(regex) >= 0;
if(replaced){
newStr = newStr.replace(regex, '!');
}
If you don't want that either, you can abuse the replace callback to achieve that in a single pass:
var replaced = false;
var newStr = str.replace(/i/g, function(token){replaced = true; return '!';});
As a workaround you can implement your own callback function that will set a flag and do the replacement. The replacement argument of replace can accept functions.
Comparing the before and after strings is the easiest way to check if it did anything, there's no intrinsic support in String.replace().
[contrived example of how '==' might fail deleted because it was wrong]
Javascript replace is defected by design. Why? It has no compatibility with string replacement in callback.
For example:
"ab".replace(/(a)(b)/, "$1$2")
> "ab"
We want to verify that replace is done in single pass. I was imagine something like:
"ab".replace(/(a)(b)/, "$1$2", function replacing() { console.log('ok'); })
> "ab"
Real variant:
"ab".replace(/(a)(b)/, function replacing() {
console.log('ok');
return "$1$2";
})
> ok
> "$1$2"
But function replacing is designed to receive $0, $1, $2, offset, string and we have to fight with replacement "$1$2". The solution is:
"ab".replace(/(a)(b)/, function replacing() {
console.log('ok');
// arguments are $0, $1, ..., offset, string
return Array.from(arguments).slice(1, -2)
.reduce(function (pattern, match, index) {
// '$1' from strings like '$11 $12' shouldn't be replaced.
return pattern.replace(
new RegExp("\\$" + (index + 1) + "(?=[^\\d]|$)", "g"),
match
);
}, "$1$2");
});
> ok
> "ab"
This solution is not perfect. String replacement itself has its own WATs. For example:
"a".replace(/(a)/, "$01")
> "a"
"a".replace(/(a)/, "$001")
> "$001"
If you want to care about compatibility you have to read spec and implement all its craziness.
If your replace has a different length from the searched text, you can check the length of the string before and after. I know, this is a partial response, valid only on a subset of the problem.
OR
You can do a search. If the search is successfull you do a replace on the substring starting with the found index and then recompose the string. This could be slower because you are generating 3 strings instead of 2.
var test = "Hellllo";
var index = test.search(/ll/);
if (index >= 0) {
test = test.substr(0, index - 1) + test.substr(index).replace(/ll/g, "tt");
}
alert(test);
While this will require multiple operations, using .test() may suffice:
const regex = /foo/;
const yourString = 'foo bar';
if (regex.test(yourString)) {
console.log('yourString contains regex');
// Go ahead and do whatever else you'd like.
}
The test() method executes a search for a match between a regular expression and a specified string. Returns true or false.
With indexOf you can check wether a string contains another string.
Seems like you might want to use that.
have a look at string.match() or string.search()
After doing any RegExp method, read RegExp.lastMatch property:
/^$/.test(''); //Clear RegExp.lastMatch first, Its value will be ''
'abcd'.replace(/bc/,'12');
if(RegExp.lastMatch !== '')
console.log('has been replaced');
else
console.log('not replaced');