JavaScript indexOf to ignore Case - javascript

I am trying to find if an image has in its source name noPic which can be in upper or lower case.
var noPic = largeSrc.indexOf("nopic");
Should I write :
var noPic = largeSrc.toLowerCase().indexOf("nopic");
But this solution doesn't work...

You can use regex with a case-insensitive modifier - admittedly not necessarily as fast as indexOf.
var noPic = largeSrc.search(/nopic/i);

No, there is no case-insensitive way to call that function. Perhaps the reason your second example doesn't work is because you are missing a call to the text() function.
Try this:
var search = "nopic";
var noPic = largeSrc.text().toLowerCase().indexOf(search.toLowerCase());

Note that if the search string is from user input you'll need to escape the special regexp characters.
Here's what it would look like:
var search = getUserInput();
/* case-insensitive search takes around 2 times more than simple indexOf() */
var regex = RegExp(search.replace(/[-\/\\^$*+?.()|[\]{}]/g, '\\$&'), "i");
var noPic = testString.search(regex);
See the updated jsperf: http://jsperf.com/regex-vs-tolowercase-then-regex/4
footnote: regexp escaping from https://stackoverflow.com/a/3561711/1333402

Try with:
var lowerCaseLargeSrc = largeSrc.toLowerCase();
var noPic = lowerCaseLargeSrc.indexOf("nopic");

Your code will only work if largeSrc is already a string. You might be getting an input that's an html element instead. So, use jQuery to resolve any potential input element into the text that's inside it. Example:
var noPic = largeSrc.text().toLowerCase().indexOf("nopic");

How about using findIndex instead that way you can do all your toLowerCase() inside the callback. Worked great for me:
// Not Supported in IE 6-11
const arr = ['HELLO', 'WORLD'];
const str = 'world';
const index = arr.findIndex(element => {
return element.toLowerCase() === str.toLowerCase();
});
console.log(index); // 👉️ 1
if (index !== -1) {
// 👉️ string is in the array
}
Credit:
https://bobbyhadz.com/blog/javascript-make-array-indexof-case-insensitive

Related

JS What's the fastest way to display one specific line of a list?

In my Javascript code, I get one very long line as a string.
This one line only has around 65'000 letters. Example:
config=123&url=http://localhost/example&path_of_code=blablaba&link=kjslfdjs...
What I have to do is replace all & with an break (\n) first and then pick only the line which starts with "path_of_code=". This line I have to write in a variable.
The part with replace & with an break (\n) I already get it, but the second task I didn't.
var obj = document.getElementById('div_content');
var contentJS= obj.value;
var splittedResult;
splittedResult = contentJS.replace(/&/g, '\n');
What is the fastest way to do it? Please note, the list is usually very long.
It sounds like you want to extract the text after &path_of_code= up until either the end of the string or the next &. That's easily done with a regular expression using a capture group, then using the value of that capture group:
var rex = /&path_of_code=([^&]+)/;
var match = rex.exec(theString);
if (match) {
var text = match[1];
}
Live Example:
var theString = "config=123&url=http://localhost/example&path_of_code=blablaba&link=kjslfdjs...";
var rex = /&path_of_code=([^&]+)/;
var match = rex.exec(theString);
if (match) {
var text = match[1];
console.log(text);
}
Use combination of String.indexOf() and String.substr()
var contentJS= "123&url=http://localhost/example&path_of_code=blablaba&link=kjslfdjs...";
var index = contentJS.indexOf("&path_of_code"),
substr = contentJS.substr(index+1),
res = substr.substr(0, substr.indexOf("&"));
console.log(res)
but the second task I didn't.
You can use filter() and startsWith()
splittedResult = splittedResult.filter(i => i.startsWith('path_of_code='));

Regex append characters to a substring

My string comes in two flavours-
var a = /aid/f82eb514073124cd10d468b74eee5663?sg=1#/propertyinfo
or
var a = /aid/f82eb514073124cd10d468b74eee5663#/propertyinfo
I want to append the content that comes after aid/ and before ? or # with "-test". In either of the above scenarios the result would be f82eb514073124cd10d468b74eee5663-test
hence
a = /aid/f82eb514073124cd10d468b74eee5663-test#/propertyinfo
or
a = = /aid/f82eb514073124cd10d468b74eee5663-test?sg=1#/propertyinfo
Seems like you're looking for something like this.
Regular expression /\/aid\/[0-9A-F]*/i and replacement expression $0-test.
JavaScript is a little bit different than just plain regular expression antics, so here you go;
var a = "/aid/f82eb514073124cd10d468b74eee5663?sg=1#/propertyinfo";
alert(a.replace(/(\/aid\/[0-9A-F]*)/i, "$1-test"));
given your examples I guess that string after /aid/ is some kind of md5 hash
this should work for you:
'/aid/f82eb514073124cd10d468b74eee5663#/propertyinfo'.replace(new RegExp('/aid/([a-f0-9]{32})'), '$1-test');
if you don't want to be that much specific about length, you can try the following:
'/aid/f82eb514073124cd10d468b74eee5663#/propertyinfo'.replace(new RegExp('/aid/([a-f0-9]+)'), '$1-test');
Simple solution using String.replace function:
var a = '/aid/f82eb514073124cd10d468b74eee5663sg=1#/propertyinfo',
result = a.replace(/aid\/([^?#]+)(?=\?|#)/, "aid/$1-test");
console.log(result); // /aid/f82eb514073124cd10d468b74eee5663-test?sg=1#/propertyinfo
I suggest replacing directly the # or ? so the regex is nice and simple. :)
var a = "/aid/f82eb514073124cd10d468b74eee5663?sg=1#/propertyinfo";
var b = "/aid/f82eb514073124cd10d468b74eee5663#/propertyinfo";
console.log(a.replace(/([\?#])/,"-test$1"));
console.log(b.replace(/([\?#])/,"-test$1"));
var a = '/aid/f82eb514073124cd10d468b74eee5663?sg=1#/propertyinfo';
a.replace(/(\/aid\/.+)(\?sg=1)(#\/propertyinfo)/,function(text,c,d,e){
return c+'-test'+e;
})
//Output: "/aid/f82eb514073124cd10d468b74eee5663-test#/propertyinfo"
a.replace(/(\/aid\/.+)(\?sg=1#\/propertyinfo)/,function(text,c,d){
return c+'-test'+d;
});
//Output: "/aid/f82eb514073124cd10d468b74eee5663-test?sg=1#/propertyinfo"

Regex: .exec() function not returning expected output

This doesn't return what I, or regex101, expects:
var myString = "Accel World|http://www.anime-planet.com/anime/accel-worldAh! My Goddess|http://www.anime-planet.com/anime/ah-my-goddess";
var reg = /[^|]*/g;
var regResponse = reg.exec(myString);
console.log(regResponse);
according to regex101, this should match everything except '|' and return it yet it only matches the first string, Accel World, as opposed to everything but '|'.
How do I fix this?
Exec will only return one result at a time (subsequent calls will return the rest, but you also need to use the + instead of *)
You could use the myString.match(reg) htough to get all results in one go.
var myString = "Accel World|http://www.anime-planet.com/anime/accel-worldAh! My Goddess|http://www.anime-planet.com/anime/ah-my-goddess";
var reg = /[^|]+/g;
var regResponse = myString.match(reg);
console.log(regResponse);
You need to loop .exec() to retrieve all matches. The documentation says
If your regular expression uses the "g" flag, you can use the exec()
method multiple times to find successive matches in the same string.
var reg = /[^|]+/g;
while(regResponse = reg.exec(myString)) {
console.log(regResponse);
}
Try a "+" instead of the "*"
So,
var reg = /[^|]+/g;

Regex remove repeated characters from a string by javascript

I have found a way to remove repeated characters from a string using regular expressions.
function RemoveDuplicates() {
var str = "aaabbbccc";
var filtered = str.replace(/[^\w\s]|(.)\1/gi, "");
alert(filtered);
}
Output: abc
this is working fine.
But if str = "aaabbbccccabbbbcccccc" then output is abcabc.
Is there any way to get only unique characters or remove all duplicates one?
Please let me know if there is any way.
A lookahead like "this, followed by something and this":
var str = "aaabbbccccabbbbcccccc";
console.log(str.replace(/(.)(?=.*\1)/g, "")); // "abc"
Note that this preserves the last occurrence of each character:
var str = "aabbccxccbbaa";
console.log(str.replace(/(.)(?=.*\1)/g, "")); // "xcba"
Without regexes, preserving order:
var str = "aabbccxccbbaa";
console.log(str.split("").filter(function(x, n, s) {
return s.indexOf(x) == n
}).join("")); // "abcx"
This is an old question, but in ES6 we can use Sets. The code looks like this:
var test = 'aaabbbcccaabbbcccaaaaaaaasa';
var result = Array.from(new Set(test)).join('');
console.log(result);

javascript get string before a character

I have a string that and I am trying to extract the characters before the quote.
Example is extract the 14 from 14' - €14.99
I am using the follwing code to acheive this.
$menuItem.text().match(/[^']*/)[0]
My problem is that if the string is something like €0.88 I wish to get an empty string returned. However I get back the full string of €0.88.
What I am I doing wrong with the match?
This is the what you should use to split:
string.slice(0, string.indexOf("'"));
And then to handle your non existant value edge case:
function split(str) {
var i = str.indexOf("'");
if(i > 0)
return str.slice(0, i);
else
return "";
}
Demo on JsFiddle
Nobody seems to have presented what seems to me as the safest and most obvious option that covers each of the cases the OP asked about so I thought I'd offer this:
function getCharsBefore(str, chr) {
var index = str.indexOf(chr);
if (index != -1) {
return(str.substring(0, index));
}
return("");
}
try this
str.substring(0,str.indexOf("'"));
Here is an underscore mixin in coffescript
_.mixin
substrBefore : ->
[char, str] = arguments
return "" unless char?
fn = (s)-> s.substr(0,s.indexOf(char)+1)
return fn(str) if str?
fn
or if you prefer raw javascript : http://jsfiddle.net/snrobot/XsuQd/
You can use this to build a partial like:
var beforeQuote = _.substrBefore("'");
var hasQuote = beforeQuote("14' - €0.88"); // hasQuote = "14'"
var noQoute = beforeQuote("14 €0.88"); // noQuote = ""
Or just call it directly with your string
var beforeQuote = _.substrBefore("'", "14' - €0.88"); // beforeQuote = "14'"
I purposely chose to leave the search character in the results to match its complement mixin substrAfter (here is a demo: http://jsfiddle.net/snrobot/SEAZr/ ). The later mixin was written as a utility to parse url queries. In some cases I am just using location.search which returns a string with the leading ?.
I use "split":
let string = "one-two-three";
let um = string.split('-')[0];
let dois = string.split('-')[1];
let tres = string.split('-')[2];
document.write(tres) //three

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