Change security number to X on specific places - javascript

I am new in replacing number into some characters in specific places. I have this set of number 123-45-6789 but I need to show only like this XXX-XX-6789. But I need to change only the number not including the "dash".
Below is my sample code
var mainStr = $("#view_ssn").text(),
vis = mainStr.slice(-4),
countNum = '';
for(var i = (mainStr.length)-4; i>0; i--){
countNum += 'X';
}
$("#view_ssn").text(countNum+vis);
Output
XXXXXXX6789

You may simply use:
var ssn = "123-45-6789";
var output = ssn.replace(/\d(?=.{5,})/g, "X");
console.log(output);
The logic behind the regex pattern \d(?=.{5,}) is to replace every digit in the input Social Security number which has at least 5 characters in front of it. This excludes the last 4 digits of the SSN.

Another option could be to match the exact allowed pattern and capture the part that you want to keep in group 1.
In the replacement use XXX-XX and concatenate it with the group 1 value.
const regex = /^\d{3}-\d{2}(-\d{4}$)/;
const str = "123-45-6789";
console.log(str.replace(regex, (m, g1) => "XXX-XX" + g1));

Related

Regex match cookie value and remove hyphens

I'm trying to extract out a group of words from a larger string/cookie that are separated by hyphens. I would like to replace the hyphens with a space and set to a variable. Javascript or jQuery.
As an example, the larger string has a name and value like this within it:
facility=34222%7CConner-Department-Store;
(notice the leading "C")
So first, I need to match()/find facility=34222%7CConner-Department-Store; with regex. Then break it down to "Conner Department Store"
var cookie = document.cookie;
var facilityValue = cookie.match( REGEX ); ??
var test = "store=874635%7Csomethingelse;facility=34222%7CConner-Department-Store;store=874635%7Csomethingelse;";
var test2 = test.replace(/^(.*)facility=([^;]+)(.*)$/, function(matchedString, match1, match2, match3){
return decodeURIComponent(match2);
});
console.log( test2 );
console.log( test2.split('|')[1].replace(/[-]/g, ' ') );
If I understood it correctly, you want to make a phrase by getting all the words between hyphens and disallowing two successive Uppercase letters in a word, so I'd prefer using Regex in that case.
This is a Regex solution, that works dynamically with any cookies in the same format and extract the wanted sentence from it:
var matches = str.match(/([A-Z][a-z]+)-?/g);
console.log(matches.map(function(m) {
return m.replace('-', '');
}).join(" "));
Demo:
var str = "facility=34222%7CConner-Department-Store;";
var matches = str.match(/([A-Z][a-z]+)-?/g);
console.log(matches.map(function(m) {
return m.replace('-', '');
}).join(" "));
Explanation:
Use this Regex (/([A-Z][a-z]+)-?/g to match the words between -.
Replace any - occurence in the matched words.
Then just join these matches array with white space.
Ok,
first, you should decode this string as follows:
var str = "facility=34222%7CConner-Department-Store;"
var decoded = decodeURIComponent(str);
// decoded = "facility=34222|Conner-Department-Store;"
Then you have multiple possibilities to split up this string.
The easiest way is to use substring()
var solution1 = decoded.substring(decoded.indexOf('|') + 1, decoded.length)
// solution1 = "Conner-Department-Store;"
solution1 = solution1.replace('-', ' ');
// solution1 = "Conner Department Store;"
As you can see, substring(arg1, arg2) returns the string, starting at index arg1 and ending at index arg2. See Full Documentation here
If you want to cut the last ; just set decoded.length - 1 as arg2 in the snippet above.
decoded.substring(decoded.indexOf('|') + 1, decoded.length - 1)
//returns "Conner-Department-Store"
or all above in just one line:
decoded.substring(decoded.indexOf('|') + 1, decoded.length - 1).replace('-', ' ')
If you want still to use a regular Expression to retrieve (perhaps more) data out of the string, you could use something similar to this snippet:
var solution2 = "";
var regEx= /([A-Za-z]*)=([0-9]*)\|(\S[^:\/?#\[\]\#\;\,']*)/;
if (regEx.test(decoded)) {
solution2 = decoded.match(regEx);
/* returns
[0:"facility=34222|Conner-Department-Store",
1:"facility",
2:"34222",
3:"Conner-Department-Store",
index:0,
input:"facility=34222|Conner-Department-Store;"
length:4] */
solution2 = solution2[3].replace('-', ' ');
// "Conner Department Store"
}
I have applied some rules for the regex to work, feel free to modify them according your needs.
facility can be any Word built with alphabetical characters lower and uppercase (no other chars) at any length
= needs to be the char =
34222 can be any number but no other characters
| needs to be the char |
Conner-Department-Store can be any characters except one of the following (reserved delimiters): :/?#[]#;,'
Hope this helps :)
edit: to find only the part
facility=34222%7CConner-Department-Store; just modify the regex to
match facility= instead of ([A-z]*)=:
/(facility)=([0-9]*)\|(\S[^:\/?#\[\]\#\;\,']*)/
You can use cookies.js, a mini framework from MDN (Mozilla Developer Network).
Simply include the cookies.js file in your application, and write:
docCookies.getItem("Connor Department Store");

how to retrieve a string between to same charecter

I know how to use substring() but here I have a problem, I'd like to retrieve a number between two "_" from a unknown string length. here is my string for example.
7_28_li
and I want to get the 28. How can I proceed to do so ?
Thanks.
Regex
'7_28_li'.match(/_(\d+)_/)[1]
The slashes inside match make it's contents regex.
_s are taken literally
( and ) are for retrieving the contents (the target number) later
\d is a digit character
+ is "one or more".
The [1] on the end is accesses what got matched from the first set of parens, the one or more (+) digits (\d).
Loop
var str = '7_28_li';
var state = 0; //How many underscores have gone by
var num = '';
for (var i = 0; i < str.length; i++) {
if (str[i] == '_') state++;
else if (state == 1) num += str[i];
};
num = parseInt(num);
Probably more efficient, but kind of long and ugly.
Split
'7_28_li'.split('_')[1]
Split it into an array, then get the second element.
IndexOf
var str = "7_28_li";
var num = str.substring(str.indexOf('_') + 1, str.indexOf('_', 2));
Get the start and end point. Uses the little-known second parameter of indexOf. This works better than lastIndexOf because it is guaranteed to give the first number between _s, even when there are more than 2 underscores.
First find the index of _, and then find the next position of _. Then get the substring between them.
var data = "7_28_li";
var idx = data.indexOf("_");
console.log(data.substring(idx + 1, data.indexOf("_", idx + 1)));
# 28
You can understand that better, like this
var data = "7_28_li";
var first = data.indexOf("_");
var next = data.indexOf("_", first + 1);
console.log(data.substring(first + 1, next));
# 28
Note: The second argument to indexOf is to specify where to start looking from.
Probably the easiest way to do it is to call split on your string, with your delimiter ("_" in this case) as the argument. It'll return an array with 7, 28, and li as elements, so you can select the middle one.
"7_28_li".split("_")[1]
This will work if it'll always be 3 elements. If it's more, divide the length property by 2 and floor it to get the right element.
var splitstring = "7_28_li".split("_")
console.log(splitstring[Math.floor(splitstring.length/2)]);
I'm not sure how you want to handle even length strings, but all you have to do is set up an if statement and then do whatever you want.
If you know there would be 2 underscore, you can use this
var str = "7_28_li";
var res = str.substring(str.indexOf("_") +1, str.lastIndexOf("_"));
If you want to find the string between first 2 underscores
var str = "7_28_li";
var firstIndex = str.indexOf("_");
var secondIndex = str.indexOf("_", firstIndex+1);
var res = str.substring(firstIndex+1, secondIndex);

Javascript Regular expression to match the length of values in both sides of a range

I'm trying to validate postal code ranges using Javascript and the framework that supplies the country specific validation rule to the method uses regular expressions.
Is there a way to validate (using regex match/test/any other js regex functions) if the number of characters on both sides of the range delimiter (its colon in this case) is same ?
e.g.
85001:85255 or A9A 9AA:A9A 9ZZ is valid (both sides have same number of characters)
Whereas,
85001:255 or A9A 9AA:9ZZ is invalid (different number of characters in x vs y for x:y)
Thanks
Here a little bit less verbose version of RegExp profided by #tak3r
var r = /^(.{1}:.{1}|.{2}:.{2}|.{3}:.{3}|.{4}:.{4}|.{5}:.{5}|.{6}:.{6}|.{7}:.{7})$/
one paradigm of regular expressions is that they cannot count.
However, you can "hack it" like so if you know the max length of the number of chars on a side. Assuming a max-length of 4:
var code = '8500:1234';
/\b((.:.)|(..:..)|(...:...)|(....:....))\b/.test(code)
if you don't want to write this by hand, you can generate the regexp like so:
var max_length = 6;
var delimiter = ':';
var regexp = [];
for (var i=1; i<=max_length; ++i) {
var side = '';
for (var j=1; j<=i; ++j) {
side += '.';
}
regexp.push('(' + side + delimiter + side + ')');
}
regexp = '\b(' + regexp.join('|') + ')\b';
regexp = new RegExp(regexp);
You could simply split the tokens and compare the lengths:
var range = "123:567";
var tokens = range.split(":");
var valid = tokens[0].length === tokens[1].length;
function compareValues(str){
var str_array = str.split(':');
return (str_array[0].length===str_array[1].length);
}
alert(compareValues('A9A 9AA:A9A 9ZZ'));
alert(compareValues('85001:255'));
jsfiddle : http://jsfiddle.net/EU7uB/4/
Just an added thought, given your two postal codes:
var s1 = "85001:85255";
var s2 = "A9A 9AA:A9A 9ZZ";
This regex will test for a pattern of either 5 digits before and after the colon, OR 3 of either a letter or number followed by a space and then another 3 of either a letter or number before and after the colon.
//var re = /(((\w|\d){3}\s(\w|\d){3}:(\w|\d){3}\s(\w|\d){3})|(\d{5}:\d{5}))/;
var re = /((\w{3}\s\w{3}:\w{3}\s\w{3})|(\d{5}:\d{5}))/;
alert(re.test(s1));
alert(re.test(s2));
You can use capture parentheses on both side of the colon and then test their
Lengths for equality.
var myString = "85001:85255";
var myRegexp = /^([\w][\w\s]+):([\w][\w\s]+)$/;
var match = myRegexp.exec(myString);
if (match[1].length === match[2].length) {
alert("There is a match!")
}
else {
alert("Not a match")
}
//There is a match!

How to remove the last matched regex pattern in javascript

I have a text which goes like this...
var string = '~a=123~b=234~c=345~b=456'
I need to extract the string such that it splits into
['~a=123~b=234~c=345','']
That is, I need to split the string with /b=.*/ pattern but it should match the last found pattern. How to achieve this using RegEx?
Note: The numbers present after the equal is randomly generated.
Edit:
The above one was just an example. I did not make the question clear I guess.
Generalized String being...
<word1>=<random_alphanumeric_word>~<word2>=<random_alphanumeric_word>..~..~..<word2>=<random_alphanumeric_word>
All have random length and all wordi are alphabets, the whole string length is not fixed. the only text known would be <word2>. Hence I needed RegEx for it and pattern being /<word2>=.*/
This doesn't sound like a job for regexen considering that you want to extract a specific piece. Instead, you can just use lastIndexOf to split the string in two:
var lio = str.lastIndexOf('b=');
var arr = [];
var arr[0] = str.substr(0, lio);
var arr[1] = str.substr(lio);
http://jsfiddle.net/NJn6j/
I don't think I'd personally use a regex for this type of problem, but you can extract the last option pair with a regex like this:
var str = '~a=123~b=234~c=345~b=456';
var matches = str.match(/^(.*)~([^=]+=[^=]+)$/);
// matches[1] = "~a=123~b=234~c=345"
// matches[2] = "b=456"
Demo: http://jsfiddle.net/jfriend00/SGMRC/
Assuming the format is (~, alphanumeric name, =, and numbers) repeated arbitrary number of times. The most important assumption here is that ~ appear once for each name-value pair, and it doesn't appear in the name.
You can remove the last token by a simple replacement:
str.replace(/(.*)~.*/, '$1')
This works by using the greedy property of * to force it to match the last ~ in the input.
This can also be achieved with lastIndexOf, since you only need to know the index of the last ~:
str.substring(0, (str.lastIndexOf('~') + 1 || str.length() + 1) - 1)
(Well, I don't know if the code above is good JS or not... I would rather write in a few lines. The above is just for showing one-liner solution).
A RegExp that will give a result that you may could use is:
string.match(/[a-z]*?=(.*?((?=~)|$))/gi);
// ["a=123", "b=234", "c=345", "b=456"]
But in your case the simplest solution is to split the string before extract the content:
var results = string.split('~'); // ["", "a=123", "b=234", "c=345", "b=456"]
Now will be easy to extract the key and result to add to an object:
var myObj = {};
results.forEach(function (item) {
if(item) {
var r = item.split('=');
if (!myObj[r[0]]) {
myObj[r[0]] = [r[1]];
} else {
myObj[r[0]].push(r[1]);
}
}
});
console.log(myObj);
Object:
a: ["123"]
b: ["234", "456"]
c: ["345"]
(?=.*(~b=[^~]*))\1
will get it done in one match, but if there are duplicate entries it will go to the first. Performance also isn't great and if you string.replace it will destroy all duplicates. It would pass your example, but against '~a=123~b=234~c=345~b=234' it would go to the first 'b=234'.
.*(~b=[^~]*)
will run a lot faster, but it requires another step because the match comes out in a group:
var re = /.*(~b=[^~]*)/.exec(string);
var result = re[1]; //~b=234
var array = string.split(re[1]);
This method will also have the with exact duplicates. Another option is:
var regex = /.*(~b=[^~]*)/g;
var re = regex.exec(string);
var result = re[1];
// if you want an array from either side of the string:
var array = [string.slice(0, regex.lastIndex - re[1].length - 1), string.slice(regex.lastIndex, string.length)];
This actually finds the exact location of the last match and removes it regex.lastIndex - re[1].length - 1 is my guess for the index to remove the ellipsis from the leading side, but I didn't test it so it might be off by 1.

How can I remove all characters up to and including the 3rd slash in a string?

I'm having trouble with removing all characters up to and including the 3 third slash in JavaScript. This is my string:
http://blablab/test
The result should be:
test
Does anybody know the correct solution?
To get the last item in a path, you can split the string on / and then pop():
var url = "http://blablab/test";
alert(url.split("/").pop());
//-> "test"
To specify an individual part of a path, split on / and use bracket notation to access the item:
var url = "http://blablab/test/page.php";
alert(url.split("/")[3]);
//-> "test"
Or, if you want everything after the third slash, split(), slice() and join():
var url = "http://blablab/test/page.php";
alert(url.split("/").slice(3).join("/"));
//-> "test/page.php"
var string = 'http://blablab/test'
string = string.replace(/[\s\S]*\//,'').replace(/[\s\S]*\//,'').replace(/[\s\S]*\//,'')
alert(string)
This is a regular expression. I will explain below
The regex is /[\s\S]*\//
/ is the start of the regex
Where [\s\S] means whitespace or non whitespace (anything), not to be confused with . which does not match line breaks (. is the same as [^\r\n]).
* means that we match anywhere from zero to unlimited number of [\s\S]
\/ Means match a slash character
The last / is the end of the regex
var str = "http://blablab/test";
var index = 0;
for(var i = 0; i < 3; i++){
index = str.indexOf("/",index)+1;
}
str = str.substr(index);
To make it a one liner you could make the following:
str = str.substr(str.indexOf("/",str.indexOf("/",str.indexOf("/")+1)+1)+1);
You can use split to split the string in parts and use slice to return all parts after the third slice.
var str = "http://blablab/test",
arr = str.split("/");
arr = arr.slice(3);
console.log(arr.join("/")); // "test"
// A longer string:
var str = "http://blablab/test/test"; // "test/test";
You could use a regular expression like this one:
'http://blablab/test'.match(/^(?:[^/]*\/){3}(.*)$/);
// -> ['http://blablab/test', 'test]
A string’s match method gives you either an array (of the whole match, in this case the whole input, and of any capture groups (and we want the first capture group)), or null. So, for general use you need to pull out the 1th element of the array, or null if a match wasn’t found:
var input = 'http://blablab/test',
re = /^(?:[^/]*\/){3}(.*)$/,
match = input.match(re),
result = match && match[1]; // With this input, result contains "test"
let str = "http://blablab/test";
let data = new URL(str).pathname.split("/").pop();
console.log(data);

Categories

Resources