['abc','xyz'] – this string I want turn into abc,xyz using regex in javascript. I want to replace both open close square bracket & single quote with empty string ie "".
Use this regular expression to match square brackets or single quotes:
/[\[\]']+/g
Replace with the empty string.
console.log("['abc','xyz']".replace(/[\[\]']+/g,''));
str.replace(/[[\]]/g,'')
here you go
var str = "['abc',['def','ghi'],'jkl']";
//'[\'abc\',[\'def\',\'ghi\'],\'jkl\']'
str.replace(/[\[\]']/g,'' );
//'abc,def,ghi,jkl'
You probably don't even need string substitution for that. If your original string is JSON, try:
js> a="['abc','xyz']"
['abc','xyz']
js> eval(a).join(",")
abc,xyz
Be careful with eval, of course.
Just here to propose an alternative that I find more readable.
/\[|\]/g
JavaScript implementation:
let reg = /\[|\]/g
str.replace(reg,'')
As other people have shown, all you have to do is list the [ and ] characters, but because they are special characters you have to escape them with \.
I personally find the character group definition using [] to be confusing because it uses the same special character you're trying to replace.
Therefore using the | (OR) operator you can more easily distinguish the special characters in the regex from the literal characters being replaced.
This should work for you.
str.replace(/[[\]]/g, "");
Related
Those two regex act the same way:
var str = "43gf\\..--.65";
console.log(str.replace(/[^\d.-]/g, ""));
console.log(str.replace(/[^\d\.-]/g, ""));
In the first regex I don't escape the dot(.) while in the second regex I do(\.).
What are the differences? Why is the result the same?
The dot operator . does not need to be escaped inside of a character class [].
Because the dot is inside character class (square brackets []).
Take a look at http://www.regular-expressions.info/reference.html, it says (under char class section):
Any character except ^-]\ add that character to the possible matches
for the character class.
If you using JavaScript to test your Regex, try \\. instead of \..
It acts on the same way because JS remove first backslash.
On regular-expressions.info, it is stated:
Remember that the dot is not a metacharacter inside a character class,
so we do not need to escape it with a backslash.
So I guess the escaping of it is unnecessary...
I have a javascript string which have a leading dot. I want to remove the leading dot using javascript replace function. I tried the following code.
var a = '.2.98»';
document.write(a.replace('/^(\.+)(.+)/',"$2"));
But this is not working. Any Idea?
The following replaces a dot in the beginning of a string with an empty string leaving the rest of the string untouched:
a.replace(/^\./, "")
Don't do regexes if you don't need to.
A simple charAt() and a substring() or a substr() (only if charAt(0) is .) will be enough.
Resources:
developer.mozilla.org: charAt()
developer.mozilla.org: substring()
developer.mozilla.org: substr()
Your regex is wrong.
var a = '.2.98»';
document.write(a.replace('/^\.(.+)/',"$1"));
You tried to match (.+) to the leading dot, but that doesn't work, you want \. instead.
Keep it simple:
if (a.charAt(0)=='.') {
document.write(a.substr(1));
} else {
document.write(a);
}
Need a function to strip off a set of illegal character in javascript: |&;$%#"<>()+,
This is a classic problem to be solved with regexes, which means now I have 2 problems.
This is what I've got so far:
var cleanString = dirtyString.replace(/\|&;\$%#"<>\(\)\+,/g, "");
I am escaping the regex special chars with a backslash but I am having a hard time trying to understand what's going on.
If I try with single literals in isolation most of them seem to work, but once I put them together in the same regex depending on the order the replace is broken.
i.e. this won't work --> dirtyString.replace(/\|<>/g, ""):
Help appreciated!
What you need are character classes. In that, you've only to worry about the ], \ and - characters (and ^ if you're placing it straight after the beginning of the character class "[" ).
Syntax: [characters] where characters is a list with characters.
Example:
var cleanString = dirtyString.replace(/[|&;$%#"<>()+,]/g, "");
I tend to look at it from the inverse perspective which may be what you intended:
What characters do I want to allow?
This is because there could be lots of characters that make in into a string somehow that blow stuff up that you wouldn't expect.
For example this one only allows for letters and numbers removing groups of invalid characters replacing them with a hypen:
"This¢£«±Ÿ÷could&*()\/<>be!##$%^bad".replace(/([^a-z0-9]+)/gi, '-');
//Result: "This-could-be-bad"
You need to wrap them all in a character class. The current version means replace this sequence of characters with an empty string. When wrapped in square brackets it means replace any of these characters with an empty string.
var cleanString = dirtyString.replace(/[\|&;\$%#"<>\(\)\+,]/g, "");
Put them in brackets []:
var cleanString = dirtyString.replace(/[\|&;\$%#"<>\(\)\+,]/g, "");
I am having trouble googling this. In some code I see
name = name.replace(/[\[]/,"\\\[").replace(/[\]]/,"\\\]");
/[\[]/ looks to be 1 parameter. What do the symbols do? It looks like it's replacing [] with \[\] but what specifically does /[\[]/ do?
The syntax /…/ is the literal regular expression syntax. And the regular expression [\[] describes a character class ([…]) that’s only character the [ is). So /[\[]/ is a regular expression that describes a single [.
But since the global flag is not set (so only the first match will be replaced), the whole thing could be replaced with this (probably easier to read):
name.replace("[", "\\[").replace("]","\\]")
But if all matches should be replaced, I would probably use this:
name.replace(/([[\]])/g, "\\$1")
It's a regular expression that matches the left square bracket character.
It's a weird way to do it; overall it looks like the code is trying to put backslashes before square brackets in a string, which you could also do like this:
var s2 = s1.replace(/\[/g, '\\[').replace(/]/g, '\\]');
I think.
/[[]/ defined a character range which includes only the ']' character (escaped), you are correct that is replaced [] with [].
The [] is in regex itself used to denote a collection of to-be-matched characters. If you want to represent the actual [ or ] in regex, then you need to escape it by \, hence the [\[] and [\]]. The leading and trailing / are just part of the standard JS syntax to to denote a regex pattern.
After all, it replaces [ by \[ and then replaces ] by \].
I have a string and I want to validate that string so that it must not contain certain characters like '/' '\' '&' ';' etc... How can I validate all that at once?
You can solve this with regular expressions!
mystring = "hello"
yourstring = "bad & string"
validRegEx = /^[^\\\/&]*$/
alert(mystring.match(validRegEx))
alert(yourstring.match(validRegEx))
matching against the regex returns the string if it is ok, or null if its invalid!
Explanation:
JavaScript RegEx Literals are delimited like strings, but with slashes (/'s) instead of quotes ("'s).
The first and last characters of the validRegEx cause it to match against the whole string, instead of just part, the carat anchors it to the beginning, and the dollar sign to the end.
The part between the brackets ([ and ]) are a character class, which matches any character so long as it's in the class. The first character inside that, a carat, means that the class is negated, to match the characters not mentioned in the character class. If it had been omited, the class would match the characters it specifies.
The next two sequences, \\ and \/ are backslash escaped because the backslash by itself would be an escape sequence for something else, and the forward slash would confuse the parser into thinking that it had reached the end of the regex, (exactly similar to escaping quotes in strings).
The ampersand (&) has no special meaning and is unescaped.
The remaining character, the kleene star, (*) means that whatever preceeded it should be matched zero or more times, so that the character class will eat as many characters that are not forward or backward slashes or ampersands, including none if it cant find any. If you wanted to make sure the matched string was non-empty, you can replace it with a plus (+).
I would use regular expressions.
See this guide from Mozillla.org. This article does also give a good introduction to regular expressions in JavaScript.
Here is a good article on Javascript validation. Remember you will need to validate on the server side too. Javascript validation can easily be circumvented, so it should never be used for security reasons such as preventing SQL Injection or XSS attacks.
You could learn regular expressions, or (probably simpler if you only check for one character at a time) you could have a list of characters and then some kind of sanitize function to remove each one from the string.
var myString = "An /invalid &string;";
var charList = ['/', '\\', '&', ';']; // etc...
function sanitize(input, list) {
for (char in list) {
input = input.replace(char, '');
}
return input
}
So then:
sanitize(myString, charList) // returns "An invalid string"
You can use the test method, with regular expressions:
function validString(input){
return !(/[\\/&;]/.test(input));
}
validString('test;') //false
You can use regex. For example if your string matches:
[\\/&;]+
then it is not valid. Look at:
http://www.regular-expressions.info/javascriptexample.html
You could probably use a regular expression.
As the others have answered you can solve this with regexp but remember to also check the value server-side. There is no guarantee that the user has JavaScript activated. Never trust user input!