RegEx pattern test is failing in javascript

Question

I have below RegEx to validate a string..

var str = "Thebestthingsinlifearefree";
var patt = /[^0-9A-Za-z !\\#$%&()*+,\-.\/:;<=>?@\[\]^_`{|}~]*/g;
var res = patt.test(str);

the result will always give true but I thought it would give false.. because I checking any pattern which is not in the patt variable...

The given string is valid and it contains only Alphabets with capital and small case letters. Not sure what is wrong with the pattern.

checking any occurrence of characters which is not included patt variable. — PrabaharanKathiresan, Sep 13 '16 at 07:41
@ThomasAyoub, Sorry for delayed response... Its working fine but I have a scenario to capture non-keyboard characters(like copyright symbol or any other language characters), In that case, the regex pattern is failing... Ex.. if I entered "Ä/ä" this, the regex pattern is failing to catch this characters. any Idea on this... It will be great help to me.. — PrabaharanKathiresan, Sep 14 '16 at 06:42

Thomas Ayoub · Accepted Answer · 2016-09-13T07:48:23.913

Here's your code:

var str = "Thebestthingsinlifearefree";
var patt = /[^0-9A-Za-z !\\#$%&()*+,\-.\/:;<=>?@\[\]^_`{|}~]*/g;
console.log(patt.test(str));

The regex

/[^0-9A-Za-z !\\#$%&()*+,\-.\/:;<=>?@\[\]^_`{|}~]*/g

will match anything since it accepts match of length 0 due to the quantifier *.

Just add anchors:

var str = "Thebestthingsinlifearefree";
var patt = /^[^0-9A-Za-z !\\#$%&()*+,\-.\/:;<=>?@\[\]^_`{|}~]*$/;
console.log(patt.test(str));

Here's an explanation or your regex:

[^0-9A-Za-z !\\#$%&()*+,\-.\/:;<=>?@\[\]^_`{|}~]* match a single character not present in the list below

    Quantifier: * Between zero and unlimited times, as many times as possible, giving back as needed [greedy]
    0-9 a single character in the range between 0 and 9
    A-Z a single character in the range between A and Z (case sensitive)
    a-z a single character in the range between a and z (case sensitive)
     ! a single character in the list  ! literally
    \\ matches the character \ literally
    #$%&()*+, a single character in the list #$%&()*+, literally (case sensitive)
    \- matches the character - literally
    . the literal character .
    \/ matches the character / literally
    :;<=>?@ a single character in the list :;<=>?@ literally (case sensitive)
    \[ matches the character [ literally
    \] matches the character ] literally
    ^_`{|}~ a single character in the list ^_`{|}~ literally

The `/g` modifier [should not be used with a `RegExp#test()`](http://stackoverflow.com/questions/1520800/why-regexp-with-global-flag-in-javascript-give-wrong-results). — Wiktor Stribiżew, Sep 13 '16 at 07:45
@WiktorStribiżew Thanks for sharing, I'll take a look, learn & correct my answer — Thomas Ayoub, Sep 13 '16 at 07:47
Pretty sure OP does not want `''` to pass the expression. Of course, I could be wrong. If not, change `*` to `+`. — StackSlave, Sep 13 '16 at 07:48

score 0 · Answer 2 · answered Sep 13 '16 at 07:52

Note that:

A search for a missing pattern is better expressed by a negative condition in code (!patt.test...).
You need to escape certain characters like ., (, ), ?, etc. by prefixing them with a backslash (\).

var str = "Thebestthingsinlifearefree";
var patt = /[0-9A-Za-z !\\#$%&\(\)*+,\-\.\/:;<=>\?@\[\]^_`\{|\}~]/;
var res = !patt.test(str);
console.log(res);

This will print false, as expected.

RegEx pattern test is failing in javascript

2 Answers2