I need a regular expression to test string , either string do not contain parentheses or it contain balanced parentheses. The expression I am using only check it contain balanced parentheses or not . If I give string without parentheses it return false.
I want it return true if it do not contain any parentheses
function func(str) {
let reg = /\([^()]*\)/;
console.log(reg3.test(str))
console.log(str.match(reg3))
}
func("33")
func("1(1)6")
func("11)6")
The problem is it give false if I do not give any parentheses , I need to write that it return true if string do not contain parentheses nd false if string contain unbalanced parentheses
You can remove all substrings between closest parentheses, and then check if a string contains a ( or ) char.
function remove_nested_parens(s) {
let new_s = s.replace(/\s*\([^()]*\)/g, "");
return new_s == s ? s : remove_nested_parens(new_s);
}
const texts = ["no parens", "XX(DD(D)DDDD)XXX", "(DD(D)DDDD))XX"];
for (const text of texts) {
console.log(text, '=>', !/[()]/.test(remove_nested_parens(text)) )
}
The recursive remove_nested_parens function removes parenthesized substrings and !/[()]/.test(remove_nested_parens(text)) checks if the result contains ( or ).
Related
I'm trying to solve a string replacement problem; A function that takes in a string and object of values.
It's a 2 part algorithm:
Replace values in the string that are inside of braces.
If it's in double braces, remove the outer braces; these are escaped sequence.
Example test case:
expect(replaceValue('The sky is [condition] and the color is [[blue]]', {'condition':'clear')).toBe('The sky is clear and the color is [blue]');
I solved the first part:
function replaceValue(input, replacementValue){
let copied = input;
Object.keys(replacementValue).forEach(ele => {
copied = copied.replace(`[${ele}]`, replacementValue[ele]);
})
return copied;
}
But I'm not sure how to do the 2nd part; remove the outer braces.
You could look for square brackets and replace only with knowing properties or take the returned value.
/\[([^\[\]]+)\]/g complete regular experession
\[ \] outer brackets
( ) group for content
[^\[\]]+ inside all characters but not brackets
g search global
function replaceValue(string, replacements) {
return string.replace(/\[([^\[\]]+)\]/g, (_, s) => replacements[s] || s);
}
console.log(replaceValue('The sky is [condition] and the color is [[blue]]', { condition: 'clear' }));
A negative lookbehind helps us to avoid replacing keywords enclosed in double brackets ([[ and ]]):
function repl(str,obj) {
return str.replace(/(?<!\[)\[([^\]]+)\]/g, (_, s) => obj[s] || s);
}
console.log(repl('The sky is [condition] and the color is [[blue]]', { condition: 'clear' , blue:'#0000ff'}));
Explanation:
The regular expression
(?<!\[)\[([^\]]+)\]/g
consists of
(?<!\[): a negative lookbehind, demanding the previous character is never a [
\[: a single [
([^\]]+): a group consisting of at least one non-[ character. This group will be used in a callback function of the .replace() method
\]/g: a final ] character and the end of the regular expression with the qualifier g for global, allowing this pattern to be matched multiple times.
Differently to Nina's solution mine will not replace the key [[blue]] in the string, although a replacement value would have been available.
use a regular expression, match starts with two opening brackets [[ and ends with two closing brackets ]], replace it with the string inside the brackets with one pair of brackets
function replaceValue(input, replacementValue) {
let copied = input;
Object.keys(replacementValue).forEach(ele => {
copied = copied.replace(`[${ele}]`, replacementValue[ele]);
});
copied = copied.replace(/\[\[(.*?)\]\]/g, '[$1]');
return copied;
}
Using a negative lookahead, you can ensure that one bracket is not followed by another.
function replaceValue(input, replacementValue) {
return input
.replace(/\[(?!\[)([^\]]*?)\](?!\])/g, function(m, p) {
return replacementValue[p];
})
.replaceAll("[[", "[")
.replaceAll("]]", "]");
}
console.log(replaceValue('The sky is [condition] and the color is [[blue]]', {'condition':'clear'}));
I want to check if a string contains special characters like !##$%^&*.,<>/\'";:? and return true if the string contains atleast one of those chars.
I tried with the following regex,script:
var format = /^[!##$%^&*()_+\-=\[\]{};':"\\|,.<>\/?]*$/;
if( string.match(format) ){
return true;
}else{
return false;
}
If the string contains only the special characters then it returns true , but if the string contains something else like alphanumeric chars ( !example1 , .example2 ) it returns false.
I suggest using RegExp .test() function to check for a pattern match, and the only thing you need to change is remove the start/end of line anchors (and the * quantifier is also redundant) in the regex:
var format = /[ `!##$%^&*()_+\-=\[\]{};':"\\|,.<>\/?~]/;
// ^ ^
document.write(format.test("My#string-with(some%text)") + "<br/>");
document.write(format.test("My string with spaces") + "<br/>");
document.write(format.test("MyStringContainingNoSpecialChars"));
The anchors (like ^ start of string/line, $ end od string/line and \b word boundaries) can restrict matches at specific places in a string. When using ^ the regex engine checks if the next subpattern appears right at the start of the string (or line if /m modifier is declared in the regex). Same case with $: the preceding subpattern should match right at the end of the string.
In your case, you want to check the existence of the special character from the set anywhere in the string. Even if it is only one, you want to return false. Thus, you should remove the anchors, and the quantifier *. The * quantifier would match even an empty string, thus we must remove it in order to actually check for the presence of at least 1 special character (actually, without any quantifiers we check for exactly one occurrence, same as if we were using {1} limiting quantifier).
More specific solutions
What characters are "special" for you?
All chars other than ASCII chars: /[^\x00-\x7F]/ (demo)
All chars other than printable ASCII chars: /[^ -~]/ (demo)
Any printable ASCII chars other than space, letters and digits: /[!-\/:-#[-`{-~]/ (demo)
Any Unicode punctuation proper chars, the \p{P} Unicode property class:
ECMAScript 2018: /\p{P}/u
ES6+:
/[!-#%-*,-\/:;?#[-\]_{}\u00A1\u00A7\u00AB\u00B6\u00B7\u00BB\u00BF\u037E\u0387\u055A-\u055F\u0589\u058A\u05BE\u05C0\u05C3\u05C6\u05F3\u05F4\u0609\u060A\u060C\u060D\u061B\u061E\u061F\u066A-\u066D\u06D4\u0700-\u070D\u07F7-\u07F9\u0830-\u083E\u085E\u0964\u0965\u0970\u09FD\u0A76\u0AF0\u0C84\u0DF4\u0E4F\u0E5A\u0E5B\u0F04-\u0F12\u0F14\u0F3A-\u0F3D\u0F85\u0FD0-\u0FD4\u0FD9\u0FDA\u104A-\u104F\u10FB\u1360-\u1368\u1400\u166D\u166E\u169B\u169C\u16EB-\u16ED\u1735\u1736\u17D4-\u17D6\u17D8-\u17DA\u1800-\u180A\u1944\u1945\u1A1E\u1A1F\u1AA0-\u1AA6\u1AA8-\u1AAD\u1B5A-\u1B60\u1BFC-\u1BFF\u1C3B-\u1C3F\u1C7E\u1C7F\u1CC0-\u1CC7\u1CD3\u2010-\u2027\u2030-\u2043\u2045-\u2051\u2053-\u205E\u207D\u207E\u208D\u208E\u2308-\u230B\u2329\u232A\u2768-\u2775\u27C5\u27C6\u27E6-\u27EF\u2983-\u2998\u29D8-\u29DB\u29FC\u29FD\u2CF9-\u2CFC\u2CFE\u2CFF\u2D70\u2E00-\u2E2E\u2E30-\u2E4E\u3001-\u3003\u3008-\u3011\u3014-\u301F\u3030\u303D\u30A0\u30FB\uA4FE\uA4FF\uA60D-\uA60F\uA673\uA67E\uA6F2-\uA6F7\uA874-\uA877\uA8CE\uA8CF\uA8F8-\uA8FA\uA8FC\uA92E\uA92F\uA95F\uA9C1-\uA9CD\uA9DE\uA9DF\uAA5C-\uAA5F\uAADE\uAADF\uAAF0\uAAF1\uABEB\uFD3E\uFD3F\uFE10-\uFE19\uFE30-\uFE52\uFE54-\uFE61\uFE63\uFE68\uFE6A\uFE6B\uFF01-\uFF03\uFF05-\uFF0A\uFF0C-\uFF0F\uFF1A\uFF1B\uFF1F\uFF20\uFF3B-\uFF3D\uFF3F\uFF5B\uFF5D\uFF5F-\uFF65\u{10100}-\u{10102}\u{1039F}\u{103D0}\u{1056F}\u{10857}\u{1091F}\u{1093F}\u{10A50}-\u{10A58}\u{10A7F}\u{10AF0}-\u{10AF6}\u{10B39}-\u{10B3F}\u{10B99}-\u{10B9C}\u{10F55}-\u{10F59}\u{11047}-\u{1104D}\u{110BB}\u{110BC}\u{110BE}-\u{110C1}\u{11140}-\u{11143}\u{11174}\u{11175}\u{111C5}-\u{111C8}\u{111CD}\u{111DB}\u{111DD}-\u{111DF}\u{11238}-\u{1123D}\u{112A9}\u{1144B}-\u{1144F}\u{1145B}\u{1145D}\u{114C6}\u{115C1}-\u{115D7}\u{11641}-\u{11643}\u{11660}-\u{1166C}\u{1173C}-\u{1173E}\u{1183B}\u{11A3F}-\u{11A46}\u{11A9A}-\u{11A9C}\u{11A9E}-\u{11AA2}\u{11C41}-\u{11C45}\u{11C70}\u{11C71}\u{11EF7}\u{11EF8}\u{12470}-\u{12474}\u{16A6E}\u{16A6F}\u{16AF5}\u{16B37}-\u{16B3B}\u{16B44}\u{16E97}-\u{16E9A}\u{1BC9F}\u{1DA87}-\u{1DA8B}\u{1E95E}\u{1E95F}]/u
● ES5 (demo):
/(?:[!-#%-\*,-\/:;\?#\[-\]_\{\}\xA1\xA7\xAB\xB6\xB7\xBB\xBF\u037E\u0387\u055A-\u055F\u0589\u058A\u05BE\u05C0\u05C3\u05C6\u05F3\u05F4\u0609\u060A\u060C\u060D\u061B\u061E\u061F\u066A-\u066D\u06D4\u0700-\u070D\u07F7-\u07F9\u0830-\u083E\u085E\u0964\u0965\u0970\u09FD\u0A76\u0AF0\u0C84\u0DF4\u0E4F\u0E5A\u0E5B\u0F04-\u0F12\u0F14\u0F3A-\u0F3D\u0F85\u0FD0-\u0FD4\u0FD9\u0FDA\u104A-\u104F\u10FB\u1360-\u1368\u1400\u166D\u166E\u169B\u169C\u16EB-\u16ED\u1735\u1736\u17D4-\u17D6\u17D8-\u17DA\u1800-\u180A\u1944\u1945\u1A1E\u1A1F\u1AA0-\u1AA6\u1AA8-\u1AAD\u1B5A-\u1B60\u1BFC-\u1BFF\u1C3B-\u1C3F\u1C7E\u1C7F\u1CC0-\u1CC7\u1CD3\u2010-\u2027\u2030-\u2043\u2045-\u2051\u2053-\u205E\u207D\u207E\u208D\u208E\u2308-\u230B\u2329\u232A\u2768-\u2775\u27C5\u27C6\u27E6-\u27EF\u2983-\u2998\u29D8-\u29DB\u29FC\u29FD\u2CF9-\u2CFC\u2CFE\u2CFF\u2D70\u2E00-\u2E2E\u2E30-\u2E4E\u3001-\u3003\u3008-\u3011\u3014-\u301F\u3030\u303D\u30A0\u30FB\uA4FE\uA4FF\uA60D-\uA60F\uA673\uA67E\uA6F2-\uA6F7\uA874-\uA877\uA8CE\uA8CF\uA8F8-\uA8FA\uA8FC\uA92E\uA92F\uA95F\uA9C1-\uA9CD\uA9DE\uA9DF\uAA5C-\uAA5F\uAADE\uAADF\uAAF0\uAAF1\uABEB\uFD3E\uFD3F\uFE10-\uFE19\uFE30-\uFE52\uFE54-\uFE61\uFE63\uFE68\uFE6A\uFE6B\uFF01-\uFF03\uFF05-\uFF0A\uFF0C-\uFF0F\uFF1A\uFF1B\uFF1F\uFF20\uFF3B-\uFF3D\uFF3F\uFF5B\uFF5D\uFF5F-\uFF65]|\uD800[\uDD00-\uDD02\uDF9F\uDFD0]|\uD801\uDD6F|\uD802[\uDC57\uDD1F\uDD3F\uDE50-\uDE58\uDE7F\uDEF0-\uDEF6\uDF39-\uDF3F\uDF99-\uDF9C]|\uD803[\uDF55-\uDF59]|\uD804[\uDC47-\uDC4D\uDCBB\uDCBC\uDCBE-\uDCC1\uDD40-\uDD43\uDD74\uDD75\uDDC5-\uDDC8\uDDCD\uDDDB\uDDDD-\uDDDF\uDE38-\uDE3D\uDEA9]|\uD805[\uDC4B-\uDC4F\uDC5B\uDC5D\uDCC6\uDDC1-\uDDD7\uDE41-\uDE43\uDE60-\uDE6C\uDF3C-\uDF3E]|\uD806[\uDC3B\uDE3F-\uDE46\uDE9A-\uDE9C\uDE9E-\uDEA2]|\uD807[\uDC41-\uDC45\uDC70\uDC71\uDEF7\uDEF8]|\uD809[\uDC70-\uDC74]|\uD81A[\uDE6E\uDE6F\uDEF5\uDF37-\uDF3B\uDF44]|\uD81B[\uDE97-\uDE9A]|\uD82F\uDC9F|\uD836[\uDE87-\uDE8B]|\uD83A[\uDD5E\uDD5F])/
All Unicode symbols (not punctuation proper), \p{S}:
ECMAScript 2018: /\p{S}/u
ES6+:
/[$+^`|~\u00A2-\u00A6\u00A8\u00A9\u00AC\u00AE-\u00B1\u00B4\u00B8\u00D7\u00F7\u02C2-\u02C5\u02D2-\u02DF\u02E5-\u02EB\u02ED\u02EF-\u02FF\u0375\u0384\u0385\u03F6\u0482\u058D-\u058F\u0606-\u0608\u060B\u060E\u060F\u06DE\u06E9\u06FD\u06FE\u07F6\u07FE\u07FF\u09F2\u09F3\u09FA\u09FB\u0AF1\u0B70\u0BF3-\u0BFA\u0C7F\u0D4F\u0D79\u0E3F\u0F01-\u0F03\u0F13\u0F15-\u0F17\u0F1A-\u0F1F\u0F34\u0F36\u0F38\u0FBE-\u0FC5\u0FC7-\u0FCC\u0FCE\u0FCF\u0FD5-\u0FD8\u109E\u109F\u1390-\u1399\u17DB\u1940\u19DE-\u19FF\u1B61-\u1B6A\u1B74-\u1B7C\u1FBD\u1FBF-\u1FC1\u1FCD-\u1FCF\u1FDD-\u1FDF\u1FED-\u1FEF\u1FFD\u1FFE\u2044\u2052\u207A-\u207C\u208A-\u208C\u20A0-\u20BF\u2100\u2101\u2103-\u2106\u2108\u2109\u2114\u2116-\u2118\u211E-\u2123\u2125\u2127\u2129\u212E\u213A\u213B\u2140-\u2144\u214A-\u214D\u214F\u218A\u218B\u2190-\u2307\u230C-\u2328\u232B-\u2426\u2440-\u244A\u249C-\u24E9\u2500-\u2767\u2794-\u27C4\u27C7-\u27E5\u27F0-\u2982\u2999-\u29D7\u29DC-\u29FB\u29FE-\u2B73\u2B76-\u2B95\u2B98-\u2BC8\u2BCA-\u2BFE\u2CE5-\u2CEA\u2E80-\u2E99\u2E9B-\u2EF3\u2F00-\u2FD5\u2FF0-\u2FFB\u3004\u3012\u3013\u3020\u3036\u3037\u303E\u303F\u309B\u309C\u3190\u3191\u3196-\u319F\u31C0-\u31E3\u3200-\u321E\u322A-\u3247\u3250\u3260-\u327F\u328A-\u32B0\u32C0-\u32FE\u3300-\u33FF\u4DC0-\u4DFF\uA490-\uA4C6\uA700-\uA716\uA720\uA721\uA789\uA78A\uA828-\uA82B\uA836-\uA839\uAA77-\uAA79\uAB5B\uFB29\uFBB2-\uFBC1\uFDFC\uFDFD\uFE62\uFE64-\uFE66\uFE69\uFF04\uFF0B\uFF1C-\uFF1E\uFF3E\uFF40\uFF5C\uFF5E\uFFE0-\uFFE6\uFFE8-\uFFEE\uFFFC\uFFFD\u{10137}-\u{1013F}\u{10179}-\u{10189}\u{1018C}-\u{1018E}\u{10190}-\u{1019B}\u{101A0}\u{101D0}-\u{101FC}\u{10877}\u{10878}\u{10AC8}\u{1173F}\u{16B3C}-\u{16B3F}\u{16B45}\u{1BC9C}\u{1D000}-\u{1D0F5}\u{1D100}-\u{1D126}\u{1D129}-\u{1D164}\u{1D16A}-\u{1D16C}\u{1D183}\u{1D184}\u{1D18C}-\u{1D1A9}\u{1D1AE}-\u{1D1E8}\u{1D200}-\u{1D241}\u{1D245}\u{1D300}-\u{1D356}\u{1D6C1}\u{1D6DB}\u{1D6FB}\u{1D715}\u{1D735}\u{1D74F}\u{1D76F}\u{1D789}\u{1D7A9}\u{1D7C3}\u{1D800}-\u{1D9FF}\u{1DA37}-\u{1DA3A}\u{1DA6D}-\u{1DA74}\u{1DA76}-\u{1DA83}\u{1DA85}\u{1DA86}\u{1ECAC}\u{1ECB0}\u{1EEF0}\u{1EEF1}\u{1F000}-\u{1F02B}\u{1F030}-\u{1F093}\u{1F0A0}-\u{1F0AE}\u{1F0B1}-\u{1F0BF}\u{1F0C1}-\u{1F0CF}\u{1F0D1}-\u{1F0F5}\u{1F110}-\u{1F16B}\u{1F170}-\u{1F1AC}\u{1F1E6}-\u{1F202}\u{1F210}-\u{1F23B}\u{1F240}-\u{1F248}\u{1F250}\u{1F251}\u{1F260}-\u{1F265}\u{1F300}-\u{1F6D4}\u{1F6E0}-\u{1F6EC}\u{1F6F0}-\u{1F6F9}\u{1F700}-\u{1F773}\u{1F780}-\u{1F7D8}\u{1F800}-\u{1F80B}\u{1F810}-\u{1F847}\u{1F850}-\u{1F859}\u{1F860}-\u{1F887}\u{1F890}-\u{1F8AD}\u{1F900}-\u{1F90B}\u{1F910}-\u{1F93E}\u{1F940}-\u{1F970}\u{1F973}-\u{1F976}\u{1F97A}\u{1F97C}-\u{1F9A2}\u{1F9B0}-\u{1F9B9}\u{1F9C0}-\u{1F9C2}\u{1F9D0}-\u{1F9FF}\u{1FA60}-\u{1FA6D}]/u
● ES5 (demo):
/(?:[$+^`|~\xA2-\xA6\xA8\xA9\xAC\xAE-\xB1\xB4\xB8\xD7\xF7\u02C2-\u02C5\u02D2-\u02DF\u02E5-\u02EB\u02ED\u02EF-\u02FF\u0375\u0384\u0385\u03F6\u0482\u058D-\u058F\u0606-\u0608\u060B\u060E\u060F\u06DE\u06E9\u06FD\u06FE\u07F6\u07FE\u07FF\u09F2\u09F3\u09FA\u09FB\u0AF1\u0B70\u0BF3-\u0BFA\u0C7F\u0D4F\u0D79\u0E3F\u0F01-\u0F03\u0F13\u0F15-\u0F17\u0F1A-\u0F1F\u0F34\u0F36\u0F38\u0FBE-\u0FC5\u0FC7-\u0FCC\u0FCE\u0FCF\u0FD5-\u0FD8\u109E\u109F\u1390-\u1399\u17DB\u1940\u19DE-\u19FF\u1B61-\u1B6A\u1B74-\u1B7C\u1FBD\u1FBF-\u1FC1\u1FCD-\u1FCF\u1FDD-\u1FDF\u1FED-\u1FEF\u1FFD\u1FFE\u2044\u2052\u207A-\u207C\u208A-\u208C\u20A0-\u20BF\u2100\u2101\u2103-\u2106\u2108\u2109\u2114\u2116-\u2118\u211E-\u2123\u2125\u2127\u2129\u212E\u213A\u213B\u2140-\u2144\u214A-\u214D\u214F\u218A\u218B\u2190-\u2307\u230C-\u2328\u232B-\u2426\u2440-\u244A\u249C-\u24E9\u2500-\u2767\u2794-\u27C4\u27C7-\u27E5\u27F0-\u2982\u2999-\u29D7\u29DC-\u29FB\u29FE-\u2B73\u2B76-\u2B95\u2B98-\u2BC8\u2BCA-\u2BFE\u2CE5-\u2CEA\u2E80-\u2E99\u2E9B-\u2EF3\u2F00-\u2FD5\u2FF0-\u2FFB\u3004\u3012\u3013\u3020\u3036\u3037\u303E\u303F\u309B\u309C\u3190\u3191\u3196-\u319F\u31C0-\u31E3\u3200-\u321E\u322A-\u3247\u3250\u3260-\u327F\u328A-\u32B0\u32C0-\u32FE\u3300-\u33FF\u4DC0-\u4DFF\uA490-\uA4C6\uA700-\uA716\uA720\uA721\uA789\uA78A\uA828-\uA82B\uA836-\uA839\uAA77-\uAA79\uAB5B\uFB29\uFBB2-\uFBC1\uFDFC\uFDFD\uFE62\uFE64-\uFE66\uFE69\uFF04\uFF0B\uFF1C-\uFF1E\uFF3E\uFF40\uFF5C\uFF5E\uFFE0-\uFFE6\uFFE8-\uFFEE\uFFFC\uFFFD]|\uD800[\uDD37-\uDD3F\uDD79-\uDD89\uDD8C-\uDD8E\uDD90-\uDD9B\uDDA0\uDDD0-\uDDFC]|\uD802[\uDC77\uDC78\uDEC8]|\uD805\uDF3F|\uD81A[\uDF3C-\uDF3F\uDF45]|\uD82F\uDC9C|\uD834[\uDC00-\uDCF5\uDD00-\uDD26\uDD29-\uDD64\uDD6A-\uDD6C\uDD83\uDD84\uDD8C-\uDDA9\uDDAE-\uDDE8\uDE00-\uDE41\uDE45\uDF00-\uDF56]|\uD835[\uDEC1\uDEDB\uDEFB\uDF15\uDF35\uDF4F\uDF6F\uDF89\uDFA9\uDFC3]|\uD836[\uDC00-\uDDFF\uDE37-\uDE3A\uDE6D-\uDE74\uDE76-\uDE83\uDE85\uDE86]|\uD83B[\uDCAC\uDCB0\uDEF0\uDEF1]|\uD83C[\uDC00-\uDC2B\uDC30-\uDC93\uDCA0-\uDCAE\uDCB1-\uDCBF\uDCC1-\uDCCF\uDCD1-\uDCF5\uDD10-\uDD6B\uDD70-\uDDAC\uDDE6-\uDE02\uDE10-\uDE3B\uDE40-\uDE48\uDE50\uDE51\uDE60-\uDE65\uDF00-\uDFFF]|\uD83D[\uDC00-\uDED4\uDEE0-\uDEEC\uDEF0-\uDEF9\uDF00-\uDF73\uDF80-\uDFD8]|\uD83E[\uDC00-\uDC0B\uDC10-\uDC47\uDC50-\uDC59\uDC60-\uDC87\uDC90-\uDCAD\uDD00-\uDD0B\uDD10-\uDD3E\uDD40-\uDD70\uDD73-\uDD76\uDD7A\uDD7C-\uDDA2\uDDB0-\uDDB9\uDDC0-\uDDC2\uDDD0-\uDDFF\uDE60-\uDE6D])/
All Unicode punctuation and symbols, \p{P} and \p{S}:
ECMAScript 2018: /[\p{P}\p{S}]/u
ES6+:
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● ES5 (demo):
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Your regexp use ^ and $ so it tries to match the entire string. And if you want only a boolean as the result, use test instead of match.
var format = /[!##$%^&*()_+\-=\[\]{};':"\\|,.<>\/?]+/;
if(format.test(string)){
return true;
} else {
return false;
}
Wouldn't it be easier to negative-match alphanumerics instead?
return string.match(/^[^a-zA-Z0-9]+$/) ? true : false;
var specialChars = "<>#!#$%^&*()_+[]{}?:;|'\"\\,./~`-=";
var checkForSpecialChar = function(string){
for(i = 0; i < specialChars.length;i++){
if(string.indexOf(specialChars[i]) > -1){
return true
}
}
return false;
}
var str = "YourText";
if(checkForSpecialChar(str)){
alert("Not Valid");
} else {
alert("Valid");
}
Remove the characters ^ (start of string) and $ (end of string) from the regular expression.
var format = /[!##$%^&*()_+\-=\[\]{};':"\\|,.<>\/?]/;
var format = /[`!##$%^&*()_+\-=\[\]{};':"\\|,.<>\/?~]/;
// ^ ^
document.write(format.test("My #string-with(some%text)") + "<br/>");
document.write(format.test("My string with spaces") + "<br/>");
document.write(format.test("My StringContainingNoSpecialChars"));
Simple solution
let text = "12AB"
const specialChars = `/[!##$%^&*()_+\-=\[\]{};':"\\|,.<>\/?]+/;`
const isSpecialCharsPresent = specialChars.split('').some(char =>
text.includes(char)) // true if present and false if not
console.log(isSpecialCharsPresent) // false
let text = "12AB."
console.log(isSpecialCharsPresent) // true
Check if a string contains at least one password special character:
For reference: ASCII Table -- Printable Characters
Special character ranges in the ASCII table are:
Space to /
: to #
[ to `
{ to ~
Therefore, use this:
/[ -/:-#[-`{-~]/.test(string)
I just did this,
if(string.match(/\W/)){
return true; // Contains at least one special character or space
} else {
return false;
}
You can use this regex for specials chars : /\W/
I took an alternative approuch, without using REGEX, O(n) in performence:
function hasSpecialChars(str) {
const specialCharsSet = new Set("!##$%^&*()_");
for (let letter of str) {
if (specialCharsSet.has(letter)) {
return true;
}
}
return false;
}
You can try this:
regex = [\W_]
It will definitely help you.
I'd like to replace everything between two characters with another string. I came up with this function:
String.prototype.unformat = function() {
var s='';
for (var i=0; i<this.length;i++) s+=this[i]
return s.replace(/\$[^$]*\$/g, '')
};
Using a string like 'This is a test $33$' and unformat it with the function above, it will return 'This is a test '.
Ok-cool, but I'd like to replace all occurrences in ( $ ... $ ) with the associated char code.
In the example 'This is a test $33$', I like to replace $33$ with the result of the javascript String.fromCharCode() function to get the string 'This is a test !' as result.
How to edit the prototype function above to get the desired result?
Thanks in advance :)
You can use a callback function that returns fromCharCode() with the matched code
String.prototype.unformat = function() {
return this.replace(/\$([^$]*)\$/g, function (string, charcode) {
return String.fromCharCode(charcode);
});
};
console.log(("char: $33$").unformat());
In order to avoid any future problems, I would also adapt the regex to only match digits: /\$(\d+)\$/g
You can use a match group () and replace it with the String.fromCharCode result:
String.prototype.unformat = function() {
return this.replace(/\$(.*?)\$/g, function(match, group) { // the match is the whole match (including the $s), group is the matched group (the thing between the $s)
return String.fromCharCode(group);
});
};
Notes:
No need to copy the string as replace doesn't mutate the original string (this).
The match group (.*?) is a non-greedy one (lazy one) that matches as few characters as possible.
It is better if you don't mess around natives' prototypes (such as String, Number, ...).
Example:
String.prototype.unformat = function() {
return this.replace(/\$(.*?)\$/g, function(match, group) {
return String.fromCharCode(group);
});
};
console.log('This is a test $33$'.unformat());
I am having trouble below javaScript problem.
Question:
You have been given a string s, which is supposed to be a sentence. However, someone forgot to put spaces between the different words, and for some reason they capitalized the first letter of every word. Return the sentence after making the following amendments:
Put a single space between the words.
Convert the uppercase letters to lowercase.
Example
"CodefightsIsAwesome", the output should be "codefights is awesome";
"Hello", the output should be "hello".
My current code is:
Right now, my second for-loop just manually slices the parts from the string.
How can I make this dynamic and insert "space" in front of the Capital String?
You can use String.prototype.match() with RegExp /[A-Z][^A-Z]*/g to match A-Z followed by one or more characters which are not A-Z, or character at end of string; chain Array.prototype.map() to call .toLowerCase() on matched words, .join() with parameter " " to include space character between matches at resulting string.
var str = "CodefightsIsAwesome";
var res = str.match(/[A-Z][^A-Z]*/g).map(word => word.toLowerCase()).join(" ");
console.log(res);
Alternatively, as suggested by #FissureKing, you can use String.prototype.repalce() with .trim() and .toLowerCase() chained
var str = "CodefightsIsAwesome";
var res = str.replace(/[A-Z][^A-Z]*/g, word => word + ' ').trim().toLowerCase();
console.log(res);
Rather than coding a loop, I'd do it in one line with a (reasonably) simple string replacement:
function amendTheSentence(s) {
return s.replace(/[A-Z]/g, function(m) { return " " + m.toLowerCase() })
.replace(/^ /, "");
}
console.log(amendTheSentence("CodefightsIsAwesome"));
console.log(amendTheSentence("noCapitalOnFirstWord"));
console.log(amendTheSentence("ThereIsNobodyCrazierThanI"));
That is, match any uppercase letter with the regular expression /[A-Z]/, replace the matched letter with a space plus that letter in lowercase, then remove any space that was added at the start of the string.
Further reading:
String .replace() method
Regular expressions
We can loop through once.
The below assumes the very first character should always be capitalized in our return array. If that is not true, simply remove the first if block from below.
For each character after that, we check to see if it is capitalized. If so, we add it to our return array, prefaced with a space. If not, we add it as-is into our array.
Finally, we join the array back into a string and return it.
const sentence = "CodefightsIsAwesome";
const amend = function(s) {
ret = [];
for (let i = 0; i < s.length; i++) {
const char = s[i];
if (i === 0) {
ret.push(char.toUpperCase());
} else if (char.toUpperCase() === char) {
ret.push(` ${char.toLowerCase()}`);
} else {
ret.push(char);
}
}
return ret.join('');
};
console.log(amend(sentence));
This should pass the condition:
syntax_search = (){return 0;}
syntax_search = ( ) { fsf return 0;}
syntax_search = ( ) { return 0; }
syntax_search = (){ return; }
syntax_search = (){ if(x){ sdfsdf } return 0;}
syntax_search = (){ char x[20]; return };
It is not passing all the combinations above, What is the right way?
if( /^\s*(\s*)\s*{[\s\S]*\s+return\s*[0-9]*\s*;\s*}\s*/.test(syntax_search) )
You regular expression contains many unneeded complexities and there are some characters that need escaping such as { and }.
Anyway you can use this modified version of your regex and it should work.
^\s*\(\s*\)\s*\{(.*(return\s*\d*\s*;)\s*)\}\s*;?$
// ^
// |
// There was a ? here
Regex 101 Demo
Some issues:
As M42 pointed out, you need to escape the curly brackets
The parentheses at the begining also need to be escaped (otherwise you are defining a capture group)
"return" is required by the expression. Your first 2 test cases don't contain the word return and will fail. Is that on purpose?
Same as #3 for ;.
[\s\S]* Anything which is a space and everything which isn't. Replace by a dot .* If you need to also match a newline, use [^]*
This regex is not anchored to the end of the string so it will allow invalid strings. (You can put anything you want after the last }
/^\s*(\s*)\s*{[^]return\s\d*\s*;\s*}\s*$/