Regexpr with a string

Question

I have to get the result from this regular expression; the regular expression is a string in a variable:

const dataFileUrlRegExpr = new RegExp(            
    "\\/home-affairs\\/document\\/download\\/([\\\\w-]{1,})_en?filename=visa_statistics_for_consulates_20[0-9]{2}.xlsx"
);
href = '/home-affairs/document/download/75ecba81-12db-42b0-a628-795d3292c680_en?filename=visa_statistics_for_consulates_2020.xlsx'

xlslHrefRegExpResult = dataFileUrlRegExpr.exec(xlslHref);

but the xlslHrefRegExpResult variable is null.

If I use:

const dataFileUrlRegExpr = new RegExp(
    /\/home-affairs\/document\/download\/([\w-]{1,})_en\?filename=visa_statistics_for_consulates_20[0-9]{2}.xlsx/g
);

without the string variable containing the expression, the result is achieved.
Where is the error using a string to build the regexp?

Answer 1

The correct code should be:

const dataFileUrlRegExpr = new RegExp(            
    "\\/home-affairs\\/document\\/download\\/([\\w-]{1,})_en\\?filename=visa_statistics_for_consulates_20[0-9]{2}.xlsx", 'g'
);
href = '/home-affairs/document/download/75ecba81-12db-42b0-a628-795d3292c680_en?filename=visa_statistics_for_consulates_2020.xlsx'

xlslHrefRegExpResult = dataFileUrlRegExpr.exec(href);
console.log(xlslHrefRegExpResult)

You had too many backslashes in [\\\\w-], and you were missing the backslashes before ?./

Answer 2

• Don't escape / at all, because it's nothing special inside a non-literal regex.
• Do escape the backslash in \w, but only once.
• Do escape the ? and the backslash itself.
• Do escape the . and the backslash itself.

dataFileUrlRegExpr = new RegExp(            
    "/home-affairs/document/download/([\\w-]{1,})_en\\?filename=visa_statistics_for_consulates_20[0-9]{2}\\.xlsx"
);

In short: write your regexp as normal, but double every backslash inside it so that it doesn't get interpreted as an escape character inside the string.