html - RegEx says all my inputs are false for a float number (JavaScript) -

this question has answer here:

• why regex constructors need double escaped? 4 answers

my objective check if number has been input form via html meets following...

• is number between 0.01 , 100.0
• is float number e.g. of format 00.01

my regex - "\d{1,3}\.\d{1,2}" - when check on regexr appears correct... http://regexr.com/3eme3

however code returning false... going wrong?

(i've checked both non parsed float number , parsed float number)

function checkagegrade(){     var agegradevalue = document.getelementsbyname("agegrade")[0].value;     console.log("age grade input: " + agegradevalue);     var correctdigits = new regexp("\d{1,3}\.\d{1,2}");     console.log("correct digits: " + correctdigits);      if(!hasvalue(agegradevalue)){         document.submitrunnertime.agegrade.value = "-1";     }     console.log("has value");      if(!isnan(agegradevalue)){         console.log("passed isnan");          var parsedagegradevalue = parsefloat(agegradevalue);         console.log(parsedagegradevalue);          if(parsedagegradevalue > 0.00 && parsedagegradevalue < 100.00){             console.log("range passed, value is: " + parsedagegradevalue + ", checking regex");             console.log("parsed number: " + correctdigits.test(parsedagegradevalue));             console.log("not parsed number: " + correctdigits.test(agegradevalue));             return correctdigits.test(parsedagegradevalue);         }     }     console.log("false function");     return false; } 

the function using check see if input has value or not is...

function hasvalue(avalue) {     var whitespace = new regexp("\s+");     if (avalue.length >= 1) {         if (whitespace.test(avalue)) {             return false;         } else {             return true;         }         return false;     } 

because you're creating regular expression string (which, in case, not necessary), need double-escape \d markers. otherwise, backslash ignored:

var correctdigits = new regexp("\\d{1,3}\\.\\d{1,2}"); 

you can this:

var correctdigits = /\d{1,3}\.\d{1,2}/; 

when use native regular expression syntax, don't have double-escape when expression in string.