Javascript Regex - 9个字符长，以'SO-'开头，以6个数字结尾

Question

Regular expressions are simply evil in my mind and no matter how many times I read any documentation I just cannot seem to grasp even the simplest of expressions!

I am trying to write what must be a very simple expression to query a variable in javascript but I just cannot get it to work properly.

I am trying to validate the following:-

The string must be 9 characters long, starting with SO- (case insensitive eg So-, so-, sO- and SO-) followed by 6 numbers.

So the following should all match

SO-123456, So-123456, sO-456789, so-789123

but the following should fail

SO-12d456, SO-1234567

etc etc

I have only managed to get this far so far

var _reg = /(SO-)\d{6}/i;

var _tests = new Array();
_tests[0] = "So-123456";
_tests[1] = "SO-123456";
_tests[2] = "sO-456789";
_tests[3] = "so-789123";
_tests[4] = "QR-123456";
_tests[5] = "SO-1234567";
_tests[6] = "SO-45k789";

for(var i = 0; i < _tests.length; i++){
  var _matches = _tests[i].match(_reg);
  if(_matches && _matches.length > 0)
     $('#matches').append(i+'. '+_matches[0] + '<br/>');
}

Please see http://jsfiddle.net/TzHKd/ for above example

Test number 5 is matching although it should fail as there are 7 numbers and not 6.

Any assistance would be greatly appreciated.

Cheers

Answer 1

use this regexp instead

/^(so-)\d{6}$/i;

without ^ (string starting with) or $ (string ending with) you're looking for a generic substring match (that's the reason why when you have 7 digits your regexp return true).

Answer 2

By using the anchors ^ and $ (matching beginining of line and end of line respectively), you can make the regex match the whole line. Otherwise, the match with return true as soon as the characters in the regex are matched.

So, you will apply it like this:

var _reg = /^(so-)\d{6}$/i;