Why does't regex match all numbers instead of numbers just at the end of the string?

Question

I was just looking for a regex that would watch the last numerical (\\d or [0-9]) in a given string , strings like:

var str = "7-Dec-1985"
var str = "#scrollto-section-4"

Of-course I found an answer in the following thread on SO HERE

I am using a regex like the following:

str.match(/\d+$/)

Works fine, no issues, now I used the following tool to analysis the regex HERE ,

\d+ //matches  greedy 0 to as many 
$ - specifies that the search should start at the end of the string

But why does that above regex in the below example:

var str = "7-Dec-1985"

Match only 1985 why not 71985 ?

Answer 1

Because $ means "end of input" (or "end of line or end of input" if you specify the m flag), and \\d+ means a contiguous series of digits (not digits mixed with other things). So \\d+$ means "a contiguous series of digits right before the end."

If you want to match anywhere, remove the $ . Additionally, if you want to match more than once, you'll need a g ("global") flag on it.

Examples -- your original:

 var str = "7-Dec-1985"; document.body.innerHTML = JSON.stringify(str.match(/\\d+$/)); 

Without the $ , but no g :

 var str = "7-Dec-1985"; document.body.innerHTML = JSON.stringify(str.match(/\\d+/)); 

Without the $ and with g :

 var str = "7-Dec-1985"; document.body.innerHTML = JSON.stringify(str.match(/\\d+/g)); 

Answer 2

Sorry, but $ doesn't means start search at end of string.

Your regex \\d+$ means match the number at end of string .

To match any number use \\d+ like this .

Answer 3

Because there is -Dec- between 7 and 1985 which isn't digit. Also $ means end of line. So Your pattern just matches that number which is end of string (continuously) .