CODEWITHSUNDEEP
javascriptregex
Multitut
Multitut

Reputation: 2169

JavaScript escape stars on regular expression

I am trying to get a serial number from a zigbee packet (i.e get from 702442500 *13*32*702442500#9).

So far, I've tried this:

test = "*#*0##*13*32*702442500#9##";
test.match("\*#\*0##\*13\*32\*(.*)#9##");

And this: test.match("*#*0##*13*32*(.*)#9##");

With no luck. How do I get a valid regular expression that does what I want?

Upvotes: 0

Views: 130

Answers (3)

royhowie
royhowie

Reputation: 11171

The easiest way to grab that portion of the string would be to use

var regex = /(\*\d{3,}#)/g,
    test = "*13*32*702442500#9";

var match = test.match(regex).slice(1,-1);

This captures a * followed by 3 or more \d (numbers) until it reaches an octothorpe. Using the global (/g) modifier will cause it to return an array of matches.

For example, if

var test = "*13*32*702442500#9
            *#*0##*13*32*702442500#9##";

then, test.match(regex) will return ["*702442500#", "*702442500#"]. You can then slice the elements of this array:

var results = [],
    test = "... above ... ",
    regex = /(\*\d{3,}#)/g,
    matches = test.match(regex);

matches.forEach(function (d) {
    results.push(d.slice(1,-1));
})
// results : `["702442500", "702442500"]`

Upvotes: 1

Casimir et Hippolyte
Casimir et Hippolyte

Reputation: 89584

If you want to extract the big number, you can use:

/\*#\*0##\*13\*32\*([^#]+)#9##/

Note that I use delimiters / that are needed to write a pattern in Javascript (without the regexp object syntax). When you use this syntax, (double)? quotes are not needed. I use [^#]+ instead of .* because it is more clear and more efficent for the regex engine.

Upvotes: 1

Avinash Raj
Avinash Raj

Reputation: 174796

The below regex matches the number which has atleast three digits,

/([0-9][0-9][0-9]+)/

DEMO

Upvotes: 1

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