正则表达式匹配以排除开头的单词

Question

How can I make the following JavaScript

'id=d41c14fb-1d42&'.match(/(?![id=])(.*)[^&]/)[0]

return "d41c14fb-1d42"? It currently returns "41c14fb-1d42", without the beginning "d".

Answer 1

This should do it:

'id=d41c14fb-1d42&'.match(/^id=(.*)&$/)[1]
// "d41c14fb-1d42"

~ edit

When an original question does not give any context it is hard to guess what you need exactly, in which case I just give a solution that yields the requested output. I now understand you want to be able to parse a query string, ie, name1=value1&name2=value2... . The following regular expression yields nameX followed by an optional =valueX , as I believe it is valid to provide just the parameter in a query string without a value.

var parameters = "id=foobar&empty&p1=javascript&p2=so".match(/\w+(=([^&]+))?/g)
// ["id=foobar", "empty", "p1=javascript", "p2=so"]

You can split on "=" to obtain parameter and value separately.

var param1 = parameters[0].split("=")
// ["id", "foobar"]

For a parameter without the value that would just yield an Array with 1 value, of course.

"empty".split("=") // ["empty"]

Note this assumes a query parameter to match \\w+ , if there are other cases you have to expand the regular expression.

If you only want to match the id parameter specifically anywhere in the query string and obtain its value, as I infer from your comment below, you can use:

"id=donald".match(/(?:^|&)id=([^&]+)/)[1]
// "donald"
"param1=value1&id=donald&param2=value2".match(/(?:^|&)id=([^&]+)/)[1]
// "donald"