javascript - Regex matching to exclude beginning word -

how can make following javascript

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

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

this should it:

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

~ edit

when original question not give context hard guess need exactly, in case give solution yields requested output. understand want able parse query string, i.e., name1=value1&name2=value2.... following regular expression yields namex followed optional =valuex, believe valid provide parameter in query string without value.

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

you can split on "=" obtain parameter , value separately.

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

for parameter without value yield array 1 value, of course.

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

note assumes query parameter match \w+, if there other cases have expand regular expression.

if want match id parameter anywhere in query string , obtain value, infer comment below, can use:

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