匹配URL中的google searchterms的PHP正则表达式问题

Question

im finding searchwords from google request urls. im using

preg_match("/[q=](.*?)[&]/", $requesturl, $match);

but it fails when the 'q' parameter is the last parameter of the string.

so i need to fetch everything that comes after 'q=', but the match must stop IF it finds '&'

how to do that?

EDIT: I eventually landed on this for matching google request url: /[?&]q=([^&]+)/ Because sometimes they have a param that ends with q. like 'aq=0'

Answer 1

You need /q=([^&]+)/ . The trick is to match everything except & in the query.

To build on your query, this is a slightly modified version that will (almost) do the trick, and it's the closest to what you have there: /q=(.*?)(&|$)/ . It puts the q= out of the brackets, because inside the brackets it will match either of them, not both together, and at the end you need to match either & or the end of the string ( $ ). There are, though, a few problems with this:

1. sometimes you will have an extra & at the end of the match; you don't need it. To solve this problem you can use a lookahead query: (?=&|$)
2. it introduces an extra group at the end (not necessarily bad, but can be avoided) -- actually, this is fixed by 1.

So, if you want a slightly longer query to expand what you have there, here it is: /q=(.*?)(?=&|$)/

Answer 2

Try this:

preg_match("/q=([^&]+)/", $requesturl, $match);

A little explaining:

• [q=] will search for either q or = , but not one after another.
• [&] is not needed as there is only one character. & is fine.
• the ? operator in regex tells it to match 0 or 1 occurrences of the ** preceding** character.
• [^&] will tell it to match any character except for & . Which means you'll get all the query string until it hits &.