Php preg_match使用URL作为正则表达式

Question

i have an array of urls

[
  'http://www.example.com/eng-gb/products/test-1',
  'http://www.example.com/eng-gb/products/test-3',
  'http://www.example.com/eng-gb/about-us',
]

I need to write a regex for filter only the ones end with:

http://www.example.com/eng-gb/products/(.*)

in this case i need to exclude 'about-us'.

I need also use 'http://www.example.com/eng-gb/products/(.*)' as regex.

The best way for archive?

Answer 1

preg_grep() provides a shorter line of code, but because the substring to be matched doesn't appear to have any variable characters in it, best practice would indicate strpos() is better suited.

Code: ( Demo )

$urls=[
  'http://www.example.com/eng-gb/products/test-1',
  'http://www.example.com/eng-gb/badproducts/test-2',
  'http://www.example.com/eng-gb/products/test-3',
  'http://www.example.com/eng-gb/badproducts/products/test-4',
  'http://www.example.com/products/test-5',
  'http://www.example.com/eng-gb/about-us',
];

var_export(preg_grep('~^http://www.example\.com/eng-gb/products/[^/]*$~',$urls));
echo "\n\n";
var_export(array_filter($urls,function($v){return strpos($v,'http://www.example.com/eng-gb/products/')===0;}));

Output:

array (
  0 => 'http://www.example.com/eng-gb/products/test-1',
  2 => 'http://www.example.com/eng-gb/products/test-3',
)

array (
  0 => 'http://www.example.com/eng-gb/products/test-1',
  2 => 'http://www.example.com/eng-gb/products/test-3',
)

Some notes:

Using preg_grep() :

• Use a non-slash pattern delimiter so that you don't have to escape all of the slashes inside the pattern.
• Escape the dot at .com .
• Write the full domain and directory path with start and end anchors for tightest validation.
• Use a negated character class near the end of the pattern to ensure that no additional directories are added (unless of course you wish to include all subdirectories).
• My pattern will match a url that ends with /products/ but not /products . This is in accordance with the details in your question.

Using strpos() :

• Checking for strpos()===0 means that the substring must be found at the start of the string.
• This will allow any trailing characters at the end of the string.

Answer 2

I think you need use preg_grep cause you have array of urls and this will return array of url that match your condition

$matches = preg_grep('/products\\/.*$/', $urls);

and also you can use validate filters in php to validate urls

Answer 3

You'll need to escape the forward slashes and periods to get http:\\/\\/www\\.example\\.com\\/eng-gb\\/products\\/(.*) . After that, you could just place the URL in directly.

Alternatively (better) would be to search for \\/eng-gb\\/products\\/(.*) .

Example:

$matches = array();
preg_match('/\/eng-gb\/products\/(.*)/', $your_url, $matches);
$product = $matches[1];