preg_replace表示特定模式（如果不存在）

Question

I have the url inputs in various formats. I want to convert all into one standard format. Few example inputs are below.

• http://www.example.com
• https://in.example.com
• http://uk.example.com
• uk.example.com

From above all the inputs, I wish to convert into first one (http://www.example.com) with www. So I think we can replace the string if "www" not there.

$pattern = '/((http?|https)\:\/\/)?([a-zA-Z]+)\.example.com\/[a-z]{2}\/[a-zA-Z0-9]{5,30}/i';

I tried this pattern. I know it will match replace everything. SO please suggest me, how to do that.

Answer 1

function to_standard_format($url) {
    $pattern = '#(https?://)?\w+\.example.com(?=[/\s]|$)#i';
    return preg_replace($pattern, 'http://www.example.com', $url);
}

var_dump(to_standard_format('http://www.example.com'));
var_dump(to_standard_format('https://in.example.com'));
var_dump(to_standard_format('http://uk.example.com'));
var_dump(to_standard_format('http://uk.example.com/blah'));
var_dump(to_standard_format('uk.example.com'));
var_dump(to_standard_format('http://www.example.com.au'));
var_dump(to_standard_format('uk.another-example.com'));

prints

string(22) "http://www.example.com"
string(22) "http://www.example.com"
string(22) "http://www.example.com"
string(27) "http://www.example.com/blah"
string(22) "http://www.example.com"
string(25) "http://www.example.com.au"
string(22) "uk.another-example.com"