How to extract only part of string in PHP?

Question

I have a following string and I want to extract image123.jpg.

..here_can_be_any_length \"and_here_any_length/image123.jpg\" and_here_also_any_length

Answer 1

$string = '..here_can_be_any_length "and_here_any_length/image123.jpg" and_here_also_any_length';
$data = explode('"',$string);
$basename = basename($data[1]);

Answer 2

What about something like this :

$str = '..here_can_be_any_length "and_here_any_length/image123.jpg" and_here_also_any_length';
$m = array();
if (preg_match('#".*?/([^\.]+\.(jpg|jpeg|gif|png))"#', $str, $m)) {
    var_dump($m[1]);
}

Which, here, will give you :

string(12) "image123.jpg" 

I suppose the pattern could be a bit simpler -- you could not check the extension, for instance, and accept any kind of file ; but not sure it would suit your needs.

Basically, here, the pattern :

• starts with a "
• takes any number of characters until a / : .*?/
• then takes any number of characters that are not a . : [^\.]+
• then checks for a dot : \.
• then comes the extension -- one of those you decided to allow : (jpg|jpeg|gif|png)
• and, finally, the end of pattern, another "

And the whole portion of the pattern that corresponds to the filename is surrounded by (), so it's captured -- returned in $m

Answer 3

You can use:

if(preg_match('#".*?\/(.*?)"#',$str,$matches)) {
   $filename = $matches[1];
}

Alternatively you can extract the entire path between the double quotes using preg_match and then extract the filename from the path using the function basename:

if(preg_match('#"(.*?)"#',$str,$matches)) {
    $path = $matches[1]; // extract the entire path.
    $filename =  basename ($path); // extract file name from path.
}