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
image12开发者_开发百科3 can be any length (newimage123456 etc) and with extension of jpg, jpeg, gif or png.
I assume I need to use preg_match, but I am not really sure and like to know how to code it or if there are any other ways or function I can use.
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.
}
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
$string = '..here_can_be_any_length "and_here_any_length/image123.jpg" and_here_also_any_length';
$data = explode('"',$string);
$basename = basename($data[1]);
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