[英]How to extract a filename using regex from an absolute path
I am trying to extract just the filename from the following line: 我正在尝试从以下行中提取文件名:
"GET /mapitzend/public/components/font-awesome/css/font-awesome.css HTTP/1.1" 200 26651
In this case the expected result should be "font-awesome.css" I have gotten this far with my regex: 在这种情况下,预期的结果应该是“ font-awesome.css”。
\"(.[^\"]+)\s+HTTP
eith the regex above it returns: "GET /mapitzend/public/components/font-awesome/css/font-awesome.css HTTP 上面的正则表达式将返回:“ GET /mapitzend/public/components/font-awesome/css/font-awesome.css HTTP
What am I missing here.. 我在这里想念什么..
Try this regex: 试试这个正则表达式:
(?<=\\/)[^\\s\\/]*(?=\\s*HTTP)
Explanation: 说明:
(?<=\\/)
- positive lookbehind to find the position immediately preceded by a /
(?<=\\/)
-向后寻找正号,以找到紧跟/
的位置 [^\\s\\/]*
- matches 0+ occurrences of any character which is neither a whitespace character nor a /
[^\\s\\/]*
-匹配0+次出现的既不是空格字符也不是/
的任何字符 (?=\\s*HTTP)
- Positive lookahead to match until the position which is immediately followed by 0+ whitespaces followed by HTTP
. (?=\\s*HTTP)
-要匹配的正向超前,直到位置紧跟着0+空格和HTTP
。 While extracting the file name using regex
is possible, it's easier for start to extract the path using regex
and then get the filename using basename()
or pathinfo()
: 虽然可以使用regex
提取文件名,但是开始使用regex
提取路径然后使用basename()
或pathinfo()
获取文件名更容易:
$line = '"GET /mapitzend/public/components/font-awesome/css/font-awesome.css HTTP/1.1" 200 26651';
$m = array();
if (preg_match('/GET (.*) HTTP/', $line, $m)) {
$filename = basename($m[1]);
}
Using only regex
: 仅使用regex
:
if (preg_match('#GET /(.*/)*(.*) HTTP#', $line, $m)) {
$filename = $m[2];
}
Using properties of a filename which has certain extension like .css
, .html
you can use following regex. 使用扩展名为.css
, .html
的文件名的属性,可以使用以下正则表达式。
Regex: [^\\/]+?\\.[az]+
正则表达式: [^\\/]+?\\.[az]+
Explanation: 说明:
[^\\/]+?
matches a string whose preceding character was /
. 匹配前一个字符为/
的字符串。
\\.[az]+
matches the extension making sure it's a file. \\.[az]+
与扩展名匹配,以确保它是文件。
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