How to grab the contents of a tag that can have any characters?

Question

Java

You need to capture the file content line without capturing the end of the line and closing the
tags. So far decided with (.*), and then delete the last line. But this is not an option in terms of
extensibility . Instead of (.*), you need to compose a regular expression that ends after the ,
which comes before . It must be taken into account that there may be a line in the
content of the file itself.
//

String regex="<remote\s+host="localhost"\s+port="8080">\s+<(\w+)\s+\w+="([a-zA-Z0-9]+:[a-zA-Z\\\/]*)">(.*)"
// (.*) нужно вместо нее что-то, что остановит ее.
String test="<remote host="localhost" port="8080"> <file src="C:\www\rock">file content </file></remote>"

<b>Результат на http://baco.su/articles/regulyarnye-vyrajeniya/</b>
key = 0], array (4): 
[key = 0] =>" <remote host="localhost" port="8080"> <file src="C:\www\rock">file content </file></remote>"
[key = 1] => file
[key = 2] => C:\www\rock
[key = 3] =>file content </file></remote>  /*</file></remote>  лишнее, надо остановиться. Слишком жадно*/

PS Without escaping, because toster eats everything in <> tags, so I just wrapped it in code

Answer 1

[^<]*
PS All this is govnokod. It is more correct to use ru2.php.net/manual/ru/book.simplexml.php or a similar parser.