How to find if a Java String contains X or Y and contains Z

Question

I'm pretty sure regular expressions are the way to go, but my head hurts whenever I try to work out the specific regular expression.

What regular expression do I need to find if a Java String (contains the text "ERROR" or the text "WARNING") AND (contains the text "parsing"), where all matches are case-insensitive?

EDIT: I've presented a specific case, but my problem is more general. There may be other clauses, but they all involve matching a specific word, ignoring case. There may be 1, 2, 3 or more clauses.

Answer 1

If you're not 100% comfortable with regular expressions, don't try to use them for something like this. Just do this instead:

string s = test_string.toLowerCase();
if (s.contains("parsing") && (s.contains("error") || s.contains("warning")) {
    ....

because when you come back to your code in six months time you'll understand it at a glance.

Edit: Here's a regular expression to do it:

(?i)(?=.*parsing)(.*(error|warning).*)

but it's rather inefficient. For cases where you have an OR condition, a hybrid approach where you search for several simple regular expressions and combine the results programmatically with Java is usually best, both in terms of readability and efficiency.

Answer 2

If you really want to use regular expressions, you can use the positive lookahead operator:

(?i)(?=.*?(?:ERROR|WARNING))(?=.*?parsing).*

Examples:

Pattern p = Pattern.compile("(?=.*?(?:ERROR|WARNING))(?=.*?parsing).*", Pattern.CASE_INSENSITIVE); // you can also use (?i) at the beginning
System.out.println(p.matcher("WARNING at line X doing parsing of Y").matches()); // true
System.out.println(p.matcher("An error at line X doing parsing of Y").matches()); // true
System.out.println(p.matcher("ERROR Hello parsing world").matches()); // true       
System.out.println(p.matcher("A problem at line X doing parsing of Y").matches()); // false

Answer 3

With multiple .* constucts the parser will invoke thousands of "back off and retry" trial matches.

Never use .* at the beginning or in the middle of a RegEx pattern.

Answer 4

尝试：

 If((str.indexOf("WARNING") > -1 || str.indexOf("ERROR") > -1) && str.indexOf("parsin") > -1)

Answer 5

Regular Expressions are not needed here. Try this:

if((string1.toUpperCase().indexOf("ERROR",0) >= 0 ||  
  string1.toUpperCase().indexOf("WARNING",0) >= 0 ) &&
  string1.toUpperCase().indexOf("PARSING",0) >= 0 )

This also takes care of the case-insensitive criteria

Answer 6

I usually use this applet to experiment with reg. ex. The expression may look like this:

if (str.matches("(?i)^.*?(WARNING|ERROR).*?parsing.*$")) {
...

But as stated in above answers it's better to not use reg. ex. here.

Answer 7

我认为这个正则表达式可以解决问题（但必须有更好的方法）：

(.*(ERROR|WARNING).*parsing)|(.*parsing.*(ERROR|WARNING))

Answer 8

If you've a variable number of words that you want to match I would do something like that:

String mystring = "Text I want to match";
String[] matchings = {"warning", "error", "parse", ....}
int matches = 0;
for (int i = 0; i < matchings.length(); i++) {
  if (mystring.contains(matchings[i]) {
    matches++;
  }
}

if (matches == matchings.length) {
   System.out.println("All Matches found");
} else {
   System.out.println("Some word is not matching :(");
}

Note: I haven't compiled this code, so could contain typos.