正则表达式，用于删除字符串“ foo bar”之后的所有内容（包括多行）

Question

Need to delete everything after two words that are found together. Tried looking around for similar questions

str.replaceAll("foo bar(.*)", "")

Only seems to delete the first line.

When i add $ to the end of the regex ( foo bar.*$ ) it doesn't seem to pick anything at all? Any ideas why?

Thanks

Answer 1

Use Single Line Mode matching using flag ?s .

str = str.replaceAll("(?s)foo bar(.*)", "");

If using regex is not mandatory, you can simply use the following:

str = str.indexOf("foo bar") == -1 ? str : str.substring(0, str.indexOf("foo bar"));

Answer 2

If it's as simple as finding 2 words, you could use indexOf to find them, then use substring to trim the String to drop everything after the two words.

String keyWord = "key word";
String input = "some\ntext\nkey word is here\nand more text\nto follow";
int pos = input.indexOf(keyWord);
if (pos != -1) {
    String trimmed = input.substring(0, pos + keyWord.length());
    System.err.println("'" + trimmed + "'");
}

The code above assumes you want to keep the key word; if you don't want to keep it, just drop the + keyword.length() in the substring .

If you do want to use a regular expression, pass the Pattern.DOTALL flag to have the dot match line endings. You also need to change the regex to capture what you want to keep, not what you want to discard (as your example has it):

String keyWordRegex = "key\\s+word";
Pattern pattern = Pattern.compile("(.*\\b" + keyWordRegex + "\\b).*", Pattern.DOTALL);
Matcher matcher = pattern.matcher(input);
String replaced = matcher.replaceAll("$1");
System.err.println("'" + replaced + "'");

The regex works like this:

• first, opening brace to start capturing
• capture anything (dot) zero to infinite times (star)
• then a word boundary, so that it does only match "key word" but not "masterkey word"
• then your key word regex, in this example I made it to be "key", one or more (plus) spaces (\\s), and "word"
• word boundary again
• close the capture group
• allow any number (star) of characters (dot)

Answer 3

you can try below solution. It's delete all character after the foo bar is found.

String a = "hell fdshf fdfjk foo bar ffsd fsajkh vsfdsv" ;
String as = a.split("foo bar")[0];
System.out.println(as);

Output

hell fdshf fdfjk

Answer 4

There is no need to use regex. You can use simple string manipulation concept.

String str = "testing test foo bar  test  foo bar test \n testing all  \n foo bar(.*) tester";

        String boundaryValueText = "foo bar";
        int boundaryValueIndex = str.indexOf(boundaryValueText);
        String parsedContent = boundaryValueIndex != -1 ? new StringBuilder(str)
                .delete(boundaryValueIndex + boundaryValueText.length(), str.length()).toString() : str;