使用Java中的正则表达式从字符串中删除单词的所有独立出现

Question

Need advice on how to replace a sub-string like: @sometext , but not replace "@someothertext@somemail.com" sub-string.

For example, when I've got a string something like:

An example with @sometext and also with "@someothertext@somemail.com" sometextafter

And the result, after replacing sub-strings in string above should look like:

An example with and also with "@someothertext@somemail.com" sometextafter

After getting string from a field, I'm using:

String textMod = someText.replaceAll("( |^)[^\"]@[^@]+?( |$)","");
someText = textMod + "@\"" + someone.getEmail() + "\" ";

And then I'm setting this string into field.

Answer 1

You can do a regex on a standalone occurence this way

\b@sometext\b

Putting the \\b in front and in the back of the @sometext will make sure that it's a standalone word, not part of another word like @someothertext@sometext.com. Then if it's found the result will be put inside $match, now you can do whatever you want with $match

Hope this helps

From https://docs.oracle.com/javase/tutorial/essential/regex/bounds.html

The \\b in the pattern indicates a word boundary, so only the distinct * word "web" is matched, and not a word partial like "webbing" or "cobweb"

if (preg_match("/\bweb\b/i", "PHP is the web scripting language of choice."))      {
    echo "A match was found.";
    }

^ PHP example but you get the point

Answer 2

如果在标签之前和之后始终有一个要替换的空间，这可能就足够了。

/\s(@\w+)\s/g

Answer 3

这应符合您的需求：

str = str.replaceAll("@\w+[^@]", "");

Answer 4

Try this

(?<!\w)@[^@\s]+(?!\S)

See it here on Regexr

Match on a @ but only if there is no word character \\w before (?<!\\w) . Then match a sequence of characters that are not @ and not whitespace \\s but only if its not followed by a non whitespace \\S

(?<!\\w) is called a negative lookbehind assertion

[^@\\s] is called a negated character class , means match anything that is not part of the class

(?!\\S) is a negative lookahead assertion

Answer 5

Simply adding spaces before and after "@sometext" would not work if "@sometext" is at the start or end of a sentence. However, just adding a pattern checking for start or end of sentence would not work either, as when you match "@sometext " at the start of a sentence and leave a space " ", this will make the resulting string look strange. Same goes for the end of a sentence.

We need to split the regex replace in to two actions, and perform two seperate regex replaces:

str = str.replaceAll(" @sometext ", " ");
str = str.replaceAll("^@sometext | @sometext$|(?:@sometext ){2,}", "");

^ means start of line, $ means end of line.

EDIT: Added corner case handling of when several @sometext's are after each other.

Answer 6

(c#, regex based)

//match @xxx sequences, but only if i can look back and NOT see a @xxx immediately preceding me, and if I don't end with a @
string input = @"[An example with @hello and also with ""@@hello@somemail.com"" sometext @lastone";
 var pattern = @"(?<!@\w+)(?>@\w+)(?!@)";
 var matches = Regex.Matches(input, pattern);

Answer 7

myString = myString.replaceAll(" @hello ", " ");

If @hello is a single word, then it has spaces before and after, right? So you should find all @hello s with space before and after and replace it with a space.

If you need to remove not only @hello s and all words which are starting with @ and not containing other @ , use this:

myString = myString.replaceAll(" @[^@]+? ", " ");

[^@] is any symbol except @ . +? means match at least one character until reaching the first space.

If you want to remove words with only alphanumeric characters, use \\\\w instead of [^@]

EDIT:

Yeah, ohaal's right. To make it match at the start and the end of string use this pattern:

( |^)@[^@]+?( |$)

myString = myString.replaceAll("( |^)@hello( |$)", " ");