如何从没有空格的字符串中提取某些特殊字符之间的所有单词？

Question

I have a string which is a result fetched from a website of parsing a tweet content, here is the string:

"1\\tI\\t_\\tPRP\\tPRP\\t_\\t2\\tnsubj\\t_\\t_\\n2\\tneed\\t_\\tVB\\tVBP\\t_\\t0\\tnull\\t_\\t_\\n3\\tmore\\t_\\tJJ\\tJJR\\t_\\t4\\tamod\\t_\\t_\\n4\\twords\\t_\\tNN\\tNNS\\t_\\t2\\tdobj\\t_\\t_\\n5\\tlike\\t_\\tIN\\tIN\\t_\\t4\\tprep\\t_\\t_\\n6\\tmarvel\\t_\\tNN\\tNN\\t_\\t5\\tpobj\\t_\\t_\\n7\\tor\\t_\\tCC\\tCC\\t_\\t6\\tcc\\t_\\t_\\n8\\tcat\\t_\\tNN\\tNN\\t_\\t6\\tconj\\t_\\t_\\n9\\tor\\t_\\tCC\\tCC\\t_\\t6\\tcc\\t_\\t_\\n10\\tpancake\\t_\\tNN\\tNN\\t_\\t6\\tconj\\t_\\t_\\n11\\tor\\t_\\tCC\\tCC\\t_\\t10\\tcc\\t_\\t_\\n12\\tfrance\\t_\\tNN\\tNN\\t_\\t10\\tconj\\t_\\t_", "text": "I need more words like marvel or cat or pancake or france"

I want to get all the words who are between "\\t" and "\\t_\\tNN", in other words I want the nouns, I wanted the output to be "words", "marvel", "cat", "pancake", "france".

I tried the code below:

private void regex(String s){
        if(s.indexOf("error") >= 1){
            Toast.makeText(this, "Sorry the site failed again it's not my fault :(",
                       Toast.LENGTH_SHORT).show();
        }
        else{
            Pattern pattern = Pattern.compile("\t(.*?)\t_\tNN");
            Matcher matcher = pattern.matcher(s);
            System.out.println(s);
            if (matcher.find()) {
                String result = matcher.group(1);
                System.out.println(result);
            }
        }

    }

I am sure I got the pattern.compile string wrong.. it's not working seems it can't find the words I wanted..

Could anybody tell me how should I fix it?

PS About the tab character lookalike "/t", I actually printed the whole website as result, but when I get the result as a string I guess they become just a backslash and a "t" instead of still being tab characters.

Answer 1

You can use the following:

"\\\\t([^\\\\]*?)\\\\t_\\\\tNN"

See Ideone Demo

See RegEx Demo