替换匹配的正则表达式的 substring

Question

I fetch some html and do some string manipulation and en up with a string like

string sample = "\n    \n   2 \n      \n  \ndl. \n \n    \n flour\n\n     \n 4   \n    \n cups of    \n\nsugar\n"

I would like to find all ingredient lines and remove whitespaces and linebreaks

2 dl. flour and 4 cups of sugar

My approach so far is to the following.

Pattern p = Pattern.compile("[\\d]+[\\s\\w\\.]+");
Matcher m = p.matcher(Result);

while(m.find()) {
  // This is where i need help to remove those pesky whitespaces
}

Answer 1

sample = sample.replaceAll("[\\n ]+", " ").trim();

Output:

2 dl. flour 4 cups of sugar

With no spaces in the beginning, and no spaces at the end.

It first replaces all spaces and newlines with a single space, and then trims of the extra space from the begging / end.

Answer 2

Following code should work for you:

String sample = "\n    \n   2 \n      \n  \ndl. \n \n    \n flour\n\n     \n 4   \n    \n cups of    \n\nsugar\n";
Pattern p = Pattern.compile("(\\s+)");
Matcher m = p.matcher(sample);
sb = new StringBuffer();
while(m.find())
    m.appendReplacement(sb, " ");
m.appendTail(sb);
System.out.println("Final: [" + sb.toString().trim() + ']');

OUTPUT

Final: [2 dl. flour 4 cups of sugar]

Answer 3

I think something like this will work for you:

String test = "\n    \n   2 \n      \n  \ndl. \n \n    \n flour\n\n     \n 4   \n    \n cups of    \n\nsugar\n";

/* convert all sequences of whitespace into a single space, and trim the ends */
test = test.replaceAll("\\s+", " ");

Answer 4

I assumed that the \n are not actual line feed, but it also works with linefeeds . This should work fine:

test=test.replaceAll ("(?:\\s|\\\n)+"," ");

In case there is no textual \n it can be simpler:

test=test.replaceAll ("\\s+"," ");

An you need to trim the leading/trailing spaces.

I use the RegexBuddy tool to check any single regex, very handy in so many languages.

Answer 5

You should be able to use the standard String.replaceAll(String, String) . The first parameter will take your pattern, the second will take an empty string.

Answer 6

s/^\s+//s
s/\s+$//s
s/(\s+)/ /s

Run those three substitutions (replacing leading whitespace with nothing, replace trailing whitespace with nothing, replace multiple whitespace with a space.