使用Java Regex匹配字母字符，且不带百分号

Question

tl;dr:

I want to take a string like: ab%cde%fg hij %klm n%op

And convert it to any of (all are acceptable):

• 'ab'%c'de'%f'g hij '%k'lm n'%o'p'
• 'ab'%c'de'%f'g' 'hij' %k'lm' 'n'%o'p'
• 'a''b'%c'd''e'%f'g' 'h''i''j' %k'l''m' 'n'%o'p'

(if an alphabetical character is not preceded by a % , it needs to be within single quotes. Opening and closing extra single quotes is acceptable)

Use Case

I'm trying to take a string in C strftime format and convert it to work with Java's SimpleDateFormat . For the most part, this is pretty straight forward:

String format = "%y-%m-%d %H:%M:%S";

Map<String, String> replacements = new HashMap<String, String>() {{
    put("%a", "EEE");
    put("%A", "EEEE");
    put("%b", "MMM");
    put("%B", "MMMM");
    put("%c", "EEE MMM dd HH:mm:ss yyyy");
    // ... for each strftime token, create a mapping ...
}};

for ( String key : replacements.keySet() )
{
    // apply the mappings one at a time
    format = format.replaceAll(key, replacements.get(key));
}

// Then format
SimpleDateFormat df = new SimpleDateFormat(format, Locale.getDefault());
System.out.println(df.format(Calendar.getInstance().getTime()));

However when I introduce character literals, it runs into issues. According to the strftime documentation, all character literal not preceded by a percent sign are passed along without modification to the output string. So:

Format: "%y is a great year!"
Output: "2019 is a great year!"

However with SimpleDateFormat , all character literals are treated as tokens unless surrounded by single quotes:

Format: "yyyy 'is a great year!'"
Output: "2019 is a great year!"

Format: "yyyy is a great year!"
Output: ERROR - invalid token "i"

Desired Output

Because strftime tokens are always a single character , it shouldn't be too difficult to fix our format string. In a worst case scenario, "if a letter is not preceded by a % sign, wrap it in single quotes", which would lead to:

Format: "%y is a great year!"
Processed: "%y 'i''s' 'a' 'g''r''e''a''t' 'y''e''a''r'!"

This is ugly, but would behave as expected and is an acceptable answer. Ideally we would wrap all runs of alphabetical characters not preceded by a % , like so:

Format: "%y is a great year!"
Processed: "%y 'is' 'a' 'great' 'year'!"

Or, better yet, all runs including non-alpha and non- % characters :

Format: "%y is a great year!"
Processed: "%y' is a great year!'"

What I've tried

I started with a mindless regular expression that I was pretty sure wouldn't work, and it didn't:

format.replaceAll("[^%]([a-zA-Z]+)", "'$1'");
// Format:   "Literal %t Literal"
// Output:   "'iteral' %t'Literal'"
// Expected: "'Literal' %t 'Literal'"

I don't have a firm grasp on back-references so I gave them a whirl but messed something up there as well:

format.replaceAll("(?!%)([a-zA-Z]+)", "'$1'");
// Format:   "Literal %t Literal"
// Output:   "'Literal' %'t' 'Literal'"
// Expected: "'Literal' %t 'Literal'"

I also considered writing a very simple lexer. Something like:

StringBuffer s = new StringBuffer();
boolean inQuote = false;
for (int i = 0; i < format.length; i++)
{
    if (format[i] == '%')
    {
        i++;
        s.append(replacements.get(format[i]);
    }
    else if (inQuote)
    {
        s.append(format[i]);
    }
    else
    {
        s.append("'");
        inQuote = true;
        s.append(format[i]);
    }
}

However I learned that format[i] isn't valid Java syntax, and didn't spend much time looking into how to properly get a character from a string before I decided to just post here.

I would prefer a regular expression solution so that I can write it in a single line instead of a loop like this.

Answer 1

This has been updated to work with a single regex. Additional formats can be added to test for correctness.

      String[] formats = { "ab%cde%fg hij %klm n%op", "ab%c", "%d"
      };
      for (String f : formats) {
         String parsed = f.replaceAll("(^[a-z]+|(?<=%[a-z])([a-z ]+))", "'$1'");
         System.out.println(parsed);
      }

The two possibilities are:

• Put all characters [az]+ that follow %[az] between single quotes.
• Place any characters that precede % and not included above between single quotes.

Answer 2

Why not use several replaceAll functions since you have already considered it.

First, add single quotes to all consecutive character strings;

Then, move the single quote preceded by % by one character;

Last, remove empty quotes.

Below is my testing code in Python. I believe it works in other languages such as Java as well.

>>> str1=re.sub("([a-zA-Z]+)","'\g<1>'",input)
>>> str2=re.sub("%'([a-zA-Z])'","%\g<1>",str1)
>>> str3=re.sub("''","",str2)
>>> str1
"'Literal' %'t' 'Literal'"
>>> str2
"'Literal' %t 'Literal'"
>>> str3
"'Literal' %t 'Literal'"