Regex for removing comma in a String when it is enclosed by quotes

Question

Hi,

I need to remove commas within a String only when enclosed by quotes.

example:

String a = "123, \"Anders, Jr.\", John, [email protected],A"

after replacement should be

String a = "123, Anders Jr., John, [email protected],A"

Can you please give me sample java code to do this?

Thanks much,

Lina

Answer 1

Should work:

s/(?<="[^"]*),(?=[^"]*")//g
s/"//g

Answer 2

It also seems you need to remove the quotes, judging by your example.

You can't do that in a single regexp. You would need to match over each instance of

"[^"]*"

then strip the surrounding quotes and replace the commas. Are there any other characters which are troublesome? Can quote characters be escaped inside quotes, eg. as ‘""’?

It looks like you are trying to parse CSV. If so, regex is insufficient for the task and you should look at one of the many free Java CSV parsers.

Answer 3

This looks like a line from a CSV file, parsing it through any reasonable CSV library would automatically deal with this issue for you. At least by reading the quoted value into a single 'field'.

Answer 4

Probably grossly inefficiënt but it seems to work.

import java.util.regex.*;

StringBuffer ResultString = new StringBuffer();

try {
    Pattern regex = Pattern.compile("(.*)\"(.*),(.*)\"(.*)", Pattern.CASE_INSENSITIVE | Pattern.UNICODE_CASE);
    Matcher regexMatcher = regex.matcher(a);
    while (regexMatcher.find()) {
     try {
      // You can vary the replacement text for each match on-the-fly
      regexMatcher.appendReplacement(ResultString, "$1$2$3$4");
     } catch (IllegalStateException ex) {
      // appendReplacement() called without a prior successful call to find()
     } catch (IllegalArgumentException ex) {
      // Syntax error in the replacement text (unescaped $ signs?)
     } catch (IndexOutOfBoundsException ex) {
      // Non-existent backreference used the replacement text
     } 
    }
    regexMatcher.appendTail(ResultString);
} catch (PatternSyntaxException ex) {
    // Syntax error in the regular expression
}

Answer 5

There are two major problems with the accepted answer. First, the regex "(.*)\"(.*),(.*)\"(.*)" will match the whole string if it matches anything, so it will remove at most one comma and two quotation marks.

Second, there's nothing to ensure that the comma and quotes will all be part of the same field; given the input ("foo", "bar") it will return ("foo "bar). It also doesn't account for newlines or escaped quotation marks, both of which are permitted in quoted fields.

You can use regexes to parse CSV data, but it's much trickier than most people expect. But why bother fighting with it when, as bobince pointed out, there are several free CSV libraries out there for the downloading?

Answer 6

This works fine. '<' instead of '>'

boolean deleteCommas = false;
for(int i=0; i < text.length(); i++){
    if(text.charAt(i)=='\''){
        text = text.substring(0, i) + text.substring(i+1, text.length());
        deleteCommas = !deleteCommas;
    }
    if(text.charAt(i)==','&&deleteCommas){
        text = text.substring(0, i) + text.substring(i+1, text.length());
    }
}

Answer 7

A simpler approach would be replacing the matches of this regular expression:

("[^",]+),([^"]+")

By this:

$1$2

Answer 8

The following perl works for most cases:

open(DATA,'in/my.csv');
while(<DATA>){
  if(/(,\s*|^)"[^"]*,[^"]*"(\s*,|$)/){
    print "Before: $_";
    while(/(,\s*|^)"[^"]*,[^"]*"(\s*,|$)/){
      s/((?:^|,\s*)"[^"]*),([^"]*"(?:\s*,|$))/$1 $2/
    }
    print "After: $_";
  }
}

It's looking for:

• (comma plus optional spaces) or start of line
• a quote
• 0 or more non-quotes
• a comma
• 0 or more non-quotes
• (optional spaces plus comma) or end of line

If found, it will then keep replacing the comma with a space until it can find no more examples.

It works because of an assumption that the opening quote will be preceded by a comma plus optional spaces (or will be at the start of the line), and the closing quote will be followed by optional spaces plus a comma, or will be the end of the line.

I'm sure there are cases where it will fail - if anyone can post 'em, I'd be keen to see them...