Java RegEx for parsing the quoted arguments

Question

Need a Java regex pattern for the following scenario:

Case 1:

Input string:

"a"

Matches:

Case 2:

Input string:

"a b"

Matches:

a b

Case 3:

Input string:

"aA Bb" cCc 123 4 5 6 7xy "\"z9" "\"z9$^"

Matches:

aA Bb
cCc
123
4
5
6
7xy
"z9
"z9$^

Case 4:

Input string:

"a b c

Matches:

None - since the quotes are unbalanced, hence pattern match fails.

Case 5:

Input string:

"a b" "c

Matches:

None - since the quotes are unbalanced, hence pattern match fails.

Case 6:

Input string:

"a b" p q r "x y z"

Matches:

a b
p 
q 
r
x y z

Case 7:

Input string:

"a b" p q r "x y \"z\""

Matches:

a b
p 
q
r
x y "z"

Case 8:

Input string:

"a b" p q r "x \"y \"z\""

Matches:

a b
p 
q 
r
x "y "z"

And of course, the simplest one:

Case 9:

Input string:

a b

Matches:

a
b

Tried using a pattern, but it doesn't seem to match all above cases.

public List<String> parseArgs(String argStr) {
    List<String> params = new ArrayList<String>();
    String pattern = "\\s*(\"[^\"]+\"|[^\\s\"]+)";
    Pattern quotedParamPattern = Pattern.compile(pattern);
    Matcher matcher = quotedParamPattern.matcher(argStr);
    while (matcher.find()) {
        String param = matcher.group();
            System.out.println(param);
            params.add(param);
    }
    return params;
}

public void test(String argStr) {
    String[] testStrings = new String[]{"a", "a b", "a b \"c\"", "a b \"c"};
    for(String s: testStrings){
        parseArgs(s);
    }
}

It can be solved but you need to show some effort of solving it. At least place all the example input strings in an array of String (String[]) and place that Java code here. — anubhava
– anubhava, Commented Apr 22, 2014 at 17:37
I'm guessing your actual strings will be more than just single lowercase letters. Do they need to contain capital letters, numbers, special characters, etc? Is there a limit to how long they should be? — CAustin
– CAustin, Commented Apr 22, 2014 at 18:01
@CAustin There may be capital letters, numbers, special characters and so on, but there's no limit as such. — Amith Koujalgi
– Amith Koujalgi, Commented Apr 22, 2014 at 18:03
I'm not sure what you want with this but I'm sure the answer is way before you arrive to this. — Pedro Lobito
– Pedro Lobito, Commented Apr 22, 2014 at 18:04

Amith Koujalgi · Accepted Answer · 2014-05-21 06:55:26Z

Have written a class "CLIParser" which will give you the result.

//instantiate the CLIParser 

CLIParser parser = new CLIParser("\"a b\" p q r \"x y z\"");

//call the method getTokens which gives you the result.

ArrayList<String> resultTokens = parser.getTokens();


###################CLI Parser Class definition#################################

class CLIParser {
    private String cmdString;

    public CLIParser(String cmdString) {
        this.cmdString = cmdString;
    }

    public ArrayList<String> getTokens() throws Exception {
        ArrayList<String> finalTokens = new ArrayList<String>();
        ArrayList<StringBuffer> tokens = new ArrayList<StringBuffer>();
    char inArray[] = this.cmdString.toCharArray();
    StringBuffer token = new StringBuffer();
    int valid = checkIfTheStringIsValid(inArray);
    if (valid == -1) {
        for (int i = 0; i <= inArray.length; i++) {

            if (i != inArray.length) {
                if ((inArray[i] != ' ') && (inArray[i] != '"')) {
                    token.append(inArray[i]);
                }

                if ((inArray[i] == '"') && (inArray[i - 1] != '\\')) {
                    i = i + 1;
                    while (checkIfLastQuote(inArray, i)) {
                        token.append(inArray[i]);
                        i++;
                    }
                }
            }
            if (i == inArray.length) {
                tokens.add(token);
                token = new StringBuffer();
            } else if (inArray[i] == ' ' && inArray[i] != '"') {
                tokens.add(token);
                token = new StringBuffer();
            }
        }
    } else {
        throw new InvalidCommandException(
                "Invalid command. Couldn't identify sequence at position "
                        + valid);
    }
    for(StringBuffer tok:tokens){
        finalTokens.add(tok.toString());
    }
    return finalTokens;
}

private static int checkIfTheStringIsValid(char[] inArray) {
    Stack myStack = new Stack<Character>();
    int pos = 0;
    for (int i = 0; i < inArray.length; i++) {
        if (inArray[i] == '"' && inArray[i - 1] != '\\') {
            pos = i;
            if (myStack.isEmpty())
                myStack.push(inArray[i]);
            else
                myStack.pop();
        }
    }
    if (myStack.isEmpty())
        return -1;
    else
        return pos;
}

private static boolean checkIfLastQuote(char inArray[], int i) {
    if (inArray[i] == '"') {
        if (inArray[i - 1] == '\\') {
            return true;
        } else
            return false;
    } else
        return true;
}
}

Amith Koujalgi · Accepted Answer · 2014-04-23 08:02:37Z

2

I don't know the straight way to solve with regex.

But you can replace the inner escape sequences with some unique keyword, then you can match your strings with regex.

String[] testStrings = new String[]{
         "a", "a b", "a b \"c\"", "a b \"c", "\"a b\" p q r \"x y z\""};
Pattern parsingPattern = Pattern.compile("(\".*?\")|( [^ ^\"]+)");
for(String s: testStrings) {
   s=s.replace("(?<!\\)\\"","@@@");
}
for(String s: testStrings) {
    List<String> params = null;
    int count = StringUtils.countMatches(s, "\"");
    if(count%2==0){
    params = new ArrayList<String>();
    Matcher matcher = parsePattern.matcher(s); 
    while (matcher.find())
        params.add( matcher.group(1) != null ? matcher.group(1) : matcher.group(2));
   }
}

Once you get the matches, you can replace your unique identifier with actual keyword..

I haven't tested the code snippet, but I hope you can do some minor tweaks to make it work.

edited Apr 23, 2014 at 8:02

Amith Koujalgi

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answered Apr 23, 2014 at 7:51

santhosh

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1 Comment

Amith Koujalgi Over a year ago

Seems like a fair solution. Thanks!

CAustin · Accepted Answer · 2014-04-22 18:07:47Z

0

Give this a try:

("\S+?(?: \S+?)*"|\S+?)

See it in action: http://regex101.com/r/fA5hN0

Just run a global match and return \1. Each capture group that gets returned should contain what you want.

answered Apr 22, 2014 at 18:07

CAustin

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1 Comment

Amith Koujalgi Over a year ago

Is there a Java regex string available cuz' the above pattern needs escaping slashes and quotes and even when escaped, it doesn't seem to work for all cases. Tried for this input string: "a b" p q r "x y \"z\""

anubhava · Accepted Answer · 2014-04-22 18:30:25Z

0

To get you started you can use this Java regex based code:

public List<String> parseArgs(String argStr, Pattern validPattern, Pattern parsePattern) {
    List<String> params = null;
    if (validPattern.matcher(argStr).matches()) {
        params = new ArrayList<String>();
        Matcher matcher = parsePattern.matcher(argStr); 
        while (matcher.find())
            params.add( matcher.group(1) != null ? matcher.group(1) : matcher.group(2));
    }
    return params;
}

public void parseIt() {
    Pattern validatePattern = Pattern.compile("^(?=(?:(?:[^\"]*\"){2})*[^\"]*$).*$");
    Pattern parsingPattern = Pattern.compile("\"([^\"]*)\"|(\\w+)");

    String[] testStrings = new String[]{
             "a", "a b", "a b \"c\"", "a b \"c", "\"a b\" p q r \"x y z\""};
    for(String s: testStrings) {
        List<String> parsedList = parseArgs(s, validatePattern, parsingPattern);
        System.out.printf("input: %-30s :: parsed: %s%n", s, parsedList);
    }
}

OUTPUT:

input: a                              :: parsed: [a]
input: a b                            :: parsed: [a, b]
input: a b "c"                        :: parsed: [a, b, c]
input: a b "c                         :: parsed: null
input: "a b" p q r "x y z"            :: parsed: [a b, p, q, r, x y z]

PS: Though I have noticed you latest edits where you have added nested quotes also, this answer needs to be enhanced for that.

answered Apr 22, 2014 at 18:30

anubhava

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8 Comments

Amith Koujalgi Over a year ago

This seems to work for most cases. But I'm trying to extract the strings in which there are escaped quotes as well. Tried with this: "aA Bb" cCc 123 4 5 6 7xy "\"z9" "\"z9$^" This is supposed to give me "z9 as 8th argument and "z9$^ as the 9th. Thanks.

anubhava Over a year ago

Yes I made a note of it since i didn't notice escaped/nested quotes in your question earlier.

Amith Koujalgi Over a year ago

What regular expression would match all the above cases?

anubhava Over a year ago

There is no end to escaping actually. Even backslash can be escaped. How are you getting this input?

Amith Koujalgi Over a year ago

Yes, that's correct. That's the way it's supposed to work for the app I'm building. Even this is valid "aA Bb" cCc 123 4 5 6 7xy "\"\\z9" "\"z9$^". And the strings are being read from stdin.

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