Read out elements of a string in specific order

Question

For some reasons I have to use a specific string in my project. This is the text file (it's a JSON File):

{"algorithm": 
[
    { "key": "onGapLeft", "value" : "moveLeft" },
    { "key": "onGapFront", "value" : "moveForward" },
    { "key": "onGapRight", "value" : "moveRight" },
    { "key": "default", "value" : "moveBackward" }
]
}

I've defined it in JAVA like this:

static String input = "{\"algorithm\": \n"+
"[ \n" +
    "{ \"key\": \"onGapLeft\", \"value\" : \"moveLeft\" }, \n" +
    "{ \"key\": \"onGapFront\", \"value\" : \"moveForward\" }, \n" +
    "{ \"key\": \"onGapRight\", \"value\" : \"moveRight\" }, \n" +
    "{ \"key\": \"default\", \"value\" : \"moveBackward\" } \n" +
"] \n" +
"}";

Now I have to isolate the keys and values in an array:

key[0] = onGapLeft; value[0] = moveLeft;
key[1] = onGapFront; value[1] = moveForward;
key[2] = onGapRight; value[2] = moveRight;
key[3] = default; value[3] = moveBackward;

I'm new to JAVA and don't understand the string class very well. Is there an easy way to get to that result? You would help me really!

Thanks!

UPDATE: I didn't explained it well enough, sorry. This program will run on a LEGO NXT Robot. JSON won't work there as I want it to so I have to interpret this JSON File as a normal STRING! Hope that explains what I want :)

Welcome to Stackoverflow. If your input String is a JSON String then you should look at http://stackoverflow.com/questions/1395551/convert-a-json-string-to-object-in-java — Arnaud Denoyelle, Sep 09 '13 at 16:52
Hey, thanks! JSON is unfortunatily no option because it's not running on my Lego NXT Robot. SO I have to interpret this JSON File as a string. I'm sorry I didn't explain that well enough! :) — Jonas Zimmer, Sep 09 '13 at 17:10

Arnaud Denoyelle · Accepted Answer · 2013-09-09T17:54:23.433

I propose a solution in several step.

1) Let's get the different parts of your ~JSON String. We will use a pattern to get the different {.*} parts :

public static void main(String[] args) throws Exception{
  List<String> lines = new ArrayList<String>();

  Pattern p = Pattern.compile("\\{.*\\}");
  Matcher matcher = p.matcher(input);
  while (matcher.find()) {
    lines.add(matcher.group());
  }
}

(you should take a look at Pattern and Matcher)

Now, lines contains 4 String :

{ "key": "onGapLeft", "value" : "moveLeft" }
{ "key": "onGapFront", "value" : "moveForward" }
{ "key": "onGapRight", "value" : "moveRight" }
{ "key": "default", "value" : "moveBackward" }

Given a String like one of those, you can remove curly brackets with a call to String#replaceAll();

List<String> cleanLines = new ArrayList<String>();
for(String line : lines) {
  //replace curly brackets with... nothing.
  //added a call to trim() in order to remove whitespace characters.
  cleanLines.add(line.replaceAll("[{}]","").trim());
}

(You should take a look at String String#replaceAll(String regex))

Now, cleanLines contains :

"key": "onGapLeft", "value" : "moveLeft"
"key": "onGapFront", "value" : "moveForward"
"key": "onGapRight", "value" : "moveRight"
"key": "default", "value" : "moveBackward"

2) Let's parse one of those lines :

Given a line like :

"key": "onGapLeft", "value" : "moveLeft"

You can split it on , character using String#split(). It will give you a String[] containing 2 elements :

//parts[0] = "key": "onGapLeft"
//parts[1] = "value" : "moveLeft"
String[] parts = line.split(",");

(You should take a look at String[] String#split(String regex))

Let's clean those parts (remove "") and assign them to some variables:

String keyStr = parts[0].replaceAll("\"","").trim(); //Now, key = key: onGapLeft
String valueStr = parts[1].replaceAll("\"","").trim();//Now, value = value : moveLeft

//Then, you split `key: onGapLeft` with character `:`
String key = keyStr.split(":")[1].trim();

//And the same for `value : moveLeft` : 
String value = valueStr.split(":")[1].trim();

That's it !

You should also take a look at Oracle's tutorial on regular expressions (This one is really important and you should invest time on it).

Thank you so much! This was my last attempt to get this program to work properly and you rescued me! :) (sadly I can't upvote this solution because I need 15 rep) — Jonas Zimmer, Sep 09 '13 at 17:55
Just one more question! :) String[] parts = line.split(","); line is in your case a local String inside the for argument. Should that be cleanLines.Split(",")? — Jonas Zimmer, Sep 09 '13 at 18:53
No :) cleanLines is a `String[]`. `String#split()` is only applicable to a `String`. You need to iterate on the `String[]` to work with lines one by one. — Arnaud Denoyelle, Sep 09 '13 at 19:01
Oh my god.. I'm going crazy now. I don't want to bother you any longer but the robot does "not support regular expressions". I didn't know that, I learned it the hard way right now.. Is there any easy solution to modify your solution so it didn't use regular expressions? — Jonas Zimmer, Sep 09 '13 at 20:37
Then you have 2 solutions : 1) You parse manually : use String.toCharArray() to convert your String to a char[] and interate on it. You should also be able to reimplement the split method without regex. 2) Some people developped some minimal JSON parser that you can include or copy-paste in your project : see http://eclipsesource.com/blogs/2013/04/18/minimal-json-parser-for-java/ — Arnaud Denoyelle, Sep 09 '13 at 21:41
Thanks! I've implemented the split method now with NXT conform string operations. It's working now! You've been a great help! :) — Jonas Zimmer, Sep 10 '13 at 00:19

score 0 · Answer 2 · answered Sep 09 '13 at 16:56

You need to use a JSON parser library here. For example, with org.json you could parse it as

String input = "{\"algorithm\": \n"+
        "[ \n" +
            "{ \"key\": \"onGapLeft\", \"value\" : \"moveLeft\" }, \n" +
            "{ \"key\": \"onGapFront\", \"value\" : \"moveForward\" }, \n" +
            "{ \"key\": \"onGapRight\", \"value\" : \"moveRight\" }, \n" +
            "{ \"key\": \"default\", \"value\" : \"moveBackward\" } \n" +
        "] \n" +
        "}";

JSONObject root = new JSONObject(input);
JSONArray map = root.getJSONArray("algorithm");

for (int i = 0; i < map.length(); i++) {
    JSONObject entry = map.getJSONObject(i);
    System.out.println(entry.getString("key") + ": "
                          + entry.getString("value"));
}

Output :

onGapLeft: moveLeft
onGapFront: moveForward
onGapRight: moveRight
default: moveBackward

Hey, thanks! JSON is unfortunatily no option because it's not running on my Lego NXT Robot. SO I have to interpret this JSON File as a string. I'm sorry I didn't explain that well enough! :) — Jonas Zimmer, Sep 09 '13 at 17:09

Read out elements of a string in specific order

2 Answers2