JSON overwrite using HashMap JAVA

Question

I understand there are many stackoverflow question and answer can help to fix this. My Question is not about fixing but writing properly. I need people view and I make better decision.

Here is JSON example :

"SearchResult": {
    "usersList": [
        {
            "name": "userId",
            "value": "1009"
        },
        {
            "name": "userName",
            "value": "{userName}"
        },
        {
            "name": "userUrl",
            "value": "{userUrl}"
        }
    ]
}

We have test that overwrite this JSON and send as resultList to application.

For overwrite we have put value to HashMap:

public Map<String, String> createUserList(String userName, String userUrl) {
    Map<String, String> putsValue = new HashMap<String, String>();
    //putsValue.put("{userName}", userName);
    putsValue.put("\\{userName\\}", userName);
    putsValue.put("\\{userUrl\\}", userUrl);
    return putsValue;
}

Question is When I used

putsValue.put("{userName}", userName);

When I try to parse JSON using following code :

String jsonTemplate = "fileName.json";
JSONTokener tokener = new JSONTokener(new FileReader(jsonTemplate));
JSONObject root = new JSONObject(tokener);
parsedJson = root.toString();

for (Map.Entry<String, String> putsValue: putsValue.entrySet()) {
    parsedJson = parsedJson
                    .replaceAll(putsValue.getKey(), putsValue.getValue());
}

I am getting this error

java.util.regex.PatternSyntaxException: Illegal repetition
{userName}
    at java.util.regex.Pattern.error(Unknown Source)
    at java.util.regex.Pattern.closure(Unknown Source)
    at java.util.regex.Pattern.sequence(Unknown Source)
    at java.util.regex.Pattern.expr(Unknown Source)
    at java.util.regex.Pattern.compile(Unknown Source)
    at java.util.regex.Pattern.<init>(Unknown Source)
    at java.util.regex.Pattern.compile(Unknown Source)
    at java.lang.String.replaceAll(Unknown Source)

How I can find tags like this:

putsValue.put("{userName}", userName);

And replace value I don't want to do this way

putsValue.put("\\{userName\\}", userName);

putsValue.put("{userName}", userName); has nothing to do with PatternSyntaxException. Not clear what you are asking about. — Andremoniy
– Andremoniy, Commented Aug 4, 2016 at 13:43

Nicolas Filotto · Accepted Answer · 2016-08-04 13:56:23Z

1

You can try Pattern.quote(String) as next:

parsedJson = parsedJson
                .replaceAll(Pattern.quote(putsValue.getKey()), putsValue.getValue());

edited Aug 4, 2016 at 13:56

answered Aug 4, 2016 at 13:52

Nicolas Filotto

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2 Comments

N.. Over a year ago

I have more than 20 value. Is there way I can minimize this efforts because I don't think so if I use Pattern.quote for all put my code review pass.

Nicolas Filotto Over a year ago

I updated my answer, you can do it once when you call replaceAll instead

Sergei Rybalkin · Accepted Answer · 2016-08-04 14:07:00Z

0

No need in custom json mapper in your case. Use Jackson. You need just to provide getters/setters and annotate fields.

import com.fasterxml.jackson.databind.ObjectMapper;
import com.fasterxml.jackson.annotation.JsonInclude;

Deserialize json:

ObjectMapper mapper = new ObjectMapper();
Clazz obj = mapper.readValue(jsonString, Clazz.class);

Serialize to json:

   class Clazz {
       @JsonInclude(JsonInclude.Include.NON_NULL)
       ... field declaration;
   }
   String json = mapper.writeValueAsString(obj);

edited Aug 4, 2016 at 14:07

answered Aug 4, 2016 at 14:01

Sergei Rybalkin

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Comments

Joop Eggen · Accepted Answer · 2016-08-04 15:51:03Z

0

You do not need the regular expression replaceAll; just use replace.

parsedJson = parsedJson.replace(putsValue.getKey(), putsValue.getValue());

edited Aug 4, 2016 at 15:51

answered Aug 4, 2016 at 14:01

Joop Eggen

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1 Comment

N.. Over a year ago

I tried doesn't work. Thank you so much. I have huge respect for you.

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