Escape JSON string in Java

Question

I'm using Google's com.google.api.client.json.GenericJson and com.fasterxml.jackson.core.JsonGenerator. I would like to serialize JSON object and escape quotes and backslashes so that I can pass that string in Bash. And afterwards deserialize that string.

GenericJson.toString produces simple JSON, but \n etc. are not escaped:

{commands=ls -laF\ndu -h, id=0, timeout=0}

is there a simple way how to get something like this:

"{commands=\"ls -laF\\ndu -h\", id=0, timeout=0}"

I don't want to reinvent the wheel, so I'd like to use Jackson or an existing API, if possible.

You might be looking for [this](http://stackoverflow.com/a/16652683/1066946) — Philipp Gayret, Feb 05 '14 at 12:12
I was trying to avoid writing it from scratch, there's also Apache Commons method to escape JavaScript: http://commons.apache.org/proper/commons-lang/javadocs/api-2.6/org/apache/commons/lang/StringEscapeUtils.html#escapeJavaScript(java.lang.String) — Tombart, Feb 05 '14 at 12:22
Stick the serialized JSON into an array and run it through the serializer a second time, then strip off the `[]` characters. — Hot Licks, Feb 05 '14 at 13:33

Barett · Accepted Answer · 2018-07-02T19:54:30.300

36

No additional dependencies needed: You're looking for JsonStringEncoder#quoteAsString(String).

Click for JsonStringEncoder javadoc

Example:

import com.fasterxml.jackson.core.io.JsonStringEncoder;

JsonStringEncoder e = JsonStringEncoder.getInstance();
String commands = "ls -laF\\ndu -h";
String encCommands = new String(e.quoteAsString(commands));
String o = "{commands: \"" + encCommands + "\", id: 0, timeout: 0}"

Ref: http://fasterxml.github.io/jackson-core/javadoc/2.1.0/com/fasterxml/jackson/core/io/JsonStringEncoder.html

edited Jul 02 '18 at 19:54

answered Nov 04 '16 at 19:52

Barett

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I think you mean "quoteAsString" instead of "quoteAsJson" in that first code block? – Igor Jul 02 '18 at 18:58
Well, the example he gave was a Json block. But yeah, one of those. – Barett Jul 02 '18 at 19:14

score 4 · Answer 2 · answered Feb 06 '14 at 10:33

Using Gson for serialization proved to be quite easy and bulletproof. Afterwards Apache's commons-lang3 = 3.1 escapeEcmaScript is used. In 3.2 there's also escapeJson method.

import com.google.api.client.json.GenericJson;
import com.google.api.client.util.Key;
import com.google.gson.Gson;
import org.apache.commons.lang3.StringEscapeUtils;

public class MyJson extends GenericJson {

    @Key("commands")
    public String commands;

    public String serialize() throws IOException {
      Gson gson = new Gson();
      String g = gson.toJson(this);
      return StringEscapeUtils.escapeEcmaScript(g);
    }
}

This produces escaped JSON:

{\"commands\":\"ls -laF\\ndu -h\"}

Deserialization is then quite simple:

protected MyJson deserialize(String str) throws IOException {
    String json = StringEscapeUtils.unescapeEcmaScript(str);
    JsonObjectParser parser = (new JacksonFactory()).createJsonObjectParser();
    return parser.parseAndClose(new StringReader(json), MyJson.class);
}

The escapeEcmaScript method isn't complicated, it does following replacement:

  {"'", "\\'"},
  {"\"", "\\\""},
  {"\\", "\\\\"},
  {"/", "\\/"}

But at least is something I don't have to care about.

escapeJson is accessed from StringEscapeUtils() and this is deprecated now . — sathya, Sep 20 '19 at 14:50

Escape JSON string in Java

2 Answers2

Example:

Linked