Java escape JSON String?

Question

I have the following JSON string that i am sending to a NodeJS server:

String string = "{\"id\":\"" + userID + "\",\"type\":\"" + methoden + "\",\"msg\":\"" + msget + "\", \"name\":\"" + namnet + "\", \"channel\":\"" + activeChatChannel + "\", \"visitorNick\":\"\", \"agentID\":\" " + agentID + "\"}";

PrintWriter pw = new PrintWriter(new OutputStreamWriter(os, "utf-8"));
pw.println(string);

The problem becomes when the string msget contains the character " and '

On the NodeJS server i am parsing the JSON like this:

var obj = JSON.parse(message);

Any ideas how i can manage to send all characters without problems?

stackoverflow.com/questions/34427279/…

Karthik
– Karthik

2017-06-19 16:08:01 +00:00
Commented Jun 19, 2017 at 16:08 — Karthik
– Karthik, Commented Jun 19, 2017 at 16:08

jabclab · Accepted Answer · 2013-09-19 15:32:01Z

31

I would use a library to create your JSON String for you. Some options are:

This will make dealing with escaping much easier. An example (using org.json) would be:

JSONObject obj = new JSONObject();

obj.put("id", userID);
obj.put("type", methoden);
obj.put("msg", msget);

// etc.

final String json = obj.toString(); // <-- JSON string

answered Sep 19, 2013 at 15:32

jabclab

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1 Comment

Eaton Emmerich Over a year ago

what if I don't want a builder and just the value string not the entire json?

davnicwil · Accepted Answer · 2020-03-06 11:01:52Z

27

Apache Commons

If you're already using Apache commons, it provides a static method for this:

StringEscapeUtils.escapeJson("some string")

It converts any string into one that's properly escaped for inclusion in JSON

See the documentation here

edited Mar 6, 2020 at 11:01

answered Mar 5, 2020 at 15:10

davnicwil

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2 Comments

Alfredo Morales Over a year ago

it is deprecated

LDK Sep 25 at 15:04

It only moved to org.apache.commons.text.StringEscapeUtils so not deprecated if you use this one

Ron DeSantis · Accepted Answer · 2018-05-25 06:33:14Z

25

The JSON specification at https://www.json.org/ is very simple by design. Escaping characters in JSON strings is not hard. This code works for me:

private String escape(String raw) {
    String escaped = raw;
    escaped = escaped.replace("\\", "\\\\");
    escaped = escaped.replace("\"", "\\\"");
    escaped = escaped.replace("\b", "\\b");
    escaped = escaped.replace("\f", "\\f");
    escaped = escaped.replace("\n", "\\n");
    escaped = escaped.replace("\r", "\\r");
    escaped = escaped.replace("\t", "\\t");
    // TODO: escape other non-printing characters using uXXXX notation
    return escaped;
}

answered May 25, 2018 at 6:33

Ron DeSantis

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Comments

NoobEditor · Accepted Answer · 2018-01-24 05:31:35Z

18

org.json.simple.JSONObject.escape() escapes quotes,, /, \r, \n, \b, \f, \t and other control characters.

import org.json.simple.JSONValue;
JSONValue.escape("test string");

Add pom.xml when using maven

<dependency>
    <groupId>com.googlecode.json-simple</groupId>
    <artifactId>json-simple</artifactId>
    <scope>test</scope>
</dependency>

edited Jan 24, 2018 at 5:31

NoobEditor

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answered Feb 7, 2015 at 19:24

Yeongjun Kim

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7 Comments

IgorGanapolsky Over a year ago

There is no such method escape.

Yeongjun Kim Over a year ago

doc

IgorGanapolsky Over a year ago

See Java JSONObject doc: docs.oracle.com/javaee/7/api/javax/json/JsonObject.html

Yeongjun Kim Over a year ago

Not java jsonObject. Look import in code. json-simple: code.google.com/archive/p/json-simple

Santosh Over a year ago

JSONObject.quote(String) does the job.

|

Dirk Lachowski · Accepted Answer · 2013-09-19 17:40:37Z

12

The best method would be using some JSON library, e.g. Jackson ( http://jackson.codehaus.org ).

But if this is not an option simply escape msget before adding it to your string:

The wrong way to do this is

String msgetEscaped = msget.replaceAll("\"", "\\\"");

Either use (as recommended in the comments)

String msgetEscaped = msget.replace("\"", "\\\"");

or

String msgetEscaped = msget.replaceAll("\"", "\\\\\"");

A sample with all three variants can be found here: http://ideone.com/Nt1XzO

edited Sep 19, 2013 at 17:40

answered Sep 19, 2013 at 15:29

Dirk Lachowski

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8 Comments

Sotirios Delimanolis Over a year ago

I didn't downvote, but the replaceAll() won't work. You need to use replace.

jlordo Over a year ago

Have you tried the code you posted? It has no effect. Either you use replace(...) or replaceAll("\"", "\\\\\""), because the backslash has a special meaning in the replacement string, too. See javadoc.

Dirk Lachowski Over a year ago

I'm biased if i should delete this answer because it's wrong or if i should let it as is 'cause it shows how not to do it?

jlordo Over a year ago

@DirkLachowski: If I were you, I'd either delete it (when I'm feeling lazy) or fix it.

Dirk Lachowski Over a year ago

@jlordo: I'm afraid you are wrong. The difference between replace() and replaceAll() is that in the former the first arg is a literal string and in the later it's a regex. The replacement is no regex (but can contain some special chars in the later). So replaceAll() will work as posted by me - and yes, i've tried it now.

|

Sotirios Delimanolis · Accepted Answer · 2024-06-17 13:34:17Z

6

According to the answer here, quotes in string content need to be escaped with a JSON string. You can do that by replacing quotes with \" in Java, like so

msget = msget.replace("\"", "\\\"");

If you're worried about other special characters, you can use a dedicated JSON parser to perform the transformation for you.

Jackson, for example, has JsonStringEncoder that accounts for all special characters.

If you're building the JSON manually, as in your sample code, you could do

String msget = "...";
JsonStringEncoder encoder = JsonStringEncoder.getInstance();
msget = new String(encoder.quoteAsUTF8(msget), StandardCharsets.UTF_8);

then plug the msget in to your output.

edited Jun 17, 2024 at 13:34

answered Sep 19, 2013 at 15:28

Sotirios Delimanolis

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Comments

rhengles · Accepted Answer · 2015-10-02 18:04:17Z

3

If you want to simply escape a string, not an object or array, use this:

String escaped = JSONObject.valueToString(" Quotes \" ' ' \" ");

http://www.json.org/javadoc/org/json/JSONObject.html#valueToString(java.lang.Object)

answered Oct 2, 2015 at 18:04

rhengles

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2 Comments

Renato Over a year ago

The link is broken, here is a web-archive version of it: web.archive.org/web/20151006021420/http://www.json.org/javadoc/…

Erik van Oosten Over a year ago

Escaping only double quote is not enough.

Sagar Jadhav · Accepted Answer · 2019-11-04 08:02:48Z

2

public static String ecapse(String jsString) {
    jsString = jsString.replace("\\", "\\\\");
    jsString = jsString.replace("\"", "\\\"");
    jsString = jsString.replace("\b", "\\b");
    jsString = jsString.replace("\f", "\\f");
    jsString = jsString.replace("\n", "\\n");
    jsString = jsString.replace("\r", "\\r");
    jsString = jsString.replace("\t", "\\t");
    jsString = jsString.replace("/", "\\/");
    return jsString;
}

answered Nov 4, 2019 at 8:02

Sagar Jadhav

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Tai Meng · Accepted Answer · 2021-06-28 21:31:59Z

2

In build.gradle (if using maven, make similar change) include this dependency:

implementation group: 'org.apache.avro', name: 'avro-tools', version: '1.10.2'

Then:

import net.minidev.json.JSONValue;
String escapedString = JSONValue.escape(yourString);

answered Jun 28, 2021 at 21:31

Tai Meng

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Comments

Matthew Bloch · Accepted Answer · 2022-10-10 14:26:21Z

2

If you can use slightly more modern Java and don't want another dependency, this does the job:

    static String jsonStringQuote(String string) {
        StringBuilder sb = new StringBuilder("\"");
        for (char c : string.toCharArray())
            sb.append(switch (c) {
                case '\\', '"', '/' -> "\\"+c;
                case '\b' -> "\\b";
                case '\t' -> "\\t";
                case '\n' -> "\\n";
                case '\f' -> "\\f";
                case '\r' -> "\\r";
                default -> c < ' ' ? String.format("\\u%04x", c) : c;
            });
        return sb.append('"').toString();
    }

answered Oct 10, 2022 at 14:26

Matthew Bloch

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Comments

FredG · Accepted Answer · 2023-06-08 16:10:22Z

2

On Android (and org.json), you could use JSONObject.quote.

import org.json.JSONObject;
String stringifiedString = JSONObject.quote(data);

answered Jun 8, 2023 at 16:10

FredG

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Comments

gil.fernandes · Accepted Answer · 2021-09-26 19:02:23Z

If you want to do it yourself without any library, you can use this piece of Java code:

/**
 * Escapes a string the JSON way.
 *
 * @param text The text to be escaped.
 * @return a string escaped according to the JSON format.
 */
public static String escape(final CharSequence text) {

    if (text == null) {
        return "";
    }
    final StringWriter writer = new StringWriter();
    for (int i = 0, length = text.length(); i < length; i++) {
        final char c = text.charAt(i);
        switch (c) {
            case '"':
                writer.write("\\\"");
                break;
            case '\\':
                writer.write("\\\\");
                break;
            default:
                if (c > 0x1f) {
                    writer.write(c);
                } else {
                    writer.write("\\u");
                    final String hex = "000" + Integer.toHexString(c);
                    writer.write(hex.substring(hex.length() - 4));
                }
        }
    }
    return writer.toString();
}

orip · Accepted Answer · 2020-11-03 15:12:50Z

0

Consider Moshi's JsonWriter class (source). It has a wonderful API and it reduces copying to a minimum, everything is nicely streamed to the OutputStream.

OutputStream os = ...;
JsonWriter json = new JsonWriter(Okio.sink(os));
json
  .beginObject()
  .name("id").value(userID)
  .name("type").value(methodn)
  ...
  .endObject();

edited Nov 3, 2020 at 15:12

answered Jul 21, 2015 at 14:05

orip

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bourne2program · Accepted Answer · 2025-07-01 17:58:13Z

0

Perhaps less efficient, but concise for the basic escapes without libraries using JDK 15+

  public static String escape(String s) {
    for (char c : "\\\"bfnrt".toCharArray())
      s = s.replace(("\\" + c).translateEscapes(), "\\" + c);
    return s;
  }

answered Jul 1 at 17:58

bourne2program

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