Reputation:
I am using Jackson version 2.4.3 for converting my complex Java object into a String object, so below is what I'm getting in output. The output is like below (Fyi - I just printed some part of the output)
"{\"FirstName\":\"John \",\"LastName\":cena,\"salary\":7500,\"skills\":[\"java\",\"python\"]}";
Here is my code (PaymentTnx is a complex Java object)
ObjectMapper mapper = new ObjectMapper();
mapper.setVisibility(PropertyAccessor.FIELD, Visibility.ANY);
String lpTransactionJSON = mapper.writeValueAsString(paymentTxn);
I don't want to see \ slashes in my JSON string. What do I need to do to get a string like below:
"{"FirstName":"John ","LastName":cena,"salary":7500,"skills":["java","python"]}";
Upvotes: 12
Views: 50516
Reputation: 9498
With Jackson do:
toString(paymentTxn);
with
public String toString(Object obj) {
try (StringWriter w = new StringWriter();) {
new ObjectMapper().configure(SerializationFeature.INDENT_OUTPUT, true).writeValue(w, obj);
return w.toString();
} catch (Exception e) {
throw new RuntimeException(e);
}
}
Upvotes: 0
Reputation: 791
I have not tried Jackson. I just have similar situation.
I used org.apache.commons.text.StringEscapeUtils.unescapeJson but it's not working for malformed JSON format like {\"name\": \"john\"}
So, I used this class. Perfectly working fine.
// BSD License (http://lemurproject.org/galago-license)
package org.lemurproject.galago.utility.json;
public class JSONUtil {
public static String escape(String input) {
StringBuilder output = new StringBuilder();
for(int i=0; i<input.length(); i++) {
char ch = input.charAt(i);
int chx = (int) ch;
// let's not put any nulls in our strings
assert(chx != 0);
if(ch == '\n') {
output.append("\\n");
} else if(ch == '\t') {
output.append("\\t");
} else if(ch == '\r') {
output.append("\\r");
} else if(ch == '\\') {
output.append("\\\\");
} else if(ch == '"') {
output.append("\\\"");
} else if(ch == '\b') {
output.append("\\b");
} else if(ch == '\f') {
output.append("\\f");
} else if(chx >= 0x10000) {
assert false : "Java stores as u16, so it should never give us a character that's bigger than 2 bytes. It literally can't.";
} else if(chx > 127) {
output.append(String.format("\\u%04x", chx));
} else {
output.append(ch);
}
}
return output.toString();
}
public static String unescape(String input) {
StringBuilder builder = new StringBuilder();
int i = 0;
while (i < input.length()) {
char delimiter = input.charAt(i); i++; // consume letter or backslash
if(delimiter == '\\' && i < input.length()) {
// consume first after backslash
char ch = input.charAt(i); i++;
if(ch == '\\' || ch == '/' || ch == '"' || ch == '\'') {
builder.append(ch);
}
else if(ch == 'n') builder.append('\n');
else if(ch == 'r') builder.append('\r');
else if(ch == 't') builder.append('\t');
else if(ch == 'b') builder.append('\b');
else if(ch == 'f') builder.append('\f');
else if(ch == 'u') {
StringBuilder hex = new StringBuilder();
// expect 4 digits
if (i+4 > input.length()) {
throw new RuntimeException("Not enough unicode digits! ");
}
for (char x : input.substring(i, i + 4).toCharArray()) {
if(!Character.isLetterOrDigit(x)) {
throw new RuntimeException("Bad character in unicode escape.");
}
hex.append(Character.toLowerCase(x));
}
i+=4; // consume those four digits.
int code = Integer.parseInt(hex.toString(), 16);
builder.append((char) code);
} else {
throw new RuntimeException("Illegal escape sequence: \\"+ch);
}
} else { // it's not a backslash, or it's the last character.
builder.append(delimiter);
}
}
return builder.toString();
}
}
Upvotes: 7
Reputation: 567
String test = "{\"FirstName\":\"John \",\"LastName\":cena,\"salary\":7500,\"skills\":[\"java\",\"python\"]}";
System.out.println(StringEscapeUtils.unescapeJava(test));
This might help you.
Upvotes: 27
Reputation: 16142
This here is not valid JSON:
"{"FirstName":"John ","LastName":cena,"salary":7500,"skills":["java","python"]}";
This here is valid JSON, specifically a single string value:
"{\"FirstName\":\"John \",\"LastName\":cena,\"salary\":7500,\"skills\":[\"java\",\"python\"]}";
Given that you're calling writeValueAsString
, this is the correct behaviour. I would suggest writeValue
, perhaps?
Upvotes: -5