Question

我使用Jackson版本2.4.3将我的复杂Java对象转换为String对象，所以下面是我输出的内容。输出结果低于(Fyi - I just printed some part of output)

"{\"FirstName\":\"John \",\"LastName\":cena,\"salary\":7500,\"skills\":[\"java\",\"python\"]}";

这是我的代码（PaymentTnx是复杂的Java对象）

ObjectMapper mapper = new ObjectMapper();
mapper.setVisibility(PropertyAccessor.FIELD, Visibility.ANY);
String lpTransactionJSON = mapper.writeValueAsString(paymentTxn);

我不想在我的JSON字符串中看到\ slashesh。我需要做什么？如下所示：

"{"FirstName":"John ","LastName":cena,"salary":7500,"skills":["java","python"]}";

Answer 1

String test = "{\"FirstName\":\"John \",\"LastName\":cena,\"salary\":7500,\"skills\":[\"java\",\"python\"]}";
        System.out.println(StringEscapeUtils.unescapeJava(test));

这可能会对你有帮助。

Answer 2

我没有试过杰克逊。我只是有类似的情况。

我使用了org.apache.commons.text.StringEscapeUtils.unescapeJson，但它不适用于格式错误的JSON格式，例如{\“name \”：\“john \”}

所以，我用过这门课。完美的工作。

https://gist.githubusercontent.com/jjfiv/2ac5c081e088779f49aa/raw/8bda15d27c73047621a94359492a5a9433f497b2/JSONUtil.java

// 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();
  }
}

Answer 3

使用Jackson可以做到：

toString(paymentTxn);

使用

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);
    }
}

Answer 4

这里的JSON无效：

"{"FirstName":"John ","LastName":cena,"salary":7500,"skills":["java","python"]}";

这是有效的JSON，特别是单个字符串值：

"{\"FirstName\":\"John \",\"LastName\":cena,\"salary\":7500,\"skills\":[\"java\",\"python\"]}";

鉴于您正在调用writeValueAsString，这是正确的行为。我建议writeValue，或许？

如何使用Jackson将java对象转换为JSON字符串时使用UNESCAPE String

4 个答案: