如何让jackson使用单引号或无引号生成json字符串?

时间:2021-09-28 18:02:56

For example, I want to generate a json string for ng-style:

例如,我想为ng样式生成一个json字符串:

<th ng-style="{width:247}" data-field="code">Code</th>

But with jackson, the result is:

但对杰克逊来说,结果是:

<th ng-style="{&quot;width&quot;:247}" data-field="code">Code</th>

It's not easy to read.

读起来不容易。

So I want jackson to generate the json string with single quote or no quotes. Is it possible to do this?

因此,我希望jackson用单引号或无引号生成json字符串。有可能这样做吗?

2 个解决方案

#1


28  

If you have control over the ObjectMapper instance, then configure it to handle and generate JSON the way you want:

如果您对ObjectMapper实例有控制权,那么将其配置为按您希望的方式处理和生成JSON:

final ObjectMapper mapper = new ObjectMapper();
mapper.configure(JsonGenerator.Feature.QUOTE_FIELD_NAMES, false);
mapper.configure(JsonParser.Feature.ALLOW_UNQUOTED_FIELD_NAMES, true);

#2


-4  

The simplest and the best option is to use regular expression and update the string value.

最简单和最好的选择是使用正则表达式并更新字符串值。

The sample code is as listed below.

示例代码如下所示。

partNumberList=partNumberList.replaceAll(":", ":\"").replaceAll("}", "\"}");

partNumberList = partNumberList。replaceAll(":",":\“”)。replaceAll(“}”,“\”}”);

The complete code is as shown below

完整的代码如下所示。

public static void main(String[] args) throws JsonParseException, JsonMappingException, 
    IOException {
    TestJack obj = new TestJack();
    //var jsonString ='{"it":"Stati Uniti d'America"}';
    //            jsonString =jsonString.replace("'", "\\\\u0027")
    ObjectMapper mapper = new ObjectMapper();
    String partNumberList = "[{productId:AS101R}, {productId:09902007}, {productId:09902002}, {productId:09902005}]";
    partNumberList = partNumberList.replaceAll(":", ":\"").replaceAll("}", "\"}");
    System.out.println(partNumberList);
    mapper.configure(com.fasterxml.jackson.core.JsonParser.Feature.ALLOW_UNQUOTED_FIELD_NAMES, true);
    List<ProductDto> jsonToPersonList = null;
    jsonToPersonList = mapper.readValue(partNumberList, new TypeReference<List<ProductDto>>() {
    });
    System.out.println(jsonToPersonList);
}

#1


28  

If you have control over the ObjectMapper instance, then configure it to handle and generate JSON the way you want:

如果您对ObjectMapper实例有控制权,那么将其配置为按您希望的方式处理和生成JSON:

final ObjectMapper mapper = new ObjectMapper();
mapper.configure(JsonGenerator.Feature.QUOTE_FIELD_NAMES, false);
mapper.configure(JsonParser.Feature.ALLOW_UNQUOTED_FIELD_NAMES, true);

#2


-4  

The simplest and the best option is to use regular expression and update the string value.

最简单和最好的选择是使用正则表达式并更新字符串值。

The sample code is as listed below.

示例代码如下所示。

partNumberList=partNumberList.replaceAll(":", ":\"").replaceAll("}", "\"}");

partNumberList = partNumberList。replaceAll(":",":\“”)。replaceAll(“}”,“\”}”);

The complete code is as shown below

完整的代码如下所示。

public static void main(String[] args) throws JsonParseException, JsonMappingException, 
    IOException {
    TestJack obj = new TestJack();
    //var jsonString ='{"it":"Stati Uniti d'America"}';
    //            jsonString =jsonString.replace("'", "\\\\u0027")
    ObjectMapper mapper = new ObjectMapper();
    String partNumberList = "[{productId:AS101R}, {productId:09902007}, {productId:09902002}, {productId:09902005}]";
    partNumberList = partNumberList.replaceAll(":", ":\"").replaceAll("}", "\"}");
    System.out.println(partNumberList);
    mapper.configure(com.fasterxml.jackson.core.JsonParser.Feature.ALLOW_UNQUOTED_FIELD_NAMES, true);
    List<ProductDto> jsonToPersonList = null;
    jsonToPersonList = mapper.readValue(partNumberList, new TypeReference<List<ProductDto>>() {
    });
    System.out.println(jsonToPersonList);
}