清除字符串中的反斜杠并将结果放入数据结构中

huangapple
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英文:

Clean string from backslashs and put the results into data strucutre

问题

我有这个字符串

  1. "[{\"name\":\"john\",\"family\":[],\"status\":\"single\"},{\"name\":\"david\",\"family\":[],\"status\":\"marred\"}]"

如何清除这个字符串中的反斜杠并将实体放入数据结构中？

英文:

I have this string

  1. "[{\"name\":\"john\",\"family\":[],\"status\":\"single\"},{\"name\":\"david\",\"family\":[],\"status\":\"marred\"}]"

how can I clean up the backslashs from this string and put the entities into a data structure?

答案1

得分: 1

  1. import java.util.ArrayList;
  2. import java.util.Arrays;
  3. import java.util.List;
  5. import org.json.JSONArray;
  6. import org.json.JSONException;
  8. import com.fasterxml.jackson.core.JsonProcessingException;
  9. import com.fasterxml.jackson.core.type.TypeReference;
  10. import com.fasterxml.jackson.databind.JsonMappingException;
  11. import com.fasterxml.jackson.databind.ObjectMapper;
  13. public class SOTest {
  15.     public static void main(String[] args) throws JsonMappingException, JsonProcessingException, JSONException {
  16.         ObjectMapper mapper = new ObjectMapper();
  17.         String jsonString = "[{\"name\":\"john\",\"family\":[],\"status\":\"single\"},{\"name\":\"david\",\"family\":[],\"status\":\"marred\"}]";
  18.         List<CustomClass> datas = mapper.readValue(jsonString, new TypeReference<List<CustomClass>>() {});
  19.         System.out.println(datas);
  20.     }
  21. }
  23. class CustomClass {
  24.     public String name;
  25.     public List<CustomClass> family;
  26.     public String status;
  28.     @Override
  29.     public String toString() {
  30.         return "Data [name=" + name + ", family=" + family + ", status=" + status + "]";
  31.     }
  32. }

output

  1. [Data [name=john, family=[], status=single], Data [name=david, family=[], status=marred]]
英文: 
  1. import java.util.ArrayList;
  2. import java.util.Arrays;
  3. import java.util.List;
  5. import org.json.JSONArray;
  6. import org.json.JSONException;
  8. import com.fasterxml.jackson.core.JsonProcessingException;
  9. import com.fasterxml.jackson.core.type.TypeReference;
  10. import com.fasterxml.jackson.databind.JsonMappingException;
  11. import com.fasterxml.jackson.databind.ObjectMapper;
  13. public class SOTest {
  15. 	public static void main(String[] args) throws JsonMappingException, JsonProcessingException, JSONException {
  16. 		ObjectMapper mapper = new ObjectMapper();
  17. 		String jsonString = &quot;[{\&quot;name\&quot;:\&quot;john\&quot;,\&quot;family\&quot;:[],\&quot;status\&quot;:\&quot;single\&quot;},{\&quot;name\&quot;:\&quot;david\&quot;,\&quot;family\&quot;:[],\&quot;status\&quot;:\&quot;marred\&quot;}]&quot;;
  18. 		List&lt;CustomClass&gt; datas = mapper.readValue(jsonString, new TypeReference&lt;List&lt;CustomClass&gt;&gt;() {});
  19. 		System.out.println(datas);
  20. 	}
  21. }
  23. class CustomClass {
  24. 	public String name;
  25. 	public List&lt;CustomClass&gt; family;
  26. 	public String status;
  28. 	@Override
  29. 	public String toString() {
  30. 		return &quot;Data [name=&quot; + name + &quot;, family=&quot; + family + &quot;, status=&quot; + status + &quot;]&quot;;
  31. 	}
  32. }

output

  1. [Data [name=john, family=[], status=single], Data [name=david, family=[], status=marred]]

答案2

得分: 0

以下是翻译好的内容：

这个字符串看起来没问题。转义是为了在 Java 中使用这个字符串。

针对你的任务，可以使用像 Jackson 这样的库。

英文:

The string looks okay. The escaping is done to use the string within java.

Use a library like Jackson. For your task.

答案3

得分: -1

  1. 在jshell控制台中编写，因此根据需要添加类等。
  3. import com.fasterxml.jackson.annotation.JsonCreator;
  4. import com.fasterxml.jackson.annotation.JsonProperty;
  5. import com.fasterxml.jackson.databind.ObjectMapper;
  7. import java.io.IOException;
  9. try {
  10.     String json = "[{\"name\":\"john\",\"family\":[],\"status\":\"single\"},{\"name\":\"david\",\"family\":[],\"status\":\"marred\"}]";
  11.     json = json.replace("\\\"", "\"");
  12.     System.out.println(json);
  14.     ObjectMapper mapper = new ObjectMapper();
  15.     Person[] persons = mapper.readValue(json, Person[].class);
  16.     System.out.println(persons[0].name);
  17. } catch (IOException e) {
  18.     e.printStackTrace();
  19. }
  21. static class Person {
  23.     public final String name;
  24.     public final String[] family;
  25.     public final String status;
  27.     @JsonCreator
  28.     public Person(
  29.             @JsonProperty("name") String name,
  30.             @JsonProperty("family") String[] family,
  31.             @JsonProperty("status") String status
  32.     ) {
  33.         this.name = name;
  34.         this.family = family;
  35.         this.status = status;
  36.     }
  37. }
英文:

Written in jshell console so you need to add classes etc as needed

  1. import com.fasterxml.jackson.annotation.JsonCreator;
  2. import com.fasterxml.jackson.annotation.JsonProperty;
  3. import com.fasterxml.jackson.databind.ObjectMapper;
  5. import java.io.IOException;
  7. try {
  8.     String json = &quot;[{\&quot;name\&quot;:\&quot;john\&quot;,\&quot;family\&quot;:[],\&quot;status\&quot;:\&quot;single\&quot;},{\&quot;name\&quot;:\&quot;david\&quot;,\&quot;family\&quot;:[],\&quot;status\&quot;:\&quot;marred\&quot;}]&quot;;
  9.     json = json.replace(&quot;\\\&quot;&quot;, &quot;\&quot;&quot;);
  10.     System.out.println(json);
  12.     ObjectMapper mapper = new ObjectMapper();
  13.     Person[] persons = mapper.readValue(json, Person[].class);
  14.     System.out.println(persons[0].name);
  15. } catch (IOException e) {
  16.     e.printStackTrace();
  17. }
  19. static class Person {
  21.     public final String name;
  22.     public final String[] family;
  23.     public final String status;
  25.     @JsonCreator
  26.     public Person(
  27.             @JsonProperty(&quot;name&quot;) String name,
  28.             @JsonProperty(&quot;family&quot;) String[] family,
  29.             @JsonProperty(&quot;status&quot;) String status
  30.     ) {
  31.         this.name = name;
  32.         this.family = family;
  33.         this.status = status;
  34.     }
  35. }

huangapple
  • 本文由 发表于 2020年10月9日 15:18:22
  • 转载请务必保留本文链接：https://go.coder-hub.com/64275552.html
  • data-structures
  • java
  • string
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