从Java中的JSON数组对象获取字符串值 [英] Getting the string value from a JSON array's object in Java
问题描述
编辑:其实,我找到了答案。因为我是新的我无法关闭的问题。我能够使用Array.getString(i)返回所需的字符串值。感谢所有帮助。
I actually found the answer. I can't close the question as I am new. I was able to use Array.getString(i) to return the string value needed. Thanks for all the help.
我有JSON是这样的:
I have JSON like this:
{
"List": [
"example1",
"example2",
"example3",
"example4"
]
}
和我试图获得这些对象的字符串值,而无需使用钥匙。我该怎么办呢?在的getString()
为JSONObject的要求的关键,我没有之一。
And I am trying to get the string value of those objects without using a key. How can I do that? The getString()
for jsonObject require a key and I don't have one.
推荐答案
我假设你有一个文件: /home/user/file_001.json
I assume that you have a file :/home/user/file_001.json
该文件包含此:`
{"age":34,"name":"myName","messages":["msg 1","msg 2","msg 3"]}
现在让我们写一个程序,读取该文件: /home/user/file_001.json
其内容转换为一个java 的JSONObject
。
Now let's write a program that reads the file : /home/user/file_001.json
and converts its content to a java JSONObject
.
package org.xml.json;
import java.io.FileNotFoundException;
import java.io.FileReader;
import java.io.IOException;
import java.util.Iterator;
import org.json.simple.JSONArray;
import org.json.simple.JSONObject;
import org.json.simple.parser.JSONParser;
import org.json.simple.parser.ParseException;
public class JsonSimpleRead
{
@SuppressWarnings("unchecked")
public static void main(String[] args)
{
JSONParser parser = new JSONParser();
try
{
Object obj = parser.parse(new FileReader("/home/user/file_001.json"));
JSONObject jsonObject = (JSONObject) obj;
String name = (String) jsonObject.get("name");
System.out.println(name);
long age = (Long) jsonObject.get("age");
System.out.println(age);
JSONArray msg = (JSONArray) jsonObject.get("messages");
Iterator<String> iterator = msg.iterator();
while (iterator.hasNext())
{
System.out.println(iterator.next());
}
}
catch (FileNotFoundException e)
{
e.printStackTrace();
}
catch (IOException e)
{
e.printStackTrace();
}
catch (ParseException e)
{
e.printStackTrace();
}
}
}
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