我从服务器收到以下 JSON 响应...
[["1","1"],["2","1"],["3","1"],["4","1"],["5","1"],["6","1"],["7","1"],["8","1"],["9","2"],["10","3"],["11","3"],["12","3"],["13","3"],["14","3"],["15","3"],["16","3"],["17","3"],["18","3"],["19","3"]]
它是 JSON 格式,我将它作为 String[]
获取,类似这样......
String response = Response.getValue();
System.out.println(response) = [["1","1"],["2","1"],["3","1"],["4","1"],["5","1"],["6","1"],["7","1"],["8","1"],["9","2"],["10","3"],["11","3"],["12","3"],["13","3"],["14","3"],["15","3"],["16","3"],["17","3"],["18","3"],["19","3"]]
但是,响应是一个由 2 个值组成的矩阵 [USU_ID, DEPARTMENT]
,我需要在 String[][]
中使用它。我怎样才能做到这一点?我尝试使用 StringTokenizer
但效果不太好。
这是我写的代码...
public static String[][] Json2Matrix(String jsonStringArray) {
int i = 0;
int j = 0;
String[][] mstrJsonString = null;
StringTokenizer tokElementos, tokSubelementos, tokTemp;
//jsonArray = "[["a","b"],["c","d"],["e","f"]]";
//jsonStringArray = jsonStringArray.replace("\"", "");
//jsonArray = "[[a,b],[c,d],[e,f]]";
jsonStringArray = jsonStringArray.substring(1, jsonStringArray.length() - 2);
//jsonArray = "[a,b],[c,d],[e,f]";/
//System.out.println(jsonStringArray);
//JSONSerializer.toJSON(jsonStringArray);
//System.out.println(jsonArray.toString());
//<editor-fold defaultstate="collapsed" desc="Prueba">
tokElementos = new StringTokenizer(jsonStringArray, "[]");
tokTemp = tokElementos;
tokSubelementos = new StringTokenizer(tokTemp.nextToken(), ",");
//System.out.println(tokElementos.countTokens());
//System.out.println(tokElementos.nextToken());
//System.out.println(tokSubelementos.countTokens()/2);
mstrJsonString = new String[tokElementos.countTokens()][tokSubelementos.countTokens()];
while (tokElementos.hasMoreTokens()) {
tokSubelementos = new StringTokenizer(tokElementos.nextToken(), ",");
j = 0;
while (tokSubelementos.hasMoreTokens()) {
mstrJsonString[i][j] = tokSubelementos.nextToken();
System.out.println(i + "," + j + " " + mstrJsonString[i][j]);
j++;
}
i++;
}
//</editor-fold>
return mstrJsonString;
}
我得到这个作为输出...
run:
1,0 "2"
1,1 "1"
3,0 "3"
3,1 "1"
5,0 "4"
5,1 "1"
7,0 "5"
7,1 "1"
9,0 "6"
9,1 "1"
11,0 "7"
11,1 "1"
13,0 "8"
13,1 "1"
15,0 "9"
15,1 "2"
17,0 "10"
17,1 "3"
19,0 "11"
19,1 "3"
21,0 "12"
21,1 "3"
23,0 "13"
23,1 "3"
25,0 "14"
25,1 "3"
27,0 "15"
27,1 "3"
29,0 "16"
29,1 "3"
31,0 "17"
31,1 "3"
33,0 "18"
33,1 "3"
35,0 "19"
35,1 "3"
它得到了正确的值,但索引不正确。有人可以帮我纠正这个问题吗?
最佳答案
使用 JSON 解析器比自己编写原始解析器节省很多精力。
我使用了这些基于 Java 的 JSON 解析器:
- jackson [http://jackson.codehaus.org/]
- Google 的 GSON [http://code.google.com/p/google-gson/]
我推荐 Google 的 GSON:
- 您可以将 json 字符串序列化/反序列化为 Java 对象 [POJO],无需注释 [使用变量名称]。
- 可以轻松序列化/反序列化深层对象层次结构。
- 它对 Java 泛型有很好的支持
关于java - 从 JSON 响应中获取 String[][] 数组,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/11054099/