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java - 从php中获取json数据并将其存储在java中的数组中

转载 作者:行者123 更新时间:2023-12-01 15:14:00 25 4
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我的 JSON 数组是这样的:

{"Name_1":1,"Name_2":0,"Name_3":0}

我在 java 中获取值并将它们存储在单独的数组中的代码如下:

int[] operations= new int[3];
String result = "";
InputStream is = null;
StringBuilder sb=null;
try{
HttpClient httpclient = new DefaultHttpClient();

HttpPost httppost = new HttpPost("http://testteamgr.netau.net/parsing/test.php");
HttpResponse response = httpclient.execute(httppost);
HttpEntity entity = response.getEntity();
is = entity.getContent();
}catch(Exception e){
Log.e("log_tag", "Error in http connection "+e.toString());
}
//convert response to string
try{
BufferedReader reader = new BufferedReader(new InputStreamReader(is,"iso-8859-1"),8);
sb = new StringBuilder();
String line = null;
while ((line = reader.readLine()) != null) {
sb.append(line + "\n");
}
is.close();

result=sb.toString();
}catch(Exception e){
Log.e("log_tag", "Error converting result "+e.toString());
}
try{

JSONObject json_data = new JSONObject(result);
System.out.println("Length of json is"+jArray.length());
for(int i=0;i<jArray.length();i++){

if (i==0) operations[0]=json_data.getInt("Name_1");
else if (i==1) operations[1]=json_data.getInt("Name_2");
else if (i==2) operations[2]=json_data.getInt("Name_3"); }

我收到这些错误:

value br of type java.lang.string cannot be converted to jsonobject

如果我打印结果,我看不到 JSONobject,而是看到 html 代码。

所以我想要的是将这 3 个值放入一个单独的数组中。

最佳答案

你有一个对象,而不是一个数组。要处理结果,您可以使用以下代码:

    String json = "{\"Name_1\":1,\"Name_2\":0,\"Name_3\":0}";
JSONObject object = new JSONObject(json);
String[] propertyNames = JSONObject.getNames(object);
String[] values = new String[propertyNames.length];
for (int i = 0; i < propertyNames.length; i++) {
values[i] = String.valueOf(object.get(propertyNames[i]));
}

关于java - 从php中获取json数据并将其存储在java中的数组中,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/11906499/

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