Java 从Json字符串中仅获取Json对象的第一个对象
这是我的json对象:-Java 从Json字符串中仅获取Json对象的第一个对象,java,android,json,Java,Android,Json,这是我的json对象:- {"emp_remark":"right now busy.. please wait", "emp_loginid":"pra", "emp_name":"Pragya Patel", "emp_timein":"18:9", "emp_id":"1", "emp_timeout":"18:9", "emp_emailid":"hsjsnejw", "emp_mobno":"7879467946", "emp_desigantion":"Android", "emp_
{"emp_remark":"right now busy.. please wait",
"emp_loginid":"pra",
"emp_name":"Pragya Patel",
"emp_timein":"18:9",
"emp_id":"1",
"emp_timeout":"18:9",
"emp_emailid":"hsjsnejw",
"emp_mobno":"7879467946",
"emp_desigantion":"Android",
"emp_deviceid":"APA91bFLNsLOz2iiSw9r2NKdlnWCWtNNNb-VTVY3TwmT7Nly88NnSJjJwoLNC3qveU7LSW9QY5t71JAejnkogvQRPXA-uEtnlg-1cve00k_4UhIinUH0Lzs"
}
"emp_remark":"right now busy.. please wait",
"emp_loginid":"deepp",
"emp_name":"deepak",
"emp_timein":"18:18",
"emp_id":"2",
"emp_timeout":"18:9",
"emp_emailid":"deep@gmail.com",
"emp_mobno":"7469467946",
"emp_desigantion":"java",
"emp_deviceid":null
}
{
"emp_remark":"right now busy.. please wait",
"emp_loginid":"amu",
"emp_name":"amul",
"emp_timein":"18:18",
"emp_id":"3",
"emp_timeout":"18:9",
"emp_emailid":"amu@gmail.com",
"emp_mobno":"7469462946",
"emp_desigantion":"java",
"emp_deviceid":"APA91bFLNsLOz2iiSw9r2NKdlnWCWtNNNb-VTVY3TwmT7Nly88NnSJjJwoLNC3qveU7LSW9QY5t71JAejnkogvQRPXA-uEtnlg-1cve00k_4UhIinUH0Lzs"
}
现在在android中,我将整个对象放在一个字符串中。。但当我将其转换为json对象时。。它只得到第一个对象,即
{
"emp_remark":"right now busy.. please wait",
"emp_loginid":"pra",
"emp_name":"Pragya Patel",
"emp_timein":"18:9",
"emp_id":"1",
"emp_timeout":"18:9",
"emp_emailid":"hsjsnejw",
"emp_mobno":"7879467946",
"emp_desigantion":"Android",
"emp_deviceid":"APA91bFLNsLOz2iiSw9r2NKdlnWCWtNNNb-VTVY3TwmT7Nly88NnSJjJwoLNC3qveU7LSW9QY5t71JAejnkogvQRPXA-uEtnlg-1cve00k_4UhIinUH0Lzs"}
这是我的密码:
@Override
protected Void doInBackground(Void... urls) {
// Creating service handler class instance
ServiceHandler sh = new ServiceHandler();
Log.d(TAG, "Url : "+getUrl);
// Making a request to url and getting response
jsonStr = sh.makeServiceCall(getUrl, ServiceHandler.GET);
Log.d(TAG, "String json :" + jsonStr);
try {
employeeDetailsModel = JsonParsing.getEmployeePojo(jsonStr);
List<EmployeeDetailsModel> employeedesignationList = jsonParsing.getEmployeeDes(jsonStr);
} catch (JSONException e) {
// TODO Auto-generated catch block
Log.d(TAG, "Exception: in parsing >> " + e.getMessage());
}
return null;
}
在json对象中,仅获取第一个对象这是一个格式错误的json
,
正确的对象如下所示:
[
{
“emp_备注”:“现在正忙……请稍候”,
“emp_loginid”:“pra”,
“emp_name”:“Pragya Patel”,
“emp_timein”:“18:9”,
“emp_id”:“1”,
“emp_超时”:“18:9”,
“emp_emailid”:“hsjsnejw”,
“emp_mobno”:“7879467946”,
“emp_设计”:“Android”,
“emp_deviceid”:“APA91BFLNSLOZ2是W9R2NKDLNWCWTNNB-VTVY3TWMT7NLY88NNSJJWOLNC3QVEU7LSW9QY5T71JAEJNKOGVQRPXA-uEtnlg-1cve00k\U 4UHIINU0LZS”
},
{
“emp_备注”:“现在正忙……请稍候”,
“emp_loginid”:“deepp”,
“emp_名称”:“deepak”,
“emp_timein”:“18:18”,
“emp_id”:“2”,
“emp_超时”:“18:9”,
“emp_电子邮件ID”:deep@gmail.com",
“emp_mobno”:“7469467946”,
“环境管理设计”:“java”,
“emp_设备ID”:空
},
{
“emp_备注”:“现在正忙……请稍候”,
“emp_loginid”:“amu”,
“emp_name”:“amul”,
“emp_timein”:“18:18”,
“emp_id”:“3”,
“emp_超时”:“18:9”,
“emp_电子邮件ID”:amu@gmail.com",
“emp_mobno”:“7469462946”,
“环境管理设计”:“java”,
“emp_deviceid”:“APA91BFLNSLOZ2是W9R2NKDLNWCWTNNB-VTVY3TWMT7NLY88NNSJJWOLNC3QVEU7LSW9QY5T71JAEJNKOGVQRPXA-uEtnlg-1cve00k\U 4UHIINU0LZS”
}
]
转换JsonArray中的字符串
JSONArray jsonArray = new JSONArray(str);
然后使用下面的代码
JsonObject jsonObject;
for (int i = 0; i < jsonArray.length(); i++) {
jsonObject = jsonArray.getJSONObject(i);
}
JsonObject-JsonObject;
for(int i=0;i
我认为它不是像[{a:“1”,b:“2”},{c:“3”,d:“4”}那样的普通jsonList,所以只能得到第一个对象。
对象如下:{key,value}
如下列表:[collection,collection]JSON无效,如果要将多个对象放在一起,必须将它们分开并放入数组中。tour JSON数据格式无效..请cheak@PragyaPatel使用我已经检查过的json检查您的json。。请告诉我如何使用Java获取正确的json格式。您需要更改从服务器获取的json。Java代码中没有任何内容。请在答案中添加尽可能多的信息。不建议使用外部链接,因为它们可能会在以后被破坏或更改内容。
JSONArray jsonArray = new JSONArray(str);
JsonObject jsonObject;
for (int i = 0; i < jsonArray.length(); i++) {
jsonObject = jsonArray.getJSONObject(i);
}