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php - Android 从 php org.json.JSONArray 检索数据无法转换为 JSONObject

转载 作者:搜寻专家 更新时间:2023-11-01 08:39:46 25 4
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我正在尝试从 php 服务器获取一组数据到 android。但是在执行项目的时候出现了错误,尝试引用了几个相关的问题,还是想不出解决办法。

这是我的异步任务,它将建立连接并将 momentID 发送到 php

class GetAllComment extends AsyncTask<String, String, ArrayList<Comment>> {


protected ArrayList<Comment> doInBackground(String... params) {
List<NameValuePair> param = new ArrayList<NameValuePair>();
int success;
try {
Log.d("","@@@### " + postedID);
param.add(new BasicNameValuePair("momentid", postedID));
JSONObject json = jsonParser.makeHttpRequest(
"http://192.168.168.111:80/testmoment/getallcomment.php", "GET", param);

// check your log for json response
Log.d("Single comments Details", json.toString());

// json success tag
success = json.getInt("success");
if (success == 1) {

JSONArray productObj = json
.getJSONArray("product"); // JSON Array


try {
JSONObject comments = productObj.getJSONObject(0);
Comment comment = new Comment();
comment.setCommentContentbody(comments.getString("vcontentbody"));
comment.setCommentDate(comments.getString("vcdate"));
comment.setCommentTime(comments.getString("vctime"));
list.add(comment);

Log.d("","tryget : " + comments.getString("vcid") + " AND " + comments.getString("vcontentbody") + " AND " + comments.getString("vcdate") + " AND " + comments.getString("vctime"));

} catch (JSONException e) {
e.printStackTrace();
}

} else {

}
}catch (JSONException e) {
e.printStackTrace();
}
return list;
}

protected void onPostExecute(ArrayList<Comment> list) {
cAdapter = new CommentAdapter(getActivity().getBaseContext(), mCommentLists);
commentList.setAdapter(cAdapter);
mComment = new Comment();
cAdapter.Update();

if(list != null){
for(Comment comment: list){
cAdapter.add(comment);
}
}


cAdapter.notifyDataSetChanged();
}
}

这是我的 php 脚本:

<?php

header('Content-type: application/json');
include "db.config.php";
if (isset($_GET['momentid'])){
$momentid = $_GET['momentid'];
$result =mysql_query("SELECT * FROM commenttable WHERE vcmomentid = $momentid");

if (!empty($result)) {
// check for empty result
while($row=mysql_fetch_assoc($result))
$json_output[]=$row;

print(json_encode($json_output));

mysql_close();
echo json_encode($response);
} else {
// no product found
$response["success"] = 0;
$response["message"] = "No product found";

// echo no users JSON
echo json_encode($response);
}
} else {
// no product found
$response["success"] = 0;
$response["message"] = "No product found";

// echo no users JSON
echo json_encode($response);
}

?>

这是我执行后得到的错误:

JSON Parser: Error parsing data org.json.JSONException: Value [{"vcontentbody":"gkkgv","vcid":"1","vcmomentid":"109","vcdate":"2015-11-23","vcuserid":"1578","vctime":"17:09 PM"}] of type org.json.JSONArray cannot be converted to JSONObject

请帮我找出错误。谢谢

最佳答案

实际上您得到的是 JSONArray,您正在将其转换为 JSONObject

[ // This is Array
{ // This is Object
"vcontentbody":"gkkgv",
"vcid":"1",
"vcmomentid":"109",
"vcdate":"2015-11-23",
"vcuserid":"1578",
"vctime":"17:09 PM"
}
]

所以你必须这样写:

JSONArray jsonArray = jsonParser.makeHttpRequest("http://192.168.168.111:80/testmoment/getallcomment.php", "GET", param);

现在检查它是否为null

if(jsonArray != null)
{
// Here You may not have any key like 'success' so i don't have taken.

// You can directly get JSON Object here.. If you are getting more than one JSONObject then make loop here...
JSONObject object = jsonArray.getJSONObject(0);

// Now you can do with object that you want to do.
Comment comment = new Comment();
comment.setCommentContentbody(object.getString("vcontentbody"));
comment.setCommentDate(object.getString("vcdate"));
comment.setCommentTime(object.getString("vctime"));
list.add(comment);
}

如果有任何问题,你可以联系我。愿它对你有所帮助。谢谢。

关于php - Android 从 php org.json.JSONArray 检索数据无法转换为 JSONObject,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/33887449/

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