Php 如何获取json android中返回的多行并将其传递给sqlite
我正试图通过json将下面提到的php json结果输出到我的android应用程序中,但我只在我的android应用程序sqlite中获得第一个条目作为获取结果,其中可能是php json_encode返回的所有行。我想请求帮助,了解如何获取所有行,并将其传递给sqlitePhp 如何获取json android中返回的多行并将其传递给sqlite,php,android,mysql,json,sqlite,Php,Android,Mysql,Json,Sqlite,我正试图通过json将下面提到的php json结果输出到我的android应用程序中,但我只在我的android应用程序sqlite中获得第一个条目作为获取结果,其中可能是php json_encode返回的所有行。我想请求帮助,了解如何获取所有行,并将其传递给sqlite {“error”:false,“user”:{“id”:1,“sudentid”:1,“user_id”:“1”,“全名”:“Mugisha John”,“学校”:“APAPER I”,“level”:“level 2”,
{“error”:false,“user”:{“id”:1,“sudentid”:1,“user_id”:“1”,“全名”:“Mugisha John”,“学校”:“APAPER I”,“level”:“level 2”,“year”:18}{“error”:false,“user”:{“id”:2,“user_id”:“user_id”:“1”,“全名”:“Marie Ange Karamuzi”,“学校”:“Gs.st Joseph de Gita”,“level”:“Sinior 6”,“year”:20}{“error”:false,“user”:“user”:“user”:“3”,“sudentid”,“用户id”:“1”,“全名”:“Niragire Sangano Charles”,“学校”:“Gs.shyogwe”,“等级”:“Sinior 4”,“年份”:20}
我只在我的android应用程序sqlite中获得第一行:
{“id”:1,“sudentid”:1,“用户id”:“1”,“全名”:“Mugisha John”,“学校”:“APAPER I”,“等级”:“2级”,“年份”:18}
Main activity.java方法,我在android中用于获得上述结果:
private void GetStudentDetail(final String mobile) {
// Tag used to cancel the request
String tag_string_req = "req_Verfication";
progressBarList.setVisibility(View.VISIBLE);
// myList.setVisibility(View.GONE);
StringRequest strReq = new StringRequest(Request.Method.POST,
Config.FETCH_StudentsOfParent_URL, new Response.Listener<String>
() {
@Override
public void onResponse(String response) {
Log.d(TAG, "cerfication Response: " + response.toString());
// Parsing json
for (int i = 0; i < response.length(); i++) {
try {
JSONObject jObj = new JSONObject(response);
boolean error = jObj.getBoolean("error");
// Check for error node in json
if (!error) {
// user successfully exist in database
JSONObject user = jObj.getJSONObject("user");
String id = user.getString("id");
String sutdentId = user.getString("sutdentId");
String full_name = user.getString("full_name");
String year = user.getString("year");
String school = user.getString("school");
String level = user.getString("level");
// pass id ,sutdentId, fullname ,year ,school and level
to sqlite
db.addUser(id, sutdentId, full_name, year, school, level);
progressBarList.setVisibility(View.GONE);
myList.setVisibility(View.VISIBLE);
}
else{
// Error in login. Get the error message
// hiding the progress bar
progressBarList.setVisibility(View.GONE);
myList.setVisibility(View.VISIBLE);
String errorMsg = jObj.getString("error_msg");
Toast.makeText(getActivity(), errorMsg,
Toast.LENGTH_LONG).show();
}
} catch (JSONException e) {
// JSON error
e.printStackTrace();
Toast.makeText(getActivity(), "Json error: " +
e.getMessage(), Toast.LENGTH_LONG).show();
}
}
}
}, new Response.ErrorListener() {
@Override
public void onErrorResponse(VolleyError error) {
Log.e(TAG, "Verfication error Error: " + error.getMessage());
Toast.makeText(getActivity(),
"response error", Toast.LENGTH_LONG).show();
// Toast.makeText(getApplicationContext(),
// error.getMessage(), Toast.LENGTH_LONG).show();
// hiding the progress bar
progressBarList.setVisibility(View.GONE);
myList.setVisibility(View.VISIBLE);
}
}) {
@Override
protected Map<String, String> getParams() {
// Posting parameters to verfication url
Map<String, String> params = new HashMap<String, String>();
params.put("mobile", mobile);
return params;
}
};
// // Adding request to request queue
MyApplication.getInstance().addToRequestQueue(strReq,tag_string_req);
}
private void GetStudentDetail(最终字符串移动){
//用于取消请求的标记
字符串标记\u String\u req=“req\u Verfication”;
progressBarList.setVisibility(View.VISIBLE);
//myList.setVisibility(View.GONE);
StringRequest strReq=新的StringRequest(Request.Method.POST,
Config.FETCH\u StudentsOfParent\u URL,new Response.Listener
() {
@凌驾
公共void onResponse(字符串响应){
Log.d(标签,“Cercation响应:+Response.toString());
//解析json
对于(int i=0;i
您需要将JSON数据包装在一个数组中,然后使用JSONArray在Android中解析它
[{"error":false,"user":{"id":1,"sutdentId":1,"user_id":"1","full_name":"Mugisha John","school":"APAPER I","level":"Level 2","year":18}}{"error":false,"user":{"id":2,"sutdentId":2,"user_id":"1","full_name":"Marie Ange Karamuzi","school":"Gs.st Joseph de Gitarama","level":"Sinior 6","year":20}}{"error":false,"user":{"id":3,"sutdentId":3,"user_id":"1","full_name":"Niragire Sangano Charles","school":"Gs. shyogwe","level":"Sinior 4","year":20}}]
您需要将JSON数据包装在一个数组中,然后使用JSONArray在Android中解析它
[{"error":false,"user":{"id":1,"sutdentId":1,"user_id":"1","full_name":"Mugisha John","school":"APAPER I","level":"Level 2","year":18}}{"error":false,"user":{"id":2,"sutdentId":2,"user_id":"1","full_name":"Marie Ange Karamuzi","school":"Gs.st Joseph de Gitarama","level":"Sinior 6","year":20}}{"error":false,"user":{"id":3,"sutdentId":3,"user_id":"1","full_name":"Niragire Sangano Charles","school":"Gs. shyogwe","level":"Sinior 4","year":20}}]
编辑:
我的php代码看起来像下面的文件,但它的工作是好的,
但问题在于安卓系统
public function StudentsOfParent($mobile){
$stmt = $this->conn->prepare("SELECT
a.id,
a.name,
a.mobile,
c.id as sutdentId,
c.user_id,
c.full_name,
c.school,
c.level,
c.year,
c.id
from users a
join students c
on a.id = c.user_id where a.mobile= ?");
$stmt->bind_param("i", $mobile);
if ($stmt->execute()) {
// $user = $stmt->get_result()->fetch_assoc();
// $stmt->close();
// // return user's results
// return $user;
$result = $stmt->get_result();
$usersArr = array();
while ($user = $result->fetch_assoc()){
$usersArr[] = $user;
}
$stmt->close();
return $usersArr;
} else {
return NULL;
}
$usersArr = $db->StudentsOfParent($mobile);
if ($usersArr != false) {
// user found successfully
$response["error"] = FALSE;
foreach($usersArr as $key=>$user){
$response[$key]["user"]["id"] = $user["id"];
$response[$key]["user"]["sutdentId"] = $user["sutdentId"];
$response[$key]["user"]["user_id"] = $user["user_id"];
$response[$key]["user"]["full_name"] = $user["full_name"];
$response[$key]["user"]["school"] = $user["school"];
$response[$key]["user"]["level"] = $user["level"];
$response[$key]["user"]["year"] = $user["year"];
echo json_encode($response);
}
} else {
.........
访问上述函数:从mysql获取信息
$usersArr = $db->StudentsOfParent($mobile);
if ($usersArr != false) {
// user found successfully
$response["error"] = FALSE;
foreach($usersArr as $key=>$user){
$response[$key]["user"]["id"] = $user["id"];
$response[$key]["user"]["sutdentId"] = $user["sutdentId"];
$response[$key]["user"]["user_id"] = $user["user_id"];
$response[$key]["user"]["full_name"] = $user["full_name"];
$response[$key]["user"]["school"] = $user["school"];
$response[$key]["user"]["level"] = $user["level"];
$response[$key]["user"]["year"] = $user["year"];
echo json_encode($response);
}
} else {
.........
编辑:
我的php代码看起来像下面的文件,但它的工作是好的,
但问题在于安卓系统
public function StudentsOfParent($mobile){
$stmt = $this->conn->prepare("SELECT
a.id,
a.name,
a.mobile,
c.id as sutdentId,
c.user_id,
c.full_name,
c.school,
c.level,
c.year,
c.id
from users a
join students c
on a.id = c.user_id where a.mobile= ?");
$stmt->bind_param("i", $mobile);
if ($stmt->execute()) {
// $user = $stmt->get_result()->fetch_assoc();
// $stmt->close();
// // return user's results
// return $user;
$result = $stmt->get_result();
$usersArr = array();
while ($user = $result->fetch_assoc()){
$usersArr[] = $user;
}
$stmt->close();
return $usersArr;
} else {
return NULL;
}
$usersArr = $db->StudentsOfParent($mobile);
if ($usersArr != false) {
// user found successfully
$response["error"] = FALSE;
foreach($usersArr as $key=>$user){
$response[$key]["user"]["id"] = $user["id"];
$response[$key]["user"]["sutdentId"] = $user["sutdentId"];
$response[$key]["user"]["user_id"] = $user["user_id"];
$response[$key]["user"]["full_name"] = $user["full_name"];
$response[$key]["user"]["school"] = $user["school"];
$response[$key]["user"]["level"] = $user["level"];
$response[$key]["user"]["year"] = $user["year"];
echo json_encode($response);
}
} else {
.........
访问上述函数:从mysql获取信息
$usersArr = $db->StudentsOfParent($mobile);
if ($usersArr != false) {
// user found successfully
$response["error"] = FALSE;
foreach($usersArr as $key=>$user){
$response[$key]["user"]["id"] = $user["id"];
$response[$key]["user"]["sutdentId"] = $user["sutdentId"];
$response[$key]["user"]["user_id"] = $user["user_id"];
$response[$key]["user"]["full_name"] = $user["full_name"];
$response[$key]["user"]["school"] = $user["school"];
$response[$key]["user"]["level"] = $user["level"];
$response[$key]["user"]["year"] = $user["year"];
echo json_encode($response);
}
} else {
.........
纳迪格索,谢谢。我不熟悉杰索纳雷,你能帮我举个例子吗?纳迪格索,谢谢。我不熟悉杰索纳雷,你能帮我举个例子吗?