无法将记录插入MySQL，但未显示任何错误

Question

任务：将记录从android sqlite同步到mysql。

问题：mysql / php未将数据插入mysql中的表中。 但是没有显示错误。

DB_Connect.php：

<?php

class DB_Connect {

    // constructor
    function __construct(){

    }

    // destructor
    function __destruct(){

    }

    // connecting to database
    public function connect(){
        require_once 'config.php';  // defined DB_HOST,DB_USER,DB_PASSWORD here
        // connecting to mysql
        $con = mysqli_connect(DB_HOST,DB_USER,DB_PASSWORD);
        // selecting database
        mysqli_select_db($con,"heart");
        // return database handler
        return $con;
    }

    // closing database connection
    public function close(){
        mysqli_close($this->connect());
    }
}

?>

DB_Operations.php：

<?php
class DB_Operations {

    private $db;
    public $last_id;
    public $error;
    public $error_conn;
    public $error_no;

    // constructor
    function __construct(){
        require 'DB_Connect.php';
        $this->db = new DB_Connect();
        $this->db->connect();
    }

    // destructor
    function __destruct(){

    }

    // Storing new doctor
    public function storeDoctor($_id,$firstName,$lastName,$specialization,$licenseNumber,$clinicAddress,$email,$contactNum,$username,$password,$aboutMe){
        $result = mysqli_query($this->db->connect(),"INSERT INTO doctor(_id,first_name,last_name,specialization,license_number,clinic_address,email,contact_number,username,password,about_me) VALUES('$_id','$firstName','$lastName','$specialization','$licenseNumber','$clinicAddress','$email','$contactNum','$username','$password','$aboutMe')");

        if (mysqli_connect_errno()){
            $this->error_conn = mysqli_connect_error();
        }

        if(!$result){
            if(mysqli_errno($this->db->connect()) == 1062){
                // duplicate key - primary key violation
                return true;
            } else{
                // for other reasons
                $this->error = mysqli_error($this->db->connect());
                $this->error_no = mysqli_errno($this->db->connect());
                return false;
            }
        } else{
            $this->last_id = mysqli_insert_id($this->db->connect());
            return true;
        }
    }

    // getters
    public function getError(){
        return $this->error;
    }
    public function getError_no(){
        return $this->error_no;
    }
    public function getError_conn(){
        return $this->error_conn;
    }
    ...

insertuser.php：

<?php
include 'DB_Operations.php';
// create object for DB_Operations class
$db = new DB_Operations();
// get JSON posted by Android Application
$json = $_POST["doctorsJSON"];
// remove slashes
if(get_magic_quotes_gpc()){
    $json = stripslashes($json);
}
// decode JSON into Array
$data = json_decode($json);
// util arrays to create response JSON
$a = array();
$b = array();
// loop through an array and insert data read from JSON into MySQL DB
for($i=0; $i<count($data); $i++){
    // store doctor into MySQL DB
    $res = $db->storeDoctor($data[$i]->_id,$data[$i]->first_name,$data[$i]->last_name,$data[$i]->specialization,$data[$i]->license_number,$data[$i]->clinic_address,$data[$i]->email,$data[$i]->contact_number,$data[$i]->username,$data[$i]->password,$data[$i]->about_me);

    // based on insertion, create JSON response
    $b["local_id"] = $data[$i]->_id;
    if($res){
        $b["server_id"] = $db->last_id;
        $b["status"] = 'yes';
    }else{
        $b["status"] = 'no';
        $b["err_no"] = $db->getError_no();
        $b["error"] = $db->getError();
        $b["error_conn"] = $db->getError_conn();
    }
    array_push($a,$b);
}

// post JSON response back to Android Application
echo json_encode($a);
?>

我在Java中有一个将sqlite数据同步到mysql的函数：

public void syncSQLiteToMySQL(Context context,String selectQuery){
    dop = new DatabaseOperations(context);
    AsyncHttpClient client = new AsyncHttpClient();
    RequestParams params = new RequestParams();
    ArrayList<HashMap<String,String>> doctorList = new ArrayList<HashMap<String,String>>();
    doctorList = dop.getAllDoctors();
    if(doctorList.size()!=0){
        String json = dop.composeJSONfromSQLite(selectQuery);
        params.put("doctorsJSON", json);
        Log.i("json to pass", json);
        client.post("http://"+ip+":8080/changed_server_name/insertuser.php",params ,new AsyncHttpResponseHandler() {
            @Override
            public void onSuccess(String response) {
                Log.e("client response",response);
                try {
                    JSONArray arr = new JSONArray(response);
                    for(int i=0; i<arr.length();i++){
                        JSONObject obj = (JSONObject)arr.get(i);
                        // did something with json response here
                        dop.updateSyncStatus(obj.getString("local_id"),obj.getString("status"));
                    }
                    message = "DB Sync completed!";
                } catch (JSONException e) {
                    message = "Error Occured [Server's JSON response might be invalid]!";
                    Log.e("JSONException",e.getMessage());
                }
            }

            @Override
            public void onFailure(int statusCode, Throwable error, String content) {
                if(statusCode == 404){
                    message = "Requested resource not found";
                }else if(statusCode == 500){
                    message = "Something went wrong at server end";
                }else{
                    message = "Unexpected Error occcured! [Most common Error: Device might not be connected to Internet]";
                    Log.e("sync post failure",error.toString());
                }
            }
        });
    }
}

所以，这是响应：

[{“ local_id”：“ 0”，“ status”：“ no”，“ err_no”：0，“ error”：“”，“ error_conn”：null}]

JSON可以正常工作。 没问题。 我已经检查过并通过了正确的数据。 只是PHP和MySQL方面。 不知何故，我找不到此代码的错误。 没有错误消息，错误号为0，并且连接到DB也没有错误。 但是storeDoctor（）中的查询返回false。 怎么会这样 我已经在该站点和其他站点上阅读了有关此问题的信息，但是我并没有真正找到与我的问题接近的东西。

请赐教。 您的帮助将不胜感激。 提前致谢。

编辑：我也尝试mysqli_ping($this->db->connect()); 并返回true，表示连接正常。 那么，什么使查询失败呢？

Answer 1

您是否尝试过使用error_reporting（E_ALL）;

同样在构造函数中，您使用了$ this-> db-> connect（）;。

然后在商店中，使用$ result = mysqli_query（$ this-> db-> connect（），

您可以发布连接代码吗