通过android从mysql中搜索特定值

Question

数据库已连接，一切都很好，但它只显示结果的第一个值

PHP的代码：

$res['medicine_name'], "ID_P"=>$res['ID_Pharmacy'], "id_M"=>$res['id_M'] ) ); echo json_encode(array("result"=>$result)); mysqli_close($con); } ?>

这是android中的代码：

package net.simplifiedcoding.gettingspecificdata;

import android.app.ProgressDialog;
import android.os.Bundle;
import android.support.v7.app.AppCompatActivity;
import android.view.View;
import android.widget.Button;
import android.widget.EditText;
import android.widget.TextView;
import android.widget.Toast;

import com.android.volley.RequestQueue;
import com.android.volley.Response;
import com.android.volley.VolleyError;
import com.android.volley.toolbox.StringRequest;
import com.android.volley.toolbox.Volley;

import org.json.JSONArray;
import org.json.JSONException;
import org.json.JSONObject;

public class MainActivity extends AppCompatActivity implements View.OnClickListener {

    private EditText editTextId;
    private Button buttonGet;
    private TextView textViewResult;

    private ProgressDialog loading;

    @Override
    protected void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.activity_main);

        editTextId = (EditText) findViewById(R.id.editTextNameM);
        buttonGet = (Button) findViewById(R.id.buttonSearch);
        textViewResult = (TextView) findViewById(R.id.textViewResult);

        buttonGet.setOnClickListener(this);
    }

    private void SearchMedicine() {
        String id = editTextId.getText().toString().trim();
        if (id.equals("")) {
            Toast.makeText(this, "Please enter an medicine name", Toast.LENGTH_LONG).show();
            return;
        }
        loading = ProgressDialog.show(this,"Please wait...","Fetching...",false,false);

        String url = Config.DATA_URL +editTextId.getText().toString().trim();

        StringRequest stringRequest = new StringRequest(url, new Response.Listener<String>() {
            @Override
            public void onResponse(String response) {
                loading.dismiss();
                showJSON(response);
            }
        },
        new Response.ErrorListener() {
            @Override
            public void onErrorResponse(VolleyError error) {
                Toast.makeText(MainActivity.this,error.getMessage().toString(),Toast.LENGTH_LONG).show();
            }
        });

        RequestQueue requestQueue = Volley.newRequestQueue(this);
        requestQueue.add(stringRequest);
    }

    private void showJSON(String response){
        String nameOfMedicine="";
        String IDPharmacy="";
        String IDMedicine = "";
        try {
            JSONObject jsonObject = new JSONObject(response);
            JSONArray result = jsonObject.getJSONArray(Config.JSON_ARRAY);
            JSONObject searchData = result.getJSONObject(0);
            nameOfMedicine = searchData.getString(Config.KEY_MEDICINE_NAME);
            IDPharmacy = searchData.getString(Config.KEY_ID_PHARMACY);
            IDMedicine = searchData.getString(Config.KEY_ID_MEDICINE);
        } catch (JSONException e) {
            e.printStackTrace();
        }
        textViewResult.setText("Name of medicine:\t"+nameOfMedicine+"\nID Pharmacy:\t" +IDPharmacy+ "\nID Medicine:\t"+ IDMedicine);
    }

    @Override
    public void onClick(View v) {

        SearchMedicine();
    }
}

输出： output of code

Answer 1

这是php：

<?php 

    if($_SERVER['REQUEST_METHOD']=='GET'){

        $medicine_name  = $_GET['medicine_name'];

        //$id  = $_GET['id'];

        require_once('dbConnectp.php');

         $sql = "SELECT * FROM medicine WHERE medicine_name='".$medicine_name."' AND stuteus='avilable'";

         //SELECT * FROM `medicine` WHERE `medicine_name` LIKE 'M1'
        //$sql = "SELECT * FROM Get WHERE id='".$id."'"; WHERE medicine_name=$Name_medicine

        $r = mysqli_query($con,$sql);

        $res = mysqli_fetch_array($r);

        $result = array();

        array_push($result,array(
            "medicine_name"=>$res['medicine_name'],
            "ID_P"=>$res['ID_Pharmacy'],
            "id_M"=>$res['id_M']
            )
        );

        echo json_encode(array("result"=>$result));

        mysqli_close($con);

    }
    ?>

Answer 2

错误出现在 showJSON 方法中，您没有迭代 JSON 数组以获取每个 JSON 对象并将其附加到 TextView。只拿第一个。

正确的showJSON()

private void showJSON(String response){
    String nameOfMedicine="";
    String IDPharmacy="";
    String IDMedicine = "";
    try {
        JSONObject jsonObject = new JSONObject(response);
        JSONArray result = jsonObject.getJSONArray(Config.JSON_ARRAY);
        textViewResult.setText("");
        for(int i=0;i<result.length();i++){
                JSONObject searchData = result.getJSONObject(i);
                nameOfMedicine = searchData.getString(Config.KEY_MEDICINE_NAME);
                IDPharmacy = searchData.getString(Config.KEY_ID_PHARMACY);
                IDMedicine = searchData.getString(Config.KEY_ID_MEDICINE);
        String temp = textViewResult.getText()+" \n";
        textViewResult.setText(temp+"Name of medicine:\t"+nameOfMedicine+"\nID Pharmacy:\t" +IDPharmacy+ "\nID Medicine:\t"+ IDMedicine);
    }


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

}