Question

我正在使用Zxing框架构建一个简单的QR码阅读器，到目前为止它工作正常。这里的问题是我未能找到捕获的QR码数据的变量。如何将捕获的QR码数据发送到我的在线MySQL数据库。

以下是我的代码

Main_activity.java

    package io.github.akodiaemmanuel.mapenziscanner;

import android.app.Activity;
import android.app.AlertDialog;
import android.os.Bundle;
import android.support.annotation.Nullable;
import android.util.Log;
import android.view.View;


import com.google.zxing.Result;

import me.dm7.barcodescanner.core.ViewFinderView;
import me.dm7.barcodescanner.zxing.ZXingScannerView;

/**
 * Created by Akodia Emmanuel on 9/18/2017.
 */

public class MainActivity extends Activity implements ZXingScannerView.ResultHandler {
    private ZXingScannerView mScannerview;

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

    public void onClick(View v){
        mScannerview = new ZXingScannerView(this);
        setContentView(mScannerview);
        mScannerview.setResultHandler(this);
        mScannerview.startCamera();
    }
    @Override
    public void handleResult(Result result) {
//Do anything with the result here
        Log.w("handleResult",result.getText());
        AlertDialog.Builder builder = new AlertDialog.Builder(this);
        builder.setTitle("Scan Result");
        builder.setMessage(result.getText());
        AlertDialog alertDialog = builder.create();
        alertDialog.show();

        //resume scanning
        //mScannerview.resumeCameraPreview(this);
    }
}

Answer 1

您可以将数据发送到将在MySQL数据库中保存这些数据的PHP脚本：

要从Android向PHP发送数据，您可以使用https://github.com/koush/ion

这里我们将调用页面url_to_my_script.php并在POST中发送qrdata。当页面加载（onCompleted）时，我们得到JSON响应并根据响应做一些事情。

机器人：

Ion.with(context)
    .load("url_to_my_script.php")
    .setBodyParameter("qrdata", qrdata)
    .asJsonObject()
    .withResponse()
    .setCallback(new FutureCallback<Response<JsonObject>>() {
        @Override
        public void onCompleted(Exception e, Response<JsonObject> reponse) {
            Log.i("myactivity", "jsonresult is " + reponse);
            //if we have an error or code
            if (e != null || reponse.getHeaders().code() != 200) {
                Log.e("myactivity", "Error invalid request" + e);
                Log.e("myactivity", "response code:"+reponse.getHeaders().code());                                  
            } else {
                if (!success.equals("1")) {
                    String error = result.get("error_message").toString();
                }else{
                    //everything is ok do something
                }
            }
     }
});

Php脚本：

这是PHP脚本获取数据并返回JSON响应

<?php
header('Content-type: application/json');
require_once 'database_config.php';

function insertMyData($data){
   //do something
   // if ok return true else false
}

if($_POST){
    $qrdata= trim($_POST['qrdata']);
    if(insertMyData($qrdata)){
        echo '{"success":1}';
    }else{
        echo '{"success":0,"error_message":"Put a message her"}';
    } 
} else {
    echo '{"success":0,"error_message":"Invalid date"}';
}
?>

从我的Android应用程序zxing发送数据到我的在线MySQL数据库

1 个答案: