Question

嗨，大家搜索了2-3天后，我完全不知道出了什么问题，所以我很感激一些帮助：）

我正在使用wampserver ..我遇到问题是从android中将数据导入mysql ..每当我按下插入按钮时，我的列中都会添加一个空白行而没有任何信息。

这是我的插入

的php

<?php
$abc=$_POST['abc'];
$con = mysql_connect("localhost","root","");

if (!$con)

{

die('Could not connect: ' . mysql_error());

}

mysql_select_db("dbinsert", $con);
$sql="INSERT INTO Detail (Utext) VALUES ('".$abc."')";

if (!mysql_query($sql,$con))

{

die('Error: ' . mysql_error());

}

echo "1 record added";

mysql_close($con)

?>

for my android

  package com.project.insert;

  import java.util.ArrayList;

  import org.apache.http.HttpEntity;

  import org.apache.http.HttpResponse;

  import org.apache.http.NameValuePair;

  import org.apache.http.client.HttpClient;

  import org.apache.http.client.entity.UrlEncodedFormEntity;

  import org.apache.http.client.methods.HttpPost;

  import org.apache.http.impl.client.DefaultHttpClient;

  import org.apache.http.message.BasicNameValuePair;



  import android.app.Activity;

  import android.os.Bundle;

  import android.util.Log;

  import android.view.View;

  import android.widget.Button;

  import android.widget.EditText;



  public class PassData extends Activity {

      /** Called when the activity is first created. */

      @Override  

      public void onCreate(Bundle savedInstanceState) {

          super.onCreate(savedInstanceState);

          setContentView(R.layout.main);





          Button Insert=(Button)findViewById(R.id.button1);

          Insert.setOnClickListener(new View.OnClickListener(){

          public void onClick(View v)

          {      

          try          

               {  

                  EditText CompleteText=(EditText)findViewById(R.id.editText1);

                  ArrayList<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>();

                  nameValuePairs.add(new BasicNameValuePair("abc", "'"+CompleteText.getText().toString()+"'"));

                       Log.e(""+CompleteText.getText().toString(),"0");    

                       HttpClient httpclient = new DefaultHttpClient();

                       HttpPost httppost = new HttpPost("http://10.0.2.2/InsertStatement.php");

                       HttpResponse response = httpclient.execute(httppost);



               }  

               catch(Exception e)

               {

                       Log.e("log_tag", "Error in http connection "+e.toString());

               }

          }

      });

      }

  }

Answer 1

我没有看到您在nameValuePairs处执行任何操作以将其放入请求中。

HttpPost httppost = new HttpPost("http://10.0.2.2/InsertStatement.php");
httppost.setEntity(new UrlEncodedFormEntity(nameValuePairs, HTTP.UTF_8));
HttpResponse response = httpclient.execute(httppost);

可能会成功。

出于安全原因，您可能希望在服务器上添加'（并且不安卓代码 - 现在您使用两者都会显示mysql错误）并使用{ {1}}服务器端。

mysql插入问题并进行比较

1 个答案: