Question

我在连接到我创建的Web服务时遇到问题。当我尝试从我的Android代码调用web服务时，它不起作用，但当我手动编写请求时，地址栏中传递的变量可以正常工作。请看一下代码。我认为问题在于我在页面加载时调用的函数，当我通过Android调用它时它不起作用。

using System;
using System.Collections.Generic;
using System.Linq;
using System.Web;
using System.Web.UI;
using System.Web.UI.WebControls;
using System.Web.Script.Serialization;
using System.Web;
using System.Data.SqlClient;
using System.Collections;

namespace MyService
{
    public partial class Caller : System.Web.UI.Page
    {
        //public int a, b;
        protected void Page_Load(object sender, EventArgs e)
        {
            this.txtBox1.Text = Request.QueryString["Name"];
            this.txtBox2.Text = Request.QueryString["Price"];
            if (txtBox1.Text != "" && txtBox2.Text != "")
                Add(txtBox1.Text, Convert.ToInt32(txtBox2.Text));  

        }

        public void Add(String name, int price)
        {
            //this.txtBox1.Text = Request.QueryString["Name"];
            //this.txtBox2.Text = Request.QueryString["Price"];
            SqlConnection con = new SqlConnection(util.ConnectionString);
            con.Open();
            String query = "insert into Cart values ('" + name + "',"+ price +");";
            SqlCommand cmd = new SqlCommand(query, con);
            SqlDataReader reader = cmd.ExecuteReader();
            reader.Read();
        } 
    }
}

确定！当我在IIS上托管网站并在地址栏中输入查询时，它的工作原理为192 ........ / MyAndroid / Caller.aspx？Name = Banana＆amp; Price = 15 这工作正常，但从Android调用它不起作用。这是android代码。

try {                   
                            Connectivity c = new Connectivity();    
                            String Name = holder.txtItemName.getText().toString();
                            int Price = Integer.parseInt((String)holder.txtPrice.getText());
                            String url="http://192.168.15.2/MyAndroid/Caller.aspx?Name="+Name+"&Price="+Price; //do not use localhost
                            String response=c.callWebService(url);
                            } catch (Exception e) {
                            // TODO Auto-generated catch block
                            e.printStackTrace();
                            }

一点帮助将不胜感激。谢谢!!

Android清单文件

 <?xml version="1.0" encoding="utf-8"?>
<manifest xmlns:android="http://schemas.android.com/apk/res/android"
    package="com.quiz.activities"
    android:versionCode="1"
    android:versionName="1.0" >

    <uses-sdk android:minSdkVersion="10" />

    <application
        android:icon="@drawable/ic_launcher"
        android:label="@string/app_name" >

        <activity
            android:name=".QuizActivity"
            android:label="@string/app_name" >
            <intent-filter>
                <action android:name="android.intent.action.MAIN" />

                <category android:name="android.intent.category.LAUNCHER" />
            </intent-filter>

        </activity>

        <activity            
            android:name=".Add" 
            android:label="@string/app_name" >

            <intent-filter>
                <action android:name="android.intent.action.Add" />

                <category android:name="android.intent.category.DEFAULT" />
            </intent-filter>

        </activity>
        <activity            
            android:name=".Get" 
            android:label="@string/app_name" >

            <intent-filter>
                <action android:name="android.intent.action.Get" />

                <category android:name="android.intent.category.DEFAULT" />
            </intent-filter>

        </activity>
    </application>
    <uses-permission android:name="android.permission.INTERNET"/>
</manifest>

从Android调用webservice asp.net

0 个答案: