如何连接android和php mysql数据库

Answer 1

参考下面代码及代码中的注释即可：
PHP代码：

conn.php是连接MySQL数据库的。代码如下：
<?php
$dbhost = "localhost:3306";
$dbuser = "root"; //我的用户名
$dbpass = ""; //我的密码
$dbname = "testlogin"; //我的mysql库名
$cn = mysql_connect($dbhost,$dbuser,$dbpass) or die("connect error");
@mysql_select_db($dbname)or die("db error");
mysql_query("set names 'UTF-8'");
?>
login.php代码：
<?php
include ("conn.php");//连接数据库
$username=str_replace(" ","",$_POST['name']);//接收客户端发来的username；
$sql="select * from users where name='$username'";
$query=mysql_query($sql);
$rs = mysql_fetch_array($query);
if(is_array($rs)){
if($_POST['pwd']==$rs['password']){
echo "login succeed";
}else{
echo "error";
}
}
?>

class LoginHandler implements Runnable {
@Override
public void run() {
// TODO Auto-generated method stub
//get username and password;
userName = user_name.getText().toString().trim();
password = pass_word.getText().toString().trim();
//连接到服务器的地址，我监听的是8080端口
String connectURL="网站地址/text0/com.light.text/login.php/";
//填入用户名密码和连接地址
boolean isLoginSucceed = gotoLogin(userName, password,connectURL);
//判断返回值是否为true，若是的话就跳到主页。
if(isLoginSucceed){
Intent intent = new Intent();
intent.setClass(getApplicationContext(), HomeActivity.class);
startActivity(intent);
proDialog.dismiss();
}else{
proDialog.dismiss();
// Toast.makeText(ClientActivity.this, "登入错误", Toast.LENGTH_LONG).show();
System.out.println("登入错误");
}
}
}
//登入的方法，传入用户 密码 和连接地址
private boolean gotoLogin(String userName, String password,String connectUrl) {
String result = null; //用来取得返回的String；
boolean isLoginSucceed = false;
//test
System.out.println("username:"+userName);
System.out.println("password:"+password);
//发送post请求
HttpPost httpRequest = new HttpPost(connectUrl);
//Post运作传送变数必须用NameValuePair[]阵列储存
List params = new ArrayList();
params.add(new BasicNameValuePair("name",userName));
params.add(new BasicNameValuePair("pwd",password));
try{
//发出HTTP请求
httpRequest.setEntity(new UrlEncodedFormEntity(params,HTTP.UTF_8));
//取得HTTP response
HttpResponse httpResponse=new DefaultHttpClient().execute(httpRequest);
//若状态码为200则请求成功，取到返回数据
if(httpResponse.getStatusLine().getStatusCode()==200){
//取出字符串
result=EntityUtils.toString(httpResponse.getEntity());
ystem.out.println("result= "+result);
}
}catch(Exception e){
e.printStackTrace();
}
//判断返回的数据是否为php中成功登入是输出的
if(result.equals("login succeed")){
isLoginSucceed = true;
}
return isLoginSucceed;
}

Answer 2

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