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我是一个初学者在android.I已经建立了一个应用程序登录leave模块。为了通过PHP连接它与mySql ..我尝试了这个代码从一个网站。Android通过Java的PHP连接
package com.example.axdroid;
import java.util.ArrayList;
import java.util.List;
import org.apache.http.HttpResponse;
import org.apache.http.NameValuePair;
import org.apache.http.client.HttpClient;
import org.apache.http.client.ResponseHandler;
import org.apache.http.client.entity.UrlEncodedFormEntity;
import org.apache.http.client.methods.HttpPost;
import org.apache.http.impl.client.BasicResponseHandler;
import org.apache.http.impl.client.DefaultHttpClient;
import org.apache.http.message.BasicNameValuePair;
import android.app.Activity;
import android.app.AlertDialog;
import android.app.ProgressDialog;
import android.content.DialogInterface;
import android.content.Intent;
import android.os.Bundle;
import android.view.View;
import android.view.View.OnClickListener;
import android.widget.Button;
import android.widget.EditText;
import android.widget.TextView;
import android.widget.Toast;
public class Axdroid extends Activity {
Button b;
EditText et,pass;
TextView tv;
HttpPost httppost;
StringBuffer buffer;
HttpResponse response;
HttpClient httpclient;
List<NameValuePair> nameValuePairs;
ProgressDialog dialog = null;
@Override
public void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.activity_axdroid);
b = (Button)findViewById(R.id.Button01);
et = (EditText)findViewById(R.id.username);
pass= (EditText)findViewById(R.id.password);
tv = (TextView)findViewById(R.id.tv);
b.setOnClickListener(new OnClickListener() {
@Override
public void onClick(View v) {
dialog = ProgressDialog.show(Axdroid.this, "",
"Validating user...", true);
Intent i=new Intent(getApplicationContext(),Userpage.class);
startActivity(i);
new Thread(new Runnable() {
public void run() {
login();
}
}).start();
}
});}
void login(){
try{
httpclient=new DefaultHttpClient();
httppost= new HttpPost("http://192.168.1.222/AndroidLeave/check.php");
//add your data
nameValuePairs = new ArrayList<NameValuePair>(2);
// Always use the same variable name for posting i.e the android side variable name and php side variable name should be similar,
nameValuePairs.add(new BasicNameValuePair("username",et.getText().toString().trim())); // $Edittext_value = $_POST['Edittext_value'];
nameValuePairs.add(new BasicNameValuePair("password",pass.getText().toString().trim()));
httppost.setEntity(new UrlEncodedFormEntity(nameValuePairs));
//Execute HTTP Post Request
response=httpclient.execute(httppost);
// edited by James from coderzheaven.. from here....
ResponseHandler<String> responseHandler = new BasicResponseHandler();
final String response = httpclient.execute(httppost, responseHandler);
System.out.println("Response : " + response);
runOnUiThread(new Runnable() {
public void run() {
tv.setText("Response from PHP : " + response);
dialog.dismiss();
}
});
if(response.equalsIgnoreCase("User Found")){
runOnUiThread(new Runnable() {
public void run() {
Toast.makeText(Axdroid.this,"Login Success", Toast.LENGTH_SHORT).show();
}
});
startActivity(new Intent(getApplicationContext(), Userpage.class));
}else{
showAlert();
}
}catch(Exception e){
dialog.dismiss();
System.out.println("Exception : " + e.getMessage());
}
}
public void showAlert(){
Axdroid.this.runOnUiThread(new Runnable() {
public void run() {
AlertDialog.Builder builder = new AlertDialog.Builder(Axdroid.this);
builder.setTitle("Login Error.");
builder.setMessage("User not Found.")
.setCancelable(false)
.setPositiveButton("OK", new DialogInterface.OnClickListener() {
public void onClick(DialogInterface dialog, int id) {
}
});
AlertDialog alert = builder.create();
alert.show();
}
});
}
}
我无法使这项工作在调试代码...控制从HttpClient的跳跃赶code.Would欣赏这个排序任何帮助。
这是PHP文件。我已经单独检查过它，它正在工作。
<?php
$hostname_localhost ="localhost";
$database_localhost ="mydatabase";
$username_localhost ="root";
$password_localhost ="";
$localhost = mysql_connect($hostname_localhost,$username_localhost,$password_localhost)
or
trigger_error(mysql_error(),E_USER_ERROR);
mysql_select_db($database_localhost, $localhost);
$username = $_POST['username'];
$password = $_POST['password'];
$query_search = "select * from tbl_user where username = '".$username."' AND password = '".$password. "'";
$query_exec = mysql_query($query_search) or die(mysql_error());
$rows = mysql_num_rows($query_exec);
//echo $rows;
if($rows == 0) {
echo "No Such User Found";
}
else {
echo "User Found";
}
?>
@ashwani 这是本Axdroid.java页（主要活动页）
public class Axdroid extends Activity {
Button b;
EditText et,pass;
TextView tv;
HttpPost httppost;
StringBuffer buffer;
HttpResponse response;
HttpClient httpclient;
List<NameValuePair> nameValuePairs;
ProgressDialog dialog = null;
@Override
public void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.activity_axdroid);
b = (Button)findViewById(R.id.Button01);
et = (EditText)findViewById(R.id.username);
pass= (EditText)findViewById(R.id.password);
tv = (TextView)findViewById(R.id.tv);
b.setOnClickListener(new OnClickListener() {
@Override
public void onClick(View v) {
String url="http://192.168.1.222/AndroidLeave/check.php";
JSONParser jparser= new JSONParser();
ArrayList<NameValuePair> params = new ArrayList<NameValuePair>();
params.add(new BasicNameValuePair("tag", "getcategory"));
params.add(new BasicNameValuePair("username", et.getText().toString()));
params.add(new BasicNameValuePair("password", pass.getText().toString()));
JSONObject jObj= jparser.makeHttpRequest(url, "POST", params);try {
String success = jObj.get("success").toString();
} catch (JSONException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
}
});}'
请把你的PHP文件也。所以，它能够更好地发现错误 –
这是PHP file..I已分别检查它和它的工作。 – tanmayee
是否在控制台上打印？ System.out.println（“Response：”+ response）;然后打印然后打印哪个字符串？ –