我拥有将我的应用程序链接到我的sql数据库的所有代码logcat中没有错误,应用程序不崩溃,并且在“数据输入”的吐司消息在成功出现所有捕获条款,但在数据库表中没有任何类型的数据输入。在mysql数据库中没有输入数据
这里是MainActivity代码:
public class MainActivity extends Activity {
EditText name,age,email;
Button b;
@Override
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
StrictMode.ThreadPolicy policy = new StrictMode.ThreadPolicy.Builder().permitAll().build();
StrictMode.setThreadPolicy(policy);
setContentView(R.layout.activity_main);
name = (EditText) findViewById(R.id.name);
age = (EditText) findViewById(R.id.age);
email = (EditText) findViewById(R.id.email);
b = (Button) findViewById(R.id.done);
b.setOnClickListener(new View.OnClickListener() {
InputStream is = null;
@Override
public void onClick(View v) {
String username = name.getText().toString();
String userage = age.getText().toString();
String useremail = email.getText().toString();
//setting the nameValuePair
List<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>(1);
nameValuePairs.add(new BasicNameValuePair("username",username));
nameValuePairs.add(new BasicNameValuePair("userage",userage));
nameValuePairs.add(new BasicNameValuePair("useremail",useremail));
//
try{
HttpClient httpClient = new DefaultHttpClient();
HttpPost httpPost = new HttpPost("http://sql17.000webhost.com/phpMyAdmin/index.php?db=a6923033_hamsau&lang=en-utf-8&token=4abc43994f0264c44701d85da5b524ee&phpMyAdmin=qsZJnJbsiQaPRx-aDIDtwuxOx5f,new.php");
httpPost.setEntity(new UrlEncodedFormEntity(nameValuePairs));
HttpResponse httpResponse = httpClient.execute(httpPost);
HttpEntity entity = httpResponse.getEntity();
is = entity.getContent();
}
catch (UnsupportedEncodingException e) {
Log.i("Hammas","UnsupportedEncodingException "+e);
} catch (ClientProtocolException e) {
Log.i("Hammas", "ClientProtocolException " + e);
} catch (IOException e) {
Log.i("Hammas", "IOException " + e);
}
Toast.makeText(getApplicationContext(),"Data is entered",Toast.LENGTH_SHORT).show();
}
});
}
}
这里是phpscript
enter code here
<?php
$mysql_host = "mysql17.000webhost.com";
$mysql_database = "a6923033_hamsau";
$mysql_user = "a6923033_hamsau";
$mysql_password = "123master95";
$name=$_POST['name'];
$age=$_POST['age'];
$email=$_POST['email'];
mysql_query("insert into users(name,age,email)values('{$name}','{$age}', {$email}')");
?>
我认为有错误的HTTP链接或phpscript,因为这些都只是一些我没了解清楚
我当然希望那些不是你的实际DB凭证。如果他们是,你最好现在就去改变他们。 –
你能否更清楚你的答案? – Usman
你基本上只是把整个世界的钥匙给了你的数据库。你可能有一天醒来,发现一切已被删除 – Osuwariboy