我已经创建了"id,name,username,email,age,password“列的数据库。如下所示的php文件和android活动。我总是以返回空值结束。至于我的android活动没有给出任何错误,但我发现只有php文件有问题。
我的php文件:
<?php
define('HOST','mysql.hostinger.in');
define('USER','u115908902_willi');
define('PASS','123');
define('DB','u115908902_prac');
$con = mysqli_connect(HOST,USER,PASS,DB)
if($_SERVER['REQUEST_METHOD']=='GET'){
$id = $_GET['id'];
$sql = "SELECT * FROM user WHERE id='".$id."'";
$r = mysqli_query($con,$sql);
$res = mysqli_fetch_array($r);
$result = array();
array_push($result,array(
"name"=>$res['name'],
"username"=>$res['username'],
"email"=>$res['email'],
"age"=>$res['age'],
"password"=>$res['password'],
)
);
echo json_encode(array("result"=>$result));
mysqli_close($con);
}那么我的MainActivity代码是:
import android.app.ProgressDialog;
import android.os.Bundle;
import android.support.v7.app.AppCompatActivity;
import android.view.View;
import android.widget.Button;
import android.widget.EditText;
import android.widget.TextView;
import android.widget.Toast;
import com.android.volley.RequestQueue;
import com.android.volley.Response;
import com.android.volley.VolleyError;
import com.android.volley.toolbox.StringRequest;
import com.android.volley.toolbox.Volley;
import org.json.JSONArray;
import org.json.JSONException;
import org.json.JSONObject;
public class SortId extends AppCompatActivity implements View.OnClickListener {
private EditText editTextId;
private TextView textViewResult;
private ProgressDialog loading;
@Override
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.activity_sort_id);
editTextId = (EditText) findViewById(R.id.editTextId);
Button buttonGet = (Button) findViewById(R.id.buttonGet);
textViewResult = (TextView) findViewById(R.id.textViewResult);
assert buttonGet != null;
buttonGet.setOnClickListener(this);
}
private void getData() {
String id = editTextId.getText().toString().trim();
if (id.equals("")) {
Toast.makeText(this, "Please enter an id", Toast.LENGTH_LONG).show();
return;
}
loading = ProgressDialog.show(this,"Please wait...","Fetching...",false,false);
String url = Config.DATA_URL+editTextId.getText().toString().trim();
StringRequest stringRequest = new StringRequest(url, new Response.Listener<String>() {
@Override
public void onResponse(String response) {
loading.dismiss();
showJSON(response);
}
},
new Response.ErrorListener() {
@Override
public void onErrorResponse(VolleyError error) {
Toast.makeText(SortId.this,error.getMessage().toString(),Toast.LENGTH_LONG).show();
}
});
RequestQueue requestQueue = Volley.newRequestQueue(this);
requestQueue.add(stringRequest);
}
private void showJSON(String response){
String name="";
String username="";
String email="";
String age="";
String password="";
try {
JSONObject jsonObject = new JSONObject(response);
JSONArray result = jsonObject.getJSONArray(Config.JSON_ARRAY);
JSONObject collegeData = result.getJSONObject(0);
name = collegeData.getString(Config.KEY_NAME);
username = collegeData.getString(Config.KEY_USERNAME);
email = collegeData.getString(Config.KEY_EMAIL);
age = collegeData.getString(Config.KEY_AGE);
password = collegeData.getString(Config.KEY_PASSWORD);
} catch (JSONException e) {
e.printStackTrace();
}
textViewResult.setText("name:\t" +name+ "\n username:\t" +username+"\n email:\t" +email+ "\n age:\t" +age+ "\n password:\t" +password);
}
@Override
public void onClick(View v) {
getData();
}
}发布于 2016-06-24 20:53:11
尝试以下操作(基于http://php.net/manual/en/mysqli.query.php):
<?php
define('HOST','mysql.hostinger.in');
define('USER','u115908902_willi');
define('PASS','123');
define('DB','u115908902_prac');
$con = mysqli_connect(HOST,USER,PASS,DB)
/* check connection */
if (mysqli_connect_errno()) {
printf("Connect failed: %s\n", mysqli_connect_error());
exit();
}
if($_SERVER['REQUEST_METHOD']=='GET'){
$id = $_GET['id'];
$sql = "SELECT * FROM user WHERE id='".$id."'";
if ($result = mysqli_query($link, $sql)) {
printf("Select returned %d rows.\n", mysqli_num_rows($result));
}
$r = mysqli_query($con,$sql);
$res = mysqli_fetch_array($r);
$result = array();
array_push($result,array(
"name"=>$res['name'],
"username"=>$res['username'],
"email"=>$res['email'],
"age"=>$res['age'],
"password"=>$res['password'],
)
);
echo json_encode(array("result"=>$result));
mysqli_close($con);
}这将帮助您确定1)数据库连接是否存在问题,以及2)您的查询是否返回结果。
https://stackoverflow.com/questions/38013597
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