Question

我正在尝试从使用xampp mysql创建的localhost数据库中检索数据。

public class JASONUseActivity extends Activity {

    EditText byear; // To take birthyear as input from user
    Button submit;
    TextView tv; // TextView to show the result of MySQL query 

    String returnString; // to store the result of MySQL query after decoding JSON

    /** Called when the activity is first created. */
    @Override
    protected void onCreate(Bundle savedInstanceState) {
        StrictMode.setThreadPolicy(new StrictMode.ThreadPolicy.Builder()
            .detectDiskReads().detectDiskWrites().detectNetwork() // StrictMode is most commonly used to catch accidental disk or network access on the application's main thread
        .penaltyLog().build());
        super.onCreate(savedInstanceState);
        setContentView(R.layout.main);
        byear = (EditText) findViewById(R.id.editText1);
        submit = (Button) findViewById(R.id.submitbutton);
        tv = (TextView) findViewById(R.id.showresult);

        // define the action when user clicks on submit button
        submit.setOnClickListener(new View.OnClickListener() {
            public void onClick(View v) {

                // declare parameters that are passed to PHP script i.e. the name birth year and its value submitted by user   
                ArrayList < NameValuePair > postParameters = new ArrayList < NameValuePair > ();

                // define the parameter
                postParameters.add(new BasicNameValuePair("birthyear", byear.getText().toString()));
                String response = null;

                // call executeHttpPost method passing necessary parameters 
                try {
                    response = CustomHttpClient.executeHttpPost("http://localhost/jasonscript.php", postParameters);

                    // store the result returned by PHP script that runs MySQL query
                    String result = response.toString();

我无法在textview中查看数据。网址是否正确？

这是位于xampp / htdocs

中的jasonscript.php

<?php
$db_host = "localhost";

$db_uid = "root";

$db_pass = "";

$db_name = "";

$db_con = mysql_connect($db_host, $db_uid, $db_pass) or die('could not connect');
mysql_select_db($db_name);
$sql    = "SELECT * FROM people WHERE birthyear > '" . $_POST["birthyear"] . "'";
$result = mysql_query($sql);
while ($row = mysql_fetch_assoc($result))
    $output[] = $row;
print(json_encode($output));
mysql_close();
?>

我已在清单文件中提供了互联网权限。

提前致谢。

Answer 1

设备上的

localhost是指其环回接口，不您托管要连接的数据库的开发计算机。要使虚拟设备连接到开发机器的环回接口，您需要使用10.0.2.2。您可以将URL更改为这样的内容，以便点击您的开发机器：

CustomHttpClient.executeHttpPost("http://10.0.2.2/jasonscript.php", postParameters);

有关模拟器网络如何工作的更多详细信息，请参阅我的回答here。

Answer 2

当您在Android设备上使用localhost时，您正在尝试使用设备本身的服务。使用服务器的IP /主机名来获取数据。

旁注：另请查看ASyncTask here，因为在onCreate中下载数据会锁定您的UI-Thread（如果需要很长时间）并且可能会导致App Not Responding或{{ 1}}。即使它没有发生崩溃，用户体验也会因为应用程序没有响应而降级。

url将android连接到mysql数据库

2 个答案: