php - 无法从android应用程序将数据插入mysql

Question

我想从我的 Android 应用程序将一些数据插入 mysql 数据库，但它不起作用。我尝试从模拟器和手机插入数据，但数据库中没有存储任何内容。我不知道为什么。有人可以帮我吗？这是我的 php 代码

<?php
$mysqli = new mysqli("127.0.0.1", "root", "", "food");
if ($mysqli->connect_errno) {
echo "Failed to connect to MySQL: (" . $mysqli->connect_errno . ") " . $mysqli-    >connect_error;
}
     echo "Connected to db";

$FOOD_ID =  isset($_POST['FOOD_ID']) ? $_POST['FOOD_ID'] : null;
$FOOD_NAME = isset($_POST['FOOD_NAME']) ? $_POST['FOOD_NAME'] : null;

mysql_query ("INSERT INTO food (FOOD_ID, FOOD_NAME) VALUES('$FOOD_ID', '$FOOD_NAME')");

?>

这是java代码

public class MainActivity extends Activity {

String food_id;
String food_name;

private EditText name, id;
private Button button;

@Override
protected void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.activity_cappuccino);

    name = (EditText) findViewById(R.id.FOOD_ID);
    id = (EditText) findViewById(R.id.FOOD_NAME);


    button = (Button) findViewById(R.id.button);

    button.setOnClickListener(new View.OnClickListener() {

        @Override
        public void onClick(View v) {
           food_id = name.getText().toString();
            food_name = id.getText().toString();

            new SummaryAsyncTask().execute((Void) null);
        }
    }); 
}

class SummaryAsyncTask extends AsyncTask<Void, Void, Boolean> {

    private void postData(String idid, String namename) {

        HttpClient httpclient = new DefaultHttpClient();
        HttpPost httppost = new HttpPost("http://xxxxxx/try.php");

        try {
            ArrayList<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>(4);
            nameValuePairs.add(new BasicNameValuePair("ID", idid));
            nameValuePairs.add(new BasicNameValuePair("NAME", namename));

            httppost.setEntity(new UrlEncodedFormEntity(nameValuePairs));
            HttpResponse response = httpclient.execute(httppost);
        }
        catch(Exception e)
        {
            Log.e("log_tag", "Error:  "+e.toString());
        }
    }

    @Override
    protected Boolean doInBackground(Void... params) {
        postData(food_name,food_id);
        return null;
    }
}
}

Answer 1

您正在为 3 列插入值，但您只提供了 2 列名称，这就是查询不起作用的原因。

看这里

mysql_query ("INSERT INTO food (FOOD_ID, FOOD_NAME) VALUES('$FOOD_ID', '$FOOD_NAME,    NOW())");
                                ^//col1  ^//col2                                       //^val3 (where is its corresponding column name ?)

解决方案:为NOW()提供第三列名称，您将查询工作。

其次，您使用的是已弃用的数据库 API，即 mysql_* 函数。您需要切换到 MySQLi 或 PDO。