Question

我已经成功创建了一个mysql数据库并试图用一个php文件填充它。

php的代码如下所示：

try {
    $conn = new PDO("mysql:host=$servername;dbname=$dbname", $username, $password);
    // set the PDO error mode to exception

    $conn->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
    $userid = $_POST['userid'];
    $time = $_POST['time'];
    $button = $_POST['button'];
    $weight = $_POST['weight'];
    $sex = $_POST['sex'];

    $sql = "INSERT INTO beerconsumption (userid, time, button, weight, sex)
    VALUES ('userid', 'time', 'button', 'weight', 'sex')";
    // use exec() because no results are returned
    $conn->exec($sql);
    echo "New record created successfully";
    }
    catch(PDOException $e)
    {
    echo $sql . "<br>" . $e->getMessage();
    }

$conn = null;
?>

如果我mannualy调用php文件它确实会进行修改但只是0.如果我改变了行

VALUES ('1', 'time', 'button', 'weight', 'sex')";

它将一个放在表格的第一个字段中。但如果我使用邮递员或我的Android应用程序发布不同的值，它仍然只是发布0。

目前我只是没有弄错。

Android脚本：

 public void sendData(){

    StringRequest request = new StringRequest(Request.Method.POST, insertUrl, new Response.Listener<String>() {
        @Override
        public void onResponse(String response) {

        }
    }, new Response.ErrorListener(){
        @Override
        public void onErrorResponse(VolleyError error){

        }
    }) {
        @Override
        protected Map<String, String> getParams() throws AuthFailureError {
            Map<String, String> parameters = new HashMap<String, String>();
            parameters.put("verify", "******");
            parameters.put("userid",userID);
            parameters.put("time",Long.toString(java.lang.System.currentTimeMillis()));
            parameters.put("button", Integer.toString(button));
            parameters.put("weight",Integer.toString(calc.person.getWeight()));
            parameters.put("sex",Integer.toString(calc.person.getSex()));

            return parameters;
        }
    };
    requestQueue.add(request);
}

Answer 1

我测试你的代码是否完美，你可以查看表单提交和POST数据。

<?php
$servername="localhost";
$dbname="android";
$username="root";
$password="root";

try {
    $conn = new PDO("mysql:host=$servername;dbname=$dbname", $username, $password);
    // set the PDO error mode to exception

    $conn->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);

    $sql = "INSERT INTO so(a,b,c,d,e) VALUES ('1', 'time', 'button', 'weight', 'sex')";
    // use exec() because no results are returned
    $conn->exec($sql);
    echo "New record created successfully";
    }
    catch(PDOException $e)
    {
    echo $sql . "<br>" . $e->getMessage();
    }

$conn = null;
?>

Answer 2

我做了一些研究并将我的PHP代码更改为：

<?php
define('HOST','localhost');
define('USER','***');
define('PASS','***');
define('DB','***');
$con = mysqli_connect(HOST,USER,PASS,DB);

$userid = $_POST['userid'];
$time = $_POST['time'];
$button = $_POST['button'];
$weight = $_POST['weight'];
$sex = $_POST['sex'];

$sql = "insert into beerconsumption (userid,time,button,weight,sex) values ('$userid','$time','$button','$weight','$sex')";
if(mysqli_query($con,$sql)){
    echo 'success';
}
else{
  echo 'failure';
}
mysqli_close($con);
?>

这似乎运作良好，但我现在有一些安全疑虑。这都是纯文本转移不是吗？

android，php和mysql之间的通信失败

2 个答案: