Question

我需要在一个mysql数据库中插入一些从Android应用程序中恢复的数据并将其插入到一些不同的表中。

我创建了一个用户的注册页面/活动，它可以工作，但我试图从其他活动插入其他数据，我无法将它们插入到我的数据库中。

这是我的代码：

RegisterUserClass是翻译的类。

RegisterBody.java - ＆gt;在这种情况下，我需要将用户的id保存到表中，这就是我保存电子邮件和密码的原因。

private static final String REGISTER_URL = "http://10.0.2.2/sFitness/RegisterBody.php";

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

        etWaist =(EditText) findViewById(R.id.etWaist);
        etHips =(EditText) findViewById(R.id.etHips);
        etBreast =(EditText) findViewById(R.id.etBreast);
        etWrist=(EditText) findViewById(R.id.etWrist);
        etWeight=(EditText)findViewById(R.id.etWeight);
        etHeight=(EditText)findViewById(R.id.etHeight);

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

        bNext.setOnClickListener(this);

        Bundle b= this.getIntent().getExtras();
        array=b.getStringArray(null);
        name=array[0];
        surname=array[1];
        age=array[2];
        email=array[3];
        password=array[4];
    }

    @Override
    public void onClick(View v) {
        switch(v.getId()){
            case R.id.bNext:
                Toast.makeText(getApplicationContext(), "All ok", Toast.LENGTH_LONG).show();
                registerUser();
                break;
        }
    }

    private void registerUser() {
        String waist = etWaist.getText().toString();
        String hips = etHips.getText().toString();
        String breast = etBreast.getText().toString();
        String wrist = etWrist.getText().toString();
        String weight = etWeight.getText().toString();
        String height = etHeight.getText().toString();

        register(waist, hips, breast, wrist, weight, height);
    }

    private void register(String waist, String hips, String breast, String wrist, String weight,String height) {
        class RegisterUser extends AsyncTask<String, Void, String> {

            ProgressDialog loading;
            RegisterUserClass ruc = new RegisterUserClass();

            @Override
            protected void onPreExecute() {
                super.onPreExecute();

                loading = ProgressDialog.show(RegisterBody.this, "Please wait...", null, true, true);
            }

            @Override
            protected void onPostExecute(String s) {
                super.onPostExecute(s);
                loading.dismiss();
                Toast.makeText(getApplicationContext(), s, Toast.LENGTH_LONG).show();
            }

            @Override
            protected String doInBackground(String... params) {
                HashMap<String, String> data = new HashMap<String, String>();
                data.put("waist", params[0]);
                data.put("hips", params[1]);
                data.put("breast", params[2]);
                data.put("wrist", params[3]);
                data.put("weight", params[4]);
                data.put("height", params[5]);

                String result = ruc.sendPostRequest(REGISTER_URL, data);
                return result;
            }
        }
        RegisterUser ru = new RegisterUser();
        ru.execute(waist,hips,breast,wrist,weight,height, email, password);
        Bundle b=new Bundle();
        b.putStringArray(null, new String[]{name, surname, age, email, password,waist,hips,breast,wrist,weight,height});
        Intent in=new Intent(this,ChooseDiet.class);
        in.putExtras(b);
        startActivity(in);
    }

RegisterBody.php

if($_SERVER['REQUEST_METHOD']=='POST')
{
    $waist = $_POST["waist"];
    $hips = $_POST["hips"];
    $breast = $_POST["breast"];
    $wrist = $_POST["wrist"];
    $weight = $_POST["weight"];
    $height = $_POST["height"];
    $email = $_POST["email"];
    $password = $_POST["password"];


    $conn = mysqli_connect("localhost","root","","sfitness") or die("Error " . mysqli_error($conn));


    $sql="SELECT ID_Person FROM Person WHERE email=".$email." AND password=".$password."";
    $id_person=mysqli_query($conn,$sql);
    if (!$sql)
    {
            echo 'Could not run query: ' . mysql_error();
            exit;
    }
    $id_person = mysql_fetch_row($sql);



    $query ="INSERT INTO Body
                    (Waistline,Hips,Breast,
                     Wrist,Height,ID_person)                  
              values('$waist','$hips','$breast',
                     '$wrist','$weight','$height','$id_peroson')";

    $rows=mysql_query($query,$conn) or die("query fallita");
    mysql_close($conn);

    if($rows>0)
        echo "Registrazione avvenuta correttamente";
    else
        echo "Registrazione fallita";

}

else
{
    echo "Error!!";
}

我没有错误，但数据不会保存到表格中..为什么？

Answer 1

确定已经格式化了INSERT查询，因此它更易读，错误很明显。它们通常是在您使用代码的一些合理格式时。

INSERT查询提到了6列，设置了7个参数

您还有$id_person

拼写错误$ID_Peroson

$query ="INSERT INTO Body                 
              (Waistline,Hips,Breast,
               Wrist,Height,ID_person) 
         values('$waist','$hips','$breast',
                '$wrist','$weight','$height','$id_person')";

$res = mysqli_query($query,$conn);
if ($res === FALSE ){
    echo mysqli_error($conn);
    exit;
}

将Weight添加到列列表中，或从参数列表中删除$weight。

附加说明;另请注意@JayBlanchards关于散列密码和使用准备和参数化查询的评论

$query ="INSERT INTO Body                 
                   (Waistline,Hips,Breast,Wrist,Weight,Height,ID_person) 
             values(?,?,?,?,?,?,?)";

$stmt = mysqli_prepare($query);
if ( $stmt === FALSE ) {
    mysqli_error($conn);
    exit;
}
mysqli_stmt_bind_param('ssssssi', '$waist','$hips','$breast',
                                  '$wrist','$weight','$height','$id_person');

$result = mysqli_stmt_execute($stmt); 
if ( $result === FALSE ){
    echo $mysqli_stmt_error($stmt);
    exit;
}

Answer 2

我找到了解决问题的方法。

在 java 文件中，我修改此部分以添加：

data.put("email", email);
data.put("password", password);

除了 php 文件外，我还修改了我的代码：

$conn = mysql_connect("127.0.0.1","root","") or die("DBMS non disponibile");
mysql_select_db("sfitness") or die("Database non disponibile");

$sql="SELECT * FROM person WHERE Email='$email' AND Password='$password'";

$result=mysql_query($sql,$conn);
if (mysql_num_rows($result)==1){
    $row = mysql_fetch_array($result);
    $id_person=$row['ID_Person'];
}


$query ="INSERT INTO Body (Waistline, Hips, Breast, Wrists, Weight, Height, ID_person) values('$waist','$hips','$breast','$wrist','$weight','$height','$id_person')";

$rows=mysql_query($query,$conn) or die("query fallita");
mysql_close($conn);
if($rows>0)
     echo "Correct registration";
else
     echo "Registration failed";

现在使用此代码，感谢所有

如何将数据插入到Android应用程序的mysql数据库中？

2 个答案: