检查Android代码中是否存在用户名

Question

I am trying to check if the Text from my EditText is already in the Database. I am really new in PHP & MySQL.

I tried to receive all the Usernames and check if it equals to the EditText Text. At first it looked good. I checked it with

if(!UsernameDB.equals(editUsername.getText().toString()){
    Toast.makeText(Register.this, "Not taken", Toast.LENGTH_SHORT).show();
}

The Problem is, that it Toasts it for every Name. So i can't ask if everything is filled in right (Email, Password etc.) because it says its not matching after it passed the Name that is matching.

Does someone know how i can resolve my Problem?

Answer 1

Im not sure if you need to do it by Ajax or is it ok to refresh the site. Here is an example for refreshing: First of all you need a Formular (html):

<?php include("loginFunktion.php"); ?>
<form action="" method="Post">      
        <input name="name">
        <input type="submit" name="test" value="test">              
</form>

Thats the basic of your DB Class (new sheet):

<?php 
class DB {
protected $mysqli;

function __construct(){
    $this->host = "***";
    $this->username = "***";
    $this->database = "***";
    $this->password = "***";

}

function connect(){
    $this->mysqli = new mysqli($this->host, $this->username, $this->password, $this->database);

    if (mysqli_connect_errno()) {
        printf("Connect failed: %s<br/>", mysqli_connect_error());
        exit();
    }   
}

function disconnect() {
    // Verbindung beenden
    $this->mysqli->close();
}
}    
?>

Now you have in include the class and connect to your DB by the constructor of your class (the following code is the loginFunktion.php sheet):

<?php 
    include("dbclassname.php");
    $_DB = new DBData();
    $_DB -> connect();
?>

Now you have a connection to your DB and u can do querys:

if(isset($_POST["test"])){
$queryCALL ="SELECT user FROM t_user WHERE user_name=".$_REQUEST["name"];
$result = $this->mysqli->query($queryCALL);
$ergebnis=$result->fetch_array()[0];
if(count($ergebnis)>0)
    echo "this name is used";
}

If you are new on php ajax is a little bit to complicate for you. Nethertheless thats a good guide: https://www.w3schools.com/php/php_ajax_database.asp

Next time please add more informations.