Receive a response in php and define it in android studio

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Receive a response in php and define it in android studio

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#1 by (1 votes)
#2 by (0 votes)

1

I am creating a registration and login system, but I want to check if there is already a user with that name, for this, I am checking it with this: SELECT COUNT(*) FROM login WHERE username = '$username' , so if the answer is 1, do not allow me to create a new user with that name, and if it is 0, that allows me to access and create it, this is my php code that I have at the moment:

<?php
$host = "localhost";
$username = "---";
$password = "---";
$db = "---";

//connection
$conn = new mysqli($host, $username, $password, $db);

$username = $_POST["username"];

$sql = "SELECT COUNT(*) FROM 'login' WHERE username = '$username'";

?>

Thank you.

EDIT: I would like to be able to receive the answer in the form of string or in json, since I am going to use it in andorid studio.

php

asked by Nexobeta28 YT 03.09.2018 в 10:36

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2 answers

0

Try this

$link = new PDO('mysql:host='.$host.';dbname='.$dbname, $user, $password);
$result=$link->query("SELECT COUNT(*) FROM 'login' WHERE username LIKE '$username'");
foreach($result as $rows){}

if($rows[0]>0){
   echo "1"; //Existe
}else{
   echo "0"; //No existe
}

Greetings:)

answered by 03.09.2018 в 10:40

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score 1 · Accepted Answer

Your code is missing the execution of the SQL query, the extraction of the result and sending it to the client.

In addition, your code suffers from the serious security issues associated with SQL injection to be solved with prepared queries or using mysqli::real_escape_string() .

The query is executed with a call to mysqli::query or mysqli::prepare if you use prepared queries.

The result can be obtained by using mysqli_result::fetch_assoc() .

Finally, to send the data in JSON you have to set the necessary header ( Content-Type: application/json; charset=utf-8 ) and use the function json_encode() to obtain the value correctly generated as a JSON value (although being numeric this will be the number as such).

Here is your finished code with the suggested modifications:

<?php
$host = "localhost";
$username = "---";
$password = "---";
$db = "---";

//connection
$conn = new mysqli($host, $username, $password, $db);

/* Escapamos correctamente el usuario recibido por POST
    para evitar inyección SQL */
$username = $conn->real_escape_string($_POST["username"]);

/* Agregamos un alias al COUNT(*) para poder acceder a él fácilmente */
$sql = "SELECT COUNT(*) numero FROM 'login' WHERE username = '$username'";

/* Hacemos efectiva la consulta */
$resultado = $conn->query($sql);

/* Obtenemos el primer y único registro */
$registro = $resultado->fetch_assoc();

/* Enviamos al cliente el resultado (numérico) codificado en JSON
    (que será únicamente el número sin más) */
header('Content-Type: application/json; charset=utf-8');
echo json_encode($registro['numero']);

NOTE: This code does not handle SQL errors or any other error that occurs during calls to mysqli functions. You should check the result of calling each of them.