[Villezzai]

Need a little help with my website.

Error:

Code

PHP Warning:  mysqli::query(): Empty query in /home/psctrade/public_html/test/process.php on line 24:

Code:

Code

<?php
$koodi = $_POST['koodi'];
$arvo = $_POST['arvo'];
$currency = $_POST['currency'];
$purse_type = $_POST['purse_type'];
$purse_currency = $_POST['purse_currency'];
$osoite = $_POST['osoite'];
$sahkoposti = $_POST['sahkoposti'];

$servername = "localhost";
$username = "censored";
$password = "censored";
$dbname = "censored";

$conn = new mysqli($servername, $username, $password, $dbname);
if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
}
if ($action === "record"){
$sql = "INSERT INTO `psctrade_homo`.`koodit` (`ID`, `koodi`, `arvo`, `currency`, `purse_type`, `purse_currency`, `osoite`, `sahkoposti`);
VALUES
(NULL, $koodi, $arvo, $currency, $purse_type, $purse_currency, $osoite, $sahkoposti');";
}
if ($conn->query($sql) === TRUE) {
    echo "New record created successfully";
} else {
    echo "Error: " . $sql . "<br>" . $conn->error;
}

$conn->close();        

?>

Line 24:

Code

if ($conn->query($sql) === TRUE) {

[Water Aura]

well what happens if action isn't "record"...the variable sql is empty

Code

<?php
$koodi = $_POST['koodi'];
$arvo = $_POST['arvo'];
$currency = $_POST['currency'];
$purse_type = $_POST['purse_type'];
$purse_currency = $_POST['purse_currency'];
$osoite = $_POST['osoite'];
$sahkoposti = $_POST['sahkoposti'];

$servername = "localhost";
$username = "censored";
$password = "censored";
$dbname = "censored";

$conn = new mysqli($servername, $username, $password, $dbname);
if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
}
if ($action === "record"){
    $sql = "INSERT INTO `psctrade_homo`.`koodit` (`ID`, `koodi`, `arvo`, `currency`, `purse_type`, `purse_currency`, `osoite`, `sahkoposti`);
    VALUES
    (NULL, $koodi, $arvo, $currency, $purse_type, $purse_currency, $osoite, $sahkoposti');";
    if ($conn->query($sql) === TRUE) {
        echo "New record created successfully";
    } else {
        echo "Error: " . $sql . "<br>" . $conn->error;
    }
}

$conn->close();        

?>

---

Another solution as I guess you're adding more "actions" is to check if the variable "sql" is empty before executing the query.

---

Also seems like the variable "action" isn't even defined.

[Villezzai]

well what happens if action isn't "record"...the variable sql is empty

Code

<?php
$koodi = $_POST['koodi'];
$arvo = $_POST['arvo'];
$currency = $_POST['currency'];
$purse_type = $_POST['purse_type'];
$purse_currency = $_POST['purse_currency'];
$osoite = $_POST['osoite'];
$sahkoposti = $_POST['sahkoposti'];

$servername = "localhost";
$username = "censored";
$password = "censored";
$dbname = "censored";

$conn = new mysqli($servername, $username, $password, $dbname);
if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
}
if ($action === "record"){
    $sql = "INSERT INTO `psctrade_homo`.`koodit` (`ID`, `koodi`, `arvo`, `currency`, `purse_type`, `purse_currency`, `osoite`, `sahkoposti`);
    VALUES
    (NULL, $koodi, $arvo, $currency, $purse_type, $purse_currency, $osoite, $sahkoposti');";
    if ($conn->query($sql) === TRUE) {
        echo "New record created successfully";
    } else {
        echo "Error: " . $sql . "<br>" . $conn->error;
    }
}

$conn->close();        

?>

---

Another solution as I guess you're adding more "actions" is to check if the variable "sql" is empty before executing the query.

Thanks for the help, no more errors but nothing shows up on the database when I send the form data to database, ffs

[Water Aura]

Thanks for the help, no more errors but nothing shows up on the database when I send the form data to database, ffs

Well as I stated in my edit the variable "action" isn't even defined.

[Omniscient]

Code

if ($action === "record"){

You're missing this at your script start.

$action = $_POST['action'];

And this...

Code

if ($conn->query($sql) === TRUE) {

You need to handle the error better than that imho.

And also, your script is crazy vuln to sql injection.