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include "baglan.php";
if (isset($_POST["br"]) && isset($_POST["ad"]) && isset($_POST["yz"]) && isset($_POST["yv"]) && isset($_POST["ft"]) && isset($_POST["st"]) && isset($_POST["im"])) {
$br = $_POST["br"];
$ad = $_POST["ad"];
$yz = $_POST["yz"];
$yv = $_POST["yv"];
$ft = $_POST["ft"];
$st = $_POST["st"];
$im = $_POST["im"];

$sql = "INSERT INTO `kid`(`id`, `kimg`, `ad`, `yazar`, `yv`, `ft`, `stok` ) VALUES ('$br','$im','$ad','$yz','$yv','$ft'','$st')";

if ($conn->query($sql) === TRUE) {
echo "New record created successfully";
} else {
echo "Error:
" . $conn->error;


What I have tried:

$sql = "INSERT INTO `kid`(`id`, `kimg`, `ad`, `yazar`, `yv`, `ft`, `stok` ) VALUES ('$br','$im','$ad','$yz','$yv','$ft'','$st')";
Updated 22-Jun-22 9:24am

As a follow up to what 0x01AA has said, a even better solution is not to do that at all.
Never concatenate strings to build a SQL command. It leaves you wide open to accidental or deliberate SQL Injection attack which can destroy your entire database. Always use Parameterized queries instead.

When you concatenate strings, you cause problems because SQL receives commands like:
SELECT * FROM MyTable WHERE StreetAddress = 'Baker's Wood'
The quote the user added terminates the string as far as SQL is concerned and you get problems. But it could be worse. If I come along and type this instead: "x';DROP TABLE MyTable;--" Then SQL receives a very different command:
SELECT * FROM MyTable WHERE StreetAddress = 'x';DROP TABLE MyTable;--'
Which SQL sees as three separate commands:
SELECT * FROM MyTable WHERE StreetAddress = 'x';
A perfectly valid SELECT
A perfectly valid "delete the table" command
And everything else is a comment.
So it does: selects any matching rows, deletes the table from the DB, and ignores anything else.

And in a login system? That's just asking for trouble as I don't even have to have to register with you to do it!

So ALWAYS use parameterized queries! Or be prepared to restore your DB from backup frequently. You do take backups regularly, don't you?
Check your sql carefully...
"INSERT INTO `kid`(`id`, `kimg`, `ad`, `yazar`, `yv`, `ft`, `stok` ) VALUES ('$br','$im','$ad','$yz','$yv','$ft'','$st')";

Focus on
and here the

What I mean, it should be like this:
0x01AA 22-Jun-22 15:19pm
Thank you very much. see you for the next typo ;)

This content, along with any associated source code and files, is licensed under The Code Project Open License (CPOL)

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