Php mysql sorgusu calışmıyor

Question

1.00/5 (1 vote)

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$sql = "UPDATE `kid` SET `kimg`='$im',`ad`='$ad',`yazar`='$yz',`yv`='$yv,`ft`='$ft',`stok`='$st' WHERE id=$br";

if ($conn->query($sql) === TRUE) {
echo "günçeleme başarılı";
} else {
echo "Error: " . $sql . "
" . $conn->error;
}
}

# code...
else {
echo "Error updating record: " . $conn->error;
}
$conn->error;

$conn->close();

What I have tried:

$sql = "UPDATE `kid` SET `kimg`='$im',`ad`='$ad',`yazar`='$yz',`yv`='$yv,`ft`='$ft',`stok`='$st' WHERE id=$br";

Posted 22-Jun-22 10:03am

egemen taskol

Updated 22-Jun-22 10:25am

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Comments

Richard Deeming 23-Jun-22 4:51am

Your code is vulnerable to SQL Injection[^]. NEVER use string concatenation/interpolation to build a SQL query. ALWAYS use a parameterized query.
PHP: SQL Injection - Manual[^]
PHP: Prepared statements and stored procedures - Manual[^]

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0x01AA · Accepted Answer · 2022-06-22T10:13:00

Please.... check your SQL ;)

$sql = "UPDATE `kid` SET `kimg`='$im',`ad`='$ad',`yazar`='$yz',`yv`='$yv,`ft`='$ft',`stok`='$st' WHERE id=$br";

$sql = "UPDATE `kid` SET `kimg`='$im',`ad`='$ad',`yazar`='$yz',`yv`='$yv',`ft`='$ft',`stok`='$st' WHERE id=$br";

Focus on:

,`yv`='$yv,`ft`='$ft'
should be 
,`yv`='$yv',`ft`='$ft'
          ^